Title: Amusements in Mathematics
Author: Dudeney, Henry Ernest, 1857-1930
Language: English
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*** Start of this LibraryBlog Digital Book "Amusements in Mathematics" ***


[Transcribers note: Many of the puzzles in this book assume a
familiarity with the currency of Great Britain in the early 1900s. As
this is likely not common knowledge for those outside Britain (and
possibly many within,) I am including a chart of relative values.

The most common units used were:

    the Penny,    abbreviated: d. (from the Roman penny, denarius)
    the Shilling, abbreviated: s.
    the Pound,    abbreviated: £

There was 12 Pennies to a Shilling and 20 Shillings to a Pound, so there
was 240 Pennies in a Pound.

To further complicate things, there were many coins which were various
fractional values of Pennies, Shillings or Pounds.

    Farthing                           ¼d.

    Half-penny                         ½d.

    Penny                              1d.

    Three-penny                        3d.

    Sixpence (or tanner)               6d.

    Shilling (or bob)                  1s.

    Florin or two shilling piece       2s.

    Half-crown (or half-dollar)    2s. 6d.

    Double-florin                      4s.

    Crown (or dollar)                  5s.

    Half-Sovereign                    10s.

    Sovereign (or Pound)        £1 or 20s.

This is by no means a comprehensive list, but it should be adequate to
solve the puzzles in this book.

Exponents are represented in this text by ^, e.g. '3 squared' is 3^2.

Numbers with fractional components (other than ¼, ½ and ¾) have a +
symbol separating the whole number component from the fraction. It makes
the fraction look odd, but yeilds correct solutions no matter how it is
interpreted. E.G., 4 and eleven twenty-thirds is 4+11/23, not 411/23 or
4-11/23.

]



AMUSEMENTS IN MATHEMATICS

         by

HENRY ERNEST DUDENEY


    In Mathematicks he was greater
    Than Tycho Brahe or Erra Pater:
    For he, by geometrick scale,
    Could take the size of pots of ale;
    Resolve, by sines and tangents, straight,
    If bread or butter wanted weight;
    And wisely tell what hour o' th' day
    The clock does strike by algebra.

                  BUTLER'S _Hudibras_.


1917



PREFACE


In issuing this volume of my Mathematical Puzzles, of which some have
appeared in periodicals and others are given here for the first time, I
must acknowledge the encouragement that I have received from many
unknown correspondents, at home and abroad, who have expressed a desire
to have the problems in a collected form, with some of the solutions
given at greater length than is possible in magazines and newspapers.
Though I have included a few old puzzles that have interested the world
for generations, where I felt that there was something new to be said
about them, the problems are in the main original. It is true that some
of these have become widely known through the press, and it is possible
that the reader may be glad to know their source.

On the question of Mathematical Puzzles in general there is, perhaps,
little more to be said than I have written elsewhere. The history of the
subject entails nothing short of the actual story of the beginnings and
development of exact thinking in man. The historian must start from the
time when man first succeeded in counting his ten fingers and in
dividing an apple into two approximately equal parts. Every puzzle that
is worthy of consideration can be referred to mathematics and logic.
Every man, woman, and child who tries to "reason out" the answer to the
simplest puzzle is working, though not of necessity consciously, on
mathematical lines. Even those puzzles that we have no way of attacking
except by haphazard attempts can be brought under a method of what has
been called "glorified trial"--a system of shortening our labours by
avoiding or eliminating what our reason tells us is useless. It is, in
fact, not easy to say sometimes where the "empirical" begins and where
it ends.

When a man says, "I have never solved a puzzle in my life," it is
difficult to know exactly what he means, for every intelligent
individual is doing it every day. The unfortunate inmates of our lunatic
asylums are sent there expressly because they cannot solve
puzzles--because they have lost their powers of reason. If there were no
puzzles to solve, there would be no questions to ask; and if there were
no questions to be asked, what a world it would be! We should all be
equally omniscient, and conversation would be useless and idle.

It is possible that some few exceedingly sober-minded mathematicians,
who are impatient of any terminology in their favourite science but the
academic, and who object to the elusive x and y appearing under any
other names, will have wished that various problems had been presented
in a less popular dress and introduced with a less flippant phraseology.
I can only refer them to the first word of my title and remind them that
we are primarily out to be amused--not, it is true, without some hope of
picking up morsels of knowledge by the way. If the manner is light, I
can only say, in the words of Touchstone, that it is "an ill-favoured
thing, sir, but my own; a poor humour of mine, sir."

As for the question of difficulty, some of the puzzles, especially in
the Arithmetical and Algebraical category, are quite easy. Yet some of
those examples that look the simplest should not be passed over without
a little consideration, for now and again it will be found that there is
some more or less subtle pitfall or trap into which the reader may be
apt to fall. It is good exercise to cultivate the habit of being very
wary over the exact wording of a puzzle. It teaches exactitude and
caution. But some of the problems are very hard nuts indeed, and not
unworthy of the attention of the advanced mathematician. Readers will
doubtless select according to their individual tastes.

In many cases only the mere answers are given. This leaves the beginner
something to do on his own behalf in working out the method of solution,
and saves space that would be wasted from the point of view of the
advanced student. On the other hand, in particular cases where it seemed
likely to interest, I have given rather extensive solutions and treated
problems in a general manner. It will often be found that the notes on
one problem will serve to elucidate a good many others in the book; so
that the reader's difficulties will sometimes be found cleared up as he
advances. Where it is possible to say a thing in a manner that may be
"understanded of the people" generally, I prefer to use this simple
phraseology, and so engage the attention and interest of a larger
public. The mathematician will in such cases have no difficulty in
expressing the matter under consideration in terms of his familiar
symbols.

I have taken the greatest care in reading the proofs, and trust that any
errors that may have crept in are very few. If any such should occur, I
can only plead, in the words of Horace, that "good Homer sometimes
nods," or, as the bishop put it, "Not even the youngest curate in my
diocese is infallible."

I have to express my thanks in particular to the proprietors of _The
Strand Magazine_, _Cassell's Magazine_, _The Queen_, _Tit-Bits_, and
_The Weekly Dispatch_ for their courtesy in allowing me to reprint some
of the puzzles that have appeared in their pages.

THE AUTHORS' CLUB _March_ 25, 1917



CONTENTS


    PREFACE                                                v
    ARITHMETICAL AND ALGEBRAICAL PROBLEMS                  1
      Money Puzzles                                        1
      Age and Kinship Puzzles                              6
      Clock Puzzles                                        9
      Locomotion and Speed Puzzles                        11
      Digital Puzzles                                     13
      Various Arithmetical and Algebraical Problems       17
    GEOMETRICAL PROBLEMS                                  27
      Dissection Puzzles                                  27
      Greek Cross Puzzles                                 28
      Various Dissection Puzzles                          35
      Patchwork Puzzles                                   46
      Various Geometrical Puzzles                         49
    POINTS AND LINES PROBLEMS                             56
    MOVING COUNTER PROBLEMS                               58
    UNICURSAL AND ROUTE PROBLEMS                          68
    COMBINATION AND GROUP PROBLEMS                        76
    CHESSBOARD PROBLEMS                                   85
      The Chessboard                                      85
      Statical Chess Puzzles                              88
      The Guarded Chessboard                              95
      Dynamical Chess Puzzles                             96
      Various Chess Puzzles                              105
    MEASURING, WEIGHING, AND PACKING PUZZLES             109
    CROSSING RIVER PROBLEMS                              112
    PROBLEMS CONCERNING GAMES                            114
    PUZZLE GAMES                                         117
    MAGIC SQUARE PROBLEMS                                119
      Subtracting, Multiplying, and Dividing Magics      124
      Magic Squares of Primes                            125
    MAZES AND HOW TO THREAD THEM                         127
    THE PARADOX PARTY                                    137
    UNCLASSIFIED PROBLEMS                                142
    SOLUTIONS                                            148
    INDEX                                                253



AMUSEMENTS IN MATHEMATICS.



ARITHMETICAL AND ALGEBRAICAL PROBLEMS.

                   "And what was he?
    Forsooth, a great arithmetician."
                              _Othello_, I. i.


The puzzles in this department are roughly thrown together in classes
for the convenience of the reader. Some are very easy, others quite
difficult. But they are not arranged in any order of difficulty--and
this is intentional, for it is well that the solver should not be warned
that a puzzle is just what it seems to be. It may, therefore, prove to
be quite as simple as it looks, or it may contain some pitfall into
which, through want of care or over-confidence, we may stumble.

Also, the arithmetical and algebraical puzzles are not separated in the
manner adopted by some authors, who arbitrarily require certain problems
to be solved by one method or the other. The reader is left to make his
own choice and determine which puzzles are capable of being solved by
him on purely arithmetical lines.



MONEY PUZZLES.

    "Put not your trust in money, but put your money in trust."

    OLIVER WENDELL HOLMES.


1.--A POST-OFFICE PERPLEXITY.

In every business of life we are occasionally perplexed by some chance
question that for the moment staggers us. I quite pitied a young lady in
a branch post-office when a gentleman entered and deposited a crown on
the counter with this request: "Please give me some twopenny stamps, six
times as many penny stamps, and make up the rest of the money in
twopence-halfpenny stamps." For a moment she seemed bewildered, then her
brain cleared, and with a smile she handed over stamps in exact
fulfilment of the order. How long would it have taken you to think it
out?


2.--YOUTHFUL PRECOCITY.

The precocity of some youths is surprising. One is disposed to say on
occasion, "That boy of yours is a genius, and he is certain to do great
things when he grows up;" but past experience has taught us that he
invariably becomes quite an ordinary citizen. It is so often the case,
on the contrary, that the dull boy becomes a great man. You never can
tell. Nature loves to present to us these queer paradoxes. It is well
known that those wonderful "lightning calculators," who now and again
surprise the world by their feats, lose all their mysterious powers
directly they are taught the elementary rules of arithmetic.

A boy who was demolishing a choice banana was approached by a young
friend, who, regarding him with envious eyes, asked, "How much did you
pay for that banana, Fred?" The prompt answer was quite remarkable in
its way: "The man what I bought it of receives just half as many
sixpences for sixteen dozen dozen bananas as he gives bananas for a
fiver."

Now, how long will it take the reader to say correctly just how much
Fred paid for his rare and refreshing fruit?


3.--AT A CATTLE MARKET.

Three countrymen met at a cattle market. "Look here," said Hodge to
Jakes, "I'll give you six of my pigs for one of your horses, and then
you'll have twice as many animals here as I've got." "If that's your
way of doing business," said Durrant to Hodge, "I'll give you fourteen
of my sheep for a horse, and then you'll have three times as many
animals as I." "Well, I'll go better than that," said Jakes to Durrant;
"I'll give you four cows for a horse, and then you'll have six times as
many animals as I've got here."

No doubt this was a very primitive way of bartering animals, but it is
an interesting little puzzle to discover just how many animals Jakes,
Hodge, and Durrant must have taken to the cattle market.


4.--THE BEANFEAST PUZZLE.

A number of men went out together on a bean-feast. There were four
parties invited--namely, 25 cobblers, 20 tailors, 18 hatters, and 12
glovers. They spent altogether £6, 13s. It was found that five cobblers
spent as much as four tailors; that twelve tailors spent as much as nine
hatters; and that six hatters spent as much as eight glovers. The puzzle
is to find out how much each of the four parties spent.


5.--A QUEER COINCIDENCE.

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards,
Francis, and Gudgeon, were recently engaged in play. The name of the
particular game is of no consequence. They had agreed that whenever a
player won a game he should double the money of each of the other
players--that is, he was to give the players just as much money as they
had already in their pockets. They played seven games, and, strange to
say, each won a game in turn, in the order in which their names are
given. But a more curious coincidence is this--that when they had
finished play each of the seven men had exactly the same amount--two
shillings and eightpence--in his pocket. The puzzle is to find out how
much money each man had with him before he sat down to play.


6.--A CHARITABLE BEQUEST.

A man left instructions to his executors to distribute once a year
exactly fifty-five shillings among the poor of his parish; but they were
only to continue the gift so long as they could make it in different
ways, always giving eighteenpence each to a number of women and half a
crown each to men. During how many years could the charity be
administered? Of course, by "different ways" is meant a different number
of men and women every time.


7.--THE WIDOW'S LEGACY.

A gentleman who recently died left the sum of £8,000 to be divided among
his widow, five sons, and four daughters. He directed that every son
should receive three times as much as a daughter, and that every
daughter should have twice as much as their mother. What was the widow's
share?


8.--INDISCRIMINATE CHARITY.

A charitable gentleman, on his way home one night, was appealed to by
three needy persons in succession for assistance. To the first person he
gave one penny more than half the money he had in his pocket; to the
second person he gave twopence more than half the money he then had in
his pocket; and to the third person he handed over threepence more than
half of what he had left. On entering his house he had only one penny in
his pocket. Now, can you say exactly how much money that gentleman had
on him when he started for home?


9.--THE TWO AEROPLANES.

A man recently bought two aeroplanes, but afterwards found that they
would not answer the purpose for which he wanted them. So he sold them
for £600 each, making a loss of 20 per cent. on one machine and a profit
of 20 per cent. on the other. Did he make a profit on the whole
transaction, or a loss? And how much?


10.--BUYING PRESENTS.

"Whom do you think I met in town last week, Brother William?" said Uncle
Benjamin. "That old skinflint Jorkins. His family had been taking him
around buying Christmas presents. He said to me, 'Why cannot the
government abolish Christmas, and make the giving of presents punishable
by law? I came out this morning with a certain amount of money in my
pocket, and I find I have spent just half of it. In fact, if you will
believe me, I take home just as many shillings as I had pounds, and half
as many pounds as I had shillings. It is monstrous!'" Can you say
exactly how much money Jorkins had spent on those presents?


11.--THE CYCLISTS' FEAST.

    'Twas last Bank Holiday, so I've been told,
      Some cyclists rode abroad in glorious weather.
    Resting at noon within a tavern old,
      They all agreed to have a feast together.
    "Put it all in one bill, mine host," they said,
      "For every man an equal share will pay."
    The bill was promptly on the table laid,
      And four pounds was the reckoning that day.
    But, sad to state, when they prepared to square,
      'Twas found that two had sneaked outside and fled.
    So, for two shillings more than his due share
      Each honest man who had remained was bled.
    They settled later with those rogues, no doubt.
      How many were they when they first set out?


12.--A QUEER THING IN MONEY.

It will be found that £66, 6s. 6d. equals 15,918 pence. Now, the four
6's added together make 24, and the figures in 15,918 also add to 24. It
is a curious fact that there is only one other sum of money, in pounds,
shillings, and pence (all similarly repetitions of one figure), of which
the digits shall add up the same as the digits of the amount in pence.
What is the other sum of money?


13.--A NEW MONEY PUZZLE.

The largest sum of money that can be written in pounds, shillings,
pence, and farthings, using each of the nine digits once and only once,
is £98,765, 4s. 3½d. Now, try to discover the smallest sum of money
that can be written down under precisely the same conditions. There must
be some value given for each denomination--pounds, shillings, pence,
and farthings--and the nought may not be used. It requires just a little
judgment and thought.


14.--SQUARE MONEY.

"This is queer," said McCrank to his friend. "Twopence added to twopence
is fourpence, and twopence multiplied by twopence is also fourpence." Of
course, he was wrong in thinking you can multiply money by money. The
multiplier must be regarded as an abstract number. It is true that two
feet multiplied by two feet will make four square feet. Similarly, two
pence multiplied by two pence will produce four square pence! And it
will perplex the reader to say what a "square penny" is. But we will
assume for the purposes of our puzzle that twopence multiplied by
twopence is fourpence. Now, what two amounts of money will produce the
next smallest possible result, the same in both cases, when added or
multiplied in this manner? The two amounts need not be alike, but they
must be those that can be paid in current coins of the realm.


15.--POCKET MONEY.

What is the largest sum of money--all in current silver coins and no
four-shilling piece--that I could have in my pocket without being able
to give change for a half-sovereign?

16.--THE MILLIONAIRE'S PERPLEXITY.

Mr. Morgan G. Bloomgarten, the millionaire, known in the States as the
Clam King, had, for his sins, more money than he knew what to do with.
It bored him. So he determined to persecute some of his poor but happy
friends with it. They had never done him any harm, but he resolved to
inoculate them with the "source of all evil." He therefore proposed to
distribute a million dollars among them and watch them go rapidly to the
bad. But he was a man of strange fancies and superstitions, and it was
an inviolable rule with him never to make a gift that was not either one
dollar or some power of seven--such as 7, 49, 343, 2,401, which numbers
of dollars are produced by simply multiplying sevens together. Another
rule of his was that he would never give more than six persons exactly
the same sum. Now, how was he to distribute the 1,000,000 dollars? You
may distribute the money among as many people as you like, under the
conditions given.

17.--THE PUZZLING MONEY-BOXES.

Four brothers--named John, William, Charles, and Thomas--had each a
money-box. The boxes were all given to them on the same day, and they at
once put what money they had into them; only, as the boxes were not very
large, they first changed the money into as few coins as possible. After
they had done this, they told one another how much money they had saved,
and it was found that if John had had 2s. more in his box than at
present, if William had had 2s. less, if Charles had had twice as much,
and if Thomas had had half as much, they would all have had exactly the
same amount.

Now, when I add that all four boxes together contained 45s., and that
there were only six coins in all in them, it becomes an entertaining
puzzle to discover just what coins were in each box.


18.--THE MARKET WOMEN.

A number of market women sold their various products at a certain price
per pound (different in every case), and each received the same
amount--2s. 2½d. What is the greatest number of women there could
have been? The price per pound in every case must be such as could be
paid in current money.


19.--THE NEW YEAR'S EVE SUPPERS.

The proprietor of a small London café has given me some interesting
figures. He says that the ladies who come alone to his place for
refreshment spend each on an average eighteenpence, that the
unaccompanied men spend half a crown each, and that when a gentleman
brings in a lady he spends half a guinea. On New Year's Eve he supplied
suppers to twenty-five persons, and took five pounds in all. Now,
assuming his averages to have held good in every case, how was his
company made up on that occasion? Of course, only single gentlemen,
single ladies, and pairs (a lady and gentleman) can be supposed to have
been present, as we are not considering larger parties.


20.--BEEF AND SAUSAGES.

"A neighbour of mine," said Aunt Jane, "bought a certain quantity of
beef at two shillings a pound, and the same quantity of sausages at
eighteenpence a pound. I pointed out to her that if she had divided the
same money equally between beef and sausages she would have gained two
pounds in the total weight. Can you tell me exactly how much she spent?"

"Of course, it is no business of mine," said Mrs. Sunniborne; "but a
lady who could pay such prices must be somewhat inexperienced in
domestic economy."

"I quite agree, my dear," Aunt Jane replied, "but you see that is not
the precise point under discussion, any more than the name and morals of
the tradesman."


21.--A DEAL IN APPLES.

I paid a man a shilling for some apples, but they were so small that I
made him throw in two extra apples. I find that made them cost just a
penny a dozen less than the first price he asked. How many apples did I
get for my shilling?


22.--A DEAL IN EGGS.

A man went recently into a dairyman's shop to buy eggs. He wanted them
of various qualities. The salesman had new-laid eggs at the high price
of fivepence each, fresh eggs at one penny each, eggs at a halfpenny
each, and eggs for electioneering purposes at a greatly reduced figure,
but as there was no election on at the time the buyer had no use for the
last. However, he bought some of each of the three other kinds and
obtained exactly one hundred eggs for eight and fourpence. Now, as he
brought away exactly the same number of eggs of two of the three
qualities, it is an interesting puzzle to determine just how many he
bought at each price.


23.--THE CHRISTMAS-BOXES.

Some years ago a man told me he had spent one hundred English silver
coins in Christmas-boxes, giving every person the same amount, and it
cost him exactly £1, 10s. 1d. Can you tell just how many persons
received the present, and how he could have managed the distribution?
That odd penny looks queer, but it is all right.


24.--A SHOPPING PERPLEXITY.

Two ladies went into a shop where, through some curious eccentricity, no
change was given, and made purchases amounting together to less than
five shillings. "Do you know," said one lady, "I find I shall require no
fewer than six current coins of the realm to pay for what I have
bought." The other lady considered a moment, and then exclaimed: "By a
peculiar coincidence, I am exactly in the same dilemma." "Then we will
pay the two bills together." But, to their astonishment, they still
required six coins. What is the smallest possible amount of their
purchases--both different?


25.--CHINESE MONEY.

The Chinese are a curious people, and have strange inverted ways of
doing things. It is said that they use a saw with an upward pressure
instead of a downward one, that they plane a deal board by pulling the
tool toward them instead of pushing it, and that in building a house
they first construct the roof and, having raised that into position,
proceed to work downwards. In money the currency of the country consists
of taels of fluctuating value. The tael became thinner and thinner until
2,000 of them piled together made less than three inches in height. The
common cash consists of brass coins of varying thicknesses, with a
round, square, or triangular hole in the centre, as in our illustration.

[Illustration]

These are strung on wires like buttons. Supposing that eleven coins with
round holes are worth fifteen ching-changs, that eleven with square
holes are worth sixteen ching-changs, and that eleven with triangular
holes are worth seventeen ching-changs, how can a Chinaman give me
change for half a crown, using no coins other than the three mentioned?
A ching-chang is worth exactly twopence and four-fifteenths of a
ching-chang.


26.--THE JUNIOR CLERK'S PUZZLE.

Two youths, bearing the pleasant names of Moggs and Snoggs, were
employed as junior clerks by a merchant in Mincing Lane. They were both
engaged at the same salary--that is, commencing at the rate of £50 a
year, payable half-yearly. Moggs had a yearly rise of £10, and Snoggs
was offered the same, only he asked, for reasons that do not concern our
puzzle, that he might take his rise at £2, 10s. half-yearly, to which
his employer (not, perhaps, unnaturally!) had no objection.

Now we come to the real point of the puzzle. Moggs put regularly into
the Post Office Savings Bank a certain proportion of his salary, while
Snoggs saved twice as great a proportion of his, and at the end of five
years they had together saved £268, 15s. How much had each saved? The
question of interest can be ignored.


27.--GIVING CHANGE.

Every one is familiar with the difficulties that frequently arise over
the giving of change, and how the assistance of a third person with a
few coins in his pocket will sometimes help us to set the matter right.
Here is an example. An Englishman went into a shop in New York and
bought goods at a cost of thirty-four cents. The only money he had was a
dollar, a three-cent piece, and a two-cent piece. The tradesman had only
a half-dollar and a quarter-dollar. But another customer happened to be
present, and when asked to help produced two dimes, a five-cent piece, a
two-cent piece, and a one-cent piece. How did the tradesman manage to
give change? For the benefit of those readers who are not familiar with
the American coinage, it is only necessary to say that a dollar is a
hundred cents and a dime ten cents. A puzzle of this kind should rarely
cause any difficulty if attacked in a proper manner.


28.--DEFECTIVE OBSERVATION.

Our observation of little things is frequently defective, and our
memories very liable to lapse. A certain judge recently remarked in a
case that he had no recollection whatever of putting the wedding-ring on
his wife's finger. Can you correctly answer these questions without
having the coins in sight? On which side of a penny is the date given?
Some people are so unobservant that, although they are handling the coin
nearly every day of their lives, they are at a loss to answer this
simple question. If I lay a penny flat on the table, how many other
pennies can I place around it, every one also lying flat on the table,
so that they all touch the first one? The geometrician will, of course,
give the answer at once, and not need to make any experiment. He will
also know that, since all circles are similar, the same answer will
necessarily apply to any coin. The next question is a most interesting
one to ask a company, each person writing down his answer on a slip of
paper, so that no one shall be helped by the answers of others. What is
the greatest number of three-penny-pieces that may be laid flat on the
surface of a half-crown, so that no piece lies on another or overlaps
the surface of the half-crown? It is amazing what a variety of different
answers one gets to this question. Very few people will be found to give
the correct number. Of course the answer must be given without looking
at the coins.


29.--THE BROKEN COINS.

A man had three coins--a sovereign, a shilling, and a penny--and he
found that exactly the same fraction of each coin had been broken away.
Now, assuming that the original intrinsic value of these coins was the
same as their nominal value--that is, that the sovereign was worth a
pound, the shilling worth a shilling, and the penny worth a penny--what
proportion of each coin has been lost if the value of the three
remaining fragments is exactly one pound?


30.--TWO QUESTIONS IN PROBABILITIES.

There is perhaps no class of puzzle over which people so frequently
blunder as that which involves what is called the theory of
probabilities. I will give two simple examples of the sort of puzzle I
mean. They are really quite easy, and yet many persons are tripped up by
them. A friend recently produced five pennies and said to me: "In
throwing these five pennies at the same time, what are the chances that
at least four of the coins will turn up either all heads or all tails?"
His own solution was quite wrong, but the correct answer ought not to be
hard to discover. Another person got a wrong answer to the following
little puzzle which I heard him propound: "A man placed three sovereigns
and one shilling in a bag. How much should be paid for permission to
draw one coin from it?" It is, of course, understood that you are as
likely to draw any one of the four coins as another.


31.--DOMESTIC ECONOMY.

Young Mrs. Perkins, of Putney, writes to me as follows: "I should be
very glad if you could give me the answer to a little sum that has been
worrying me a good deal lately. Here it is: We have only been married a
short time, and now, at the end of two years from the time when we set
up housekeeping, my husband tells me that he finds we have spent a third
of his yearly income in rent, rates, and taxes, one-half in domestic
expenses, and one-ninth in other ways. He has a balance of £190
remaining in the bank. I know this last, because he accidentally left
out his pass-book the other day, and I peeped into it. Don't you think
that a husband ought to give his wife his entire confidence in his money
matters? Well, I do; and--will you believe it?--he has never told me
what his income really is, and I want, very naturally, to find out. Can
you tell me what it is from the figures I have given you?"

Yes; the answer can certainly be given from the figures contained in
Mrs. Perkins's letter. And my readers, if not warned, will be
practically unanimous in declaring the income to be--something absurdly
in excess of the correct answer!


32.--THE EXCURSION TICKET PUZZLE.

When the big flaming placards were exhibited at the little provincial
railway station, announcing that the Great ---- Company would run cheap
excursion trains to London for the Christmas holidays, the inhabitants
of Mudley-cum-Turmits were in quite a flutter of excitement. Half an
hour before the train came in the little booking office was crowded with
country passengers, all bent on visiting their friends in the great
Metropolis. The booking clerk was unaccustomed to dealing with crowds of
such a dimension, and he told me afterwards, while wiping his manly
brow, that what caused him so much trouble was the fact that these
rustics paid their fares in such a lot of small money.

He said that he had enough farthings to supply a West End draper with
change for a week, and a sufficient number of threepenny pieces for the
congregations of three parish churches. "That excursion fare," said he,
"is nineteen shillings and ninepence, and I should like to know in just
how many different ways it is possible for such an amount to be paid in
the current coin of this realm."

Here, then, is a puzzle: In how many different ways may nineteen
shillings and ninepence be paid in our current coin? Remember that the
fourpenny-piece is not now current.


33.--PUZZLE IN REVERSALS.

Most people know that if you take any sum of money in pounds, shillings,
and pence, in which the number of pounds (less than £12) exceeds that of
the pence, reverse it (calling the pounds pence and the pence pounds),
find the difference, then reverse and add this difference, the result is
always £12, 18s. 11d. But if we omit the condition, "less than £12," and
allow nought to represent shillings or pence--(1) What is the lowest
amount to which the rule will not apply? (2) What is the highest amount
to which it will apply? Of course, when reversing such a sum as £14,
15s. 3d. it may be written £3, 16s. 2d., which is the same as £3, 15s.
14d.


34.--THE GROCER AND DRAPER.

A country "grocer and draper" had two rival assistants, who prided
themselves on their rapidity in serving customers. The young man on the
grocery side could weigh up two one-pound parcels of sugar per minute,
while the drapery assistant could cut three one-yard lengths of cloth in
the same time. Their employer, one slack day, set them a race, giving
the grocer a barrel of sugar and telling him to weigh up forty-eight
one-pound parcels of sugar While the draper divided a roll of
forty-eight yards of cloth into yard pieces. The two men were
interrupted together by customers for nine minutes, but the draper was
disturbed seventeen times as long as the grocer. What was the result of
the race?

35.--JUDKINS'S CATTLE.

Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals,
consisting of oxen, pigs, and sheep, with the same number of animals in
each drove. One morning he sold all that he had to eight dealers. Each
dealer bought the same number of animals, paying seventeen dollars for
each ox, four dollars for each pig, and two dollars for each sheep; and
Hiram received in all three hundred and one dollars. What is the
greatest number of animals he could have had? And how many would there
be of each kind?

36.--BUYING APPLES.

As the purchase of apples in small quantities has always presented
considerable difficulties, I think it well to offer a few remarks on
this subject. We all know the story of the smart boy who, on being told
by the old woman that she was selling her apples at four for threepence,
said: "Let me see! Four for threepence; that's three for twopence, two
for a penny, one for nothing--I'll take _one_!"

There are similar cases of perplexity. For example, a boy once picked up
a penny apple from a stall, but when he learnt that the woman's pears
were the same price he exchanged it, and was about to walk off. "Stop!"
said the woman. "You haven't paid me for the pear!" "No," said the boy,
"of course not. I gave you the apple for it." "But you didn't pay for
the apple!" "Bless the woman! You don't expect me to pay for the apple
and the pear too!" And before the poor creature could get out of the
tangle the boy had disappeared.

Then, again, we have the case of the man who gave a boy sixpence and
promised to repeat the gift as soon as the youngster had made it into
ninepence. Five minutes later the boy returned. "I have made it into
ninepence," he said, at the same time handing his benefactor threepence.
"How do you make that out?" he was asked. "I bought threepennyworth of
apples." "But that does not make it into ninepence!" "I should rather
think it did," was the boy's reply. "The apple woman has threepence,
hasn't she? Very well, I have threepennyworth of apples, and I have just
given you the other threepence. What's that but ninepence?"

I cite these cases just to show that the small boy really stands in need
of a little instruction in the art of buying apples. So I will give a
simple poser dealing with this branch of commerce.

An old woman had apples of three sizes for sale--one a penny, two a
penny, and three a penny. Of course two of the second size and three of
the third size were respectively equal to one apple of the largest size.
Now, a gentleman who had an equal number of boys and girls gave his
children sevenpence to be spent amongst them all on these apples. The
puzzle is to give each child an equal distribution of apples. How was
the sevenpence spent, and how many children were there?


37.--BUYING CHESTNUTS.

Though the following little puzzle deals with the purchase of chestnuts,
it is not itself of the "chestnut" type. It is quite new. At first sight
it has certainly the appearance of being of the "nonsense puzzle"
character, but it is all right when properly considered.

A man went to a shop to buy chestnuts. He said he wanted a pennyworth,
and was given five chestnuts. "It is not enough; I ought to have a
sixth," he remarked! "But if I give you one chestnut more." the shopman
replied, "you will have five too many." Now, strange to say, they were
both right. How many chestnuts should the buyer receive for half a
crown?


38.--THE BICYCLE THIEF.

Here is a little tangle that is perpetually cropping up in various
guises. A cyclist bought a bicycle for £15 and gave in payment a cheque
for £25. The seller went to a neighbouring shopkeeper and got him to
change the cheque for him, and the cyclist, having received his £10
change, mounted the machine and disappeared. The cheque proved to be
valueless, and the salesman was requested by his neighbour to refund the
amount he had received. To do this, he was compelled to borrow the £25
from a friend, as the cyclist forgot to leave his address, and could not
be found. Now, as the bicycle cost the salesman £11, how much money did
he lose altogether?


39.--THE COSTERMONGER'S PUZZLE.

"How much did yer pay for them oranges, Bill?"

"I ain't a-goin' to tell yer, Jim. But I beat the old cove down
fourpence a hundred."

"What good did that do yer?"

"Well, it meant five more oranges on every ten shillin's-worth."

Now, what price did Bill actually pay for the oranges? There is only one
rate that will fit in with his statements.



AGE AND KINSHIP PUZZLES.

    "The days of our years are threescore years and ten."

    --_Psalm_ xc. 10.

For centuries it has been a favourite method of propounding arithmetical
puzzles to pose them in the form of questions as to the age of an
individual. They generally lend themselves to very easy solution by the
use of algebra, though often the difficulty lies in stating them
correctly. They may be made very complex and may demand considerable
ingenuity, but no general laws can well be laid down for their solution.
The solver must use his own sagacity. As for puzzles in relationship or
kinship, it is quite curious how bewildering many people find these
things. Even in ordinary conversation, some statement as to
relationship, which is quite clear in the mind of the speaker, will
immediately tie the brains of other people into knots. Such expressions
as "He is my uncle's son-in-law's sister" convey absolutely nothing to
some people without a detailed and laboured explanation. In such cases
the best course is to sketch a brief genealogical table, when the eye
comes immediately to the assistance of the brain. In these days, when we
have a growing lack of respect for pedigrees, most people have got out
of the habit of rapidly drawing such tables, which is to be regretted,
as they would save a lot of time and brain racking on occasions.


40.--MAMMA'S AGE.

Tommy: "How old are you, mamma?"

Mamma: "Let me think, Tommy. Well, our three ages add up to exactly
seventy years."

Tommy: "That's a lot, isn't it? And how old are you, papa?"

Papa: "Just six times as old as you, my son."

Tommy: "Shall I ever be half as old as you, papa?"

Papa: "Yes, Tommy; and when that happens our three ages will add up to
exactly twice as much as to-day."

Tommy: "And supposing I was born before you, papa; and supposing mamma
had forgot all about it, and hadn't been at home when I came; and
supposing--"

Mamma: "Supposing, Tommy, we talk about bed. Come along, darling. You'll
have a headache."

Now, if Tommy had been some years older he might have calculated the
exact ages of his parents from the information they had given him. Can
you find out the exact age of mamma?


41.--THEIR AGES.

"My husband's age," remarked a lady the other day, "is represented by
the figures of my own age reversed. He is my senior, and the difference
between our ages is one-eleventh of their sum."


42.--THE FAMILY AGES.

When the Smileys recently received a visit from the favourite uncle, the
fond parents had all the five children brought into his presence. First
came Billie and little Gertrude, and the uncle was informed that the boy
was exactly twice as old as the girl. Then Henrietta arrived, and it was
pointed out that the combined ages of herself and Gertrude equalled
twice the age of Billie. Then Charlie came running in, and somebody
remarked that now the combined ages of the two boys were exactly twice
the combined ages of the two girls. The uncle was expressing his
astonishment at these coincidences when Janet came in. "Ah! uncle," she
exclaimed, "you have actually arrived on my twenty-first birthday!" To
this Mr. Smiley added the final staggerer: "Yes, and now the combined
ages of the three girls are exactly equal to twice the combined ages of
the two boys." Can you give the age of each child?


43.--MRS. TIMPKINS'S AGE.

Edwin: "Do you know, when the Timpkinses married eighteen years ago
Timpkins was three times as old as his wife, and to-day he is just twice
as old as she?"

Angelina: "Then how old was Mrs. Timpkins on the wedding day?"

Can you answer Angelina's question?


44--A CENSUS PUZZLE.

Mr. and Mrs. Jorkins have fifteen children, all born at intervals of one
year and a half. Miss Ada Jorkins, the eldest, had an objection to state
her age to the census man, but she admitted that she was just seven
times older than little Johnnie, the youngest of all. What was Ada's
age? Do not too hastily assume that you have solved this little poser.
You may find that you have made a bad blunder!


45.--MOTHER AND DAUGHTER.

"Mother, I wish you would give me a bicycle," said a girl of twelve the
other day.

"I do not think you are old enough yet, my dear," was the reply. "When I
am only three times as old as you are you shall have one."

Now, the mother's age is forty-five years. When may the young lady
expect to receive her present?


46.--MARY AND MARMADUKE.

Marmaduke: "Do you know, dear, that in seven years' time our combined
ages will be sixty-three years?"

Mary: "Is that really so? And yet it is a fact that when you were my
present age you were twice as old as I was then. I worked it out last
night."

Now, what are the ages of Mary and Marmaduke?


47--ROVER'S AGE.

"Now, then, Tommy, how old is Rover?" Mildred's young man asked her
brother.

"Well, five years ago," was the youngster's reply, "sister was four
times older than the dog, but now she is only three times as old."

Can you tell Rover's age?


48.--CONCERNING TOMMY'S AGE.

Tommy Smart was recently sent to a new school. On the first day of his
arrival the teacher asked him his age, and this was his curious reply:
"Well, you see, it is like this. At the time I was born--I forget the
year--my only sister, Ann, happened to be just one-quarter the age of
mother, and she is now one-third the age of father." "That's all very
well," said the teacher, "but what I want is not the age of your sister
Ann, but your own age." "I was just coming to that," Tommy answered; "I
am just a quarter of mother's present age, and in four years' time I
shall be a quarter the age of father. Isn't that funny?"

This was all the information that the teacher could get out of Tommy
Smart. Could you have told, from these facts, what was his precise age?
It is certainly a little puzzling.


49.--NEXT-DOOR NEIGHBOURS.

There were two families living next door to one another at Tooting
Bec--the Jupps and the Simkins. The united ages of the four Jupps
amounted to one hundred years, and the united ages of the Simkins also
amounted to the same. It was found in the case of each family that the
sum obtained by adding the squares of each of the children's ages to the
square of the mother's age equalled the square of the father's age. In
the case of the Jupps, however, Julia was one year older than her
brother Joe, whereas Sophy Simkin was two years older than her brother
Sammy. What was the age of each of the eight individuals?


50.--THE BAG OF NUTS.

Three boys were given a bag of nuts as a Christmas present, and it was
agreed that they should be divided in proportion to their ages, which
together amounted to 17½ years. Now the bag contained 770 nuts, and
as often as Herbert took four Robert took three, and as often as Herbert
took six Christopher took seven. The puzzle is to find out how many nuts
each had, and what were the boys' respective ages.


51.--HOW OLD WAS MARY?

Here is a funny little age problem, by the late Sam Loyd, which has been
very popular in the United States. Can you unravel the mystery?

The combined ages of Mary and Ann are forty-four years, and Mary is
twice as old as Ann was when Mary was half as old as Ann will be when
Ann is three times as old as Mary was when Mary was three times as old
as Ann. How old is Mary? That is all, but can you work it out? If not,
ask your friends to help you, and watch the shadow of bewilderment creep
over their faces as they attempt to grip the intricacies of the
question.


52.--QUEER RELATIONSHIPS.

"Speaking of relationships," said the Parson at a certain dinner-party,
"our legislators are getting the marriage law into a frightful tangle,
Here, for example, is a puzzling case that has come under my notice. Two
brothers married two sisters. One man died and the other man's wife also
died. Then the survivors married."

"The man married his deceased wife's sister under the recent Act?" put
in the Lawyer.

"Exactly. And therefore, under the civil law, he is legally married and
his child is legitimate. But, you see, the man is the woman's deceased
husband's brother, and therefore, also under the civil law, she is not
married to him and her child is illegitimate."

"He is married to her and she is not married to him!" said the Doctor.

"Quite so. And the child is the legitimate son of his father, but the
illegitimate son of his mother."

"Undoubtedly 'the law is a hass,'" the Artist exclaimed, "if I may be
permitted to say so," he added, with a bow to the Lawyer.

"Certainly," was the reply. "We lawyers try our best to break in the
beast to the service of man. Our legislators are responsible for the
breed."

"And this reminds me," went on the Parson, "of a man in my parish who
married the sister of his widow. This man--"

"Stop a moment, sir," said the Professor. "Married the sister of his
widow? Do you marry dead men in your parish?"

"No; but I will explain that later. Well, this man has a sister of his
own. Their names are Stephen Brown and Jane Brown. Last week a young
fellow turned up whom Stephen introduced to me as his nephew. Naturally,
I spoke of Jane as his aunt, but, to my astonishment, the youth
corrected me, assuring me that, though he was the nephew of Stephen, he
was not the nephew of Jane, the sister of Stephen. This perplexed me a
good deal, but it is quite correct."

The Lawyer was the first to get at the heart of the mystery. What was
his solution?


53.--HEARD ON THE TUBE RAILWAY.

First Lady: "And was he related to you, dear?"

Second Lady: "Oh, yes. You see, that gentleman's mother was my mother's
mother-in-law, but he is not on speaking terms with my papa."

First Lady: "Oh, indeed!" (But you could see that she was not much
wiser.)

How was the gentleman related to the Second Lady?


54.--A FAMILY PARTY.

A certain family party consisted of 1 grandfather, 1 grandmother, 2
fathers, 2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2
sons, 2 daughters, 1 father-in-law, 1 mother-in-law, and 1
daughter-in-law. Twenty-three people, you will say. No; there were only
seven persons present. Can you show how this might be?


55.--A MIXED PEDIGREE.

Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!"

John Snoggs: "It's very simple. Listen again! You happen to be my
father's brother-in-law, my brother's father-in-law, and also my
father-in-law's brother. You see, my father was--"

But Mr. Bloggs refused to hear any more. Can the reader show how this
extraordinary triple relationship might have come about?


56.--WILSON'S POSER.

"Speaking of perplexities--" said Mr. Wilson, throwing down a magazine
on the table in the commercial room of the Railway Hotel.

"Who was speaking of perplexities?" inquired Mr. Stubbs.

"Well, then, reading about them, if you want to be exact--it just
occurred to me that perhaps you three men may be interested in a little
matter connected with myself."

It was Christmas Eve, and the four commercial travellers were spending
the holiday at Grassminster. Probably each suspected that the others had
no homes, and perhaps each was conscious of the fact that he was in that
predicament himself. In any case they seemed to be perfectly
comfortable, and as they drew round the cheerful fire the conversation
became general.

"What is the difficulty?" asked Mr. Packhurst.

"There's no difficulty in the matter, when you rightly understand it. It
is like this. A man named Parker had a flying-machine that would carry
two. He was a venturesome sort of chap--reckless, I should call him--and
he had some bother in finding a man willing to risk his life in making
an ascent with him. However, an uncle of mine thought he would chance
it, and one fine morning he took his seat in the machine and she started
off well. When they were up about a thousand feet, my nephew
suddenly--"

"Here, stop, Wilson! What was your nephew doing there? You said your
uncle," interrupted Mr. Stubbs.

"Did I? Well, it does not matter. My nephew suddenly turned to Parker
and said that the engine wasn't running well, so Parker called out to my
uncle--"

"Look here," broke in Mr. Waterson, "we are getting mixed. Was it your
uncle or your nephew? Let's have it one way or the other."

"What I said is quite right. Parker called out to my uncle to do
something or other, when my nephew--"

"There you are again, Wilson," cried Mr. Stubbs; "once for all, are we
to understand that both your uncle and your nephew were on the machine?"

"Certainly. I thought I made that clear. Where was I? Well, my nephew
shouted back to Parker--"

"Phew! I'm sorry to interrupt you again, Wilson, but we can't get on
like this. Is it true that the machine would only carry two?"

"Of course. I said at the start that it only carried two."

"Then what in the name of aerostation do you mean by saying that there
were three persons on board?" shouted Mr. Stubbs.

"Who said there were three?"

"You have told us that Parker, your uncle, and your nephew went up on
this blessed flying-machine."

"That's right."

"And the thing would only carry two!"

"Right again."

"Wilson, I have known you for some time as a truthful man and a
temperate man," said Mr. Stubbs, solemnly. "But I am afraid since you
took up that new line of goods you have overworked yourself."

"Half a minute, Stubbs," interposed Mr. Waterson. "I see clearly where
we all slipped a cog. Of course, Wilson, you meant us to understand that
Parker is either your uncle or your nephew. Now we shall be all right if
you will just tell us whether Parker is your uncle or nephew."

"He is no relation to me whatever."

The three men sighed and looked anxiously at one another. Mr. Stubbs got
up from his chair to reach the matches, Mr. Packhurst proceeded to wind
up his watch, and Mr. Waterson took up the poker to attend to the fire.
It was an awkward moment, for at the season of goodwill nobody wished to
tell Mr. Wilson exactly what was in his mind.

"It's curious," said Mr. Wilson, very deliberately, "and it's rather
sad, how thick-headed some people are. You don't seem to grip the facts.
It never seems to have occurred to either of you that my uncle and my
nephew are one and the same man."

"What!" exclaimed all three together.

"Yes; David George Linklater is my uncle, and he is also my nephew.
Consequently, I am both his uncle and nephew. Queer, isn't it? I'll
explain how it comes about."

Mr. Wilson put the case so very simply that the three men saw how it
might happen without any marriage within the prohibited degrees. Perhaps
the reader can work it out for himself.



CLOCK PUZZLES.

    "Look at the clock!"

        _Ingoldsby Legends_.


In considering a few puzzles concerning clocks and watches, and the
times recorded by their hands under given conditions, it is well that a
particular convention should always be kept in mind. It is frequently
the case that a solution requires the assumption that the hands can
actually record a time involving a minute fraction of a second. Such a
time, of course, cannot be really indicated. Is the puzzle, therefore,
impossible of solution? The conclusion deduced from a logical syllogism
depends for its truth on the two premises assumed, and it is the same in
mathematics. Certain things are antecedently assumed, and the answer
depends entirely on the truth of those assumptions.

"If two horses," says Lagrange, "can pull a load of a certain weight, it
is natural to suppose that four horses could pull a load of double that
weight, six horses a load of three times that weight. Yet, strictly
speaking, such is not the case. For the inference is based on the
assumption that the four horses pull alike in amount and direction,
which in practice can scarcely ever be the case. It so happens that we
are frequently led in our reckonings to results which diverge widely
from reality. But the fault is not the fault of mathematics; for
mathematics always gives back to us exactly what we have put into it.
The ratio was constant according to that supposition. The result is
founded upon that supposition. If the supposition is false the result is
necessarily false."

If one man can reap a field in six days, we say two men will reap it in
three days, and three men will do the work in two days. We here assume,
as in the case of Lagrange's horses, that all the men are exactly
equally capable of work. But we assume even more than this. For when
three men get together they may waste time in gossip or play; or, on the
other hand, a spirit of rivalry may spur them on to greater diligence.
We may assume any conditions we like in a problem, provided they be
clearly expressed and understood, and the answer will be in accordance
with those conditions.


57.--WHAT WAS THE TIME?

"I say, Rackbrane, what is the time?" an acquaintance asked our friend
the professor the other day. The answer was certainly curious.

"If you add one quarter of the time from noon till now to half the time
from now till noon to-morrow, you will get the time exactly."

What was the time of day when the professor spoke?


58.--A TIME PUZZLE.

How many minutes is it until six o'clock if fifty minutes ago it was
four times as many minutes past three o'clock?


59.--A PUZZLING WATCH.

A friend pulled out his watch and said, "This watch of mine does not
keep perfect time; I must have it seen to. I have noticed that the
minute hand and the hour hand are exactly together every sixty-five
minutes." Does that watch gain or lose, and how much per hour?


60.--THE WAPSHAW'S WHARF MYSTERY.

There was a great commotion in Lower Thames Street on the morning of
January 12, 1887. When the early members of the staff arrived at
Wapshaw's Wharf they found that the safe had been broken open, a
considerable sum of money removed, and the offices left in great
disorder. The night watchman was nowhere to be found, but nobody who had
been acquainted with him for one moment suspected him to be guilty of
the robbery. In this belief the proprietors were confirmed when, later
in the day, they were informed that the poor fellow's body had been
picked up by the River Police. Certain marks of violence pointed to the
fact that he had been brutally attacked and thrown into the river. A
watch found in his pocket had stopped, as is invariably the case in such
circumstances, and this was a valuable clue to the time of the outrage.
But a very stupid officer (and we invariably find one or two stupid
individuals in the most intelligent bodies of men) had actually amused
himself by turning the hands round and round, trying to set the watch
going again. After he had been severely reprimanded for this serious
indiscretion, he was asked whether he could remember the time that was
indicated by the watch when found. He replied that he could not, but he
recollected that the hour hand and minute hand were exactly together,
one above the other, and the second hand had just passed the forty-ninth
second. More than this he could not remember.

What was the exact time at which the watchman's watch stopped? The watch
is, of course, assumed to have been an accurate one.


61.--CHANGING PLACES.

[Illustration]

The above clock face indicates a little before 42 minutes past 4. The
hands will again point at exactly the same spots a little after 23
minutes past 8. In fact, the hands will have changed places. How many
times do the hands of a clock change places between three o'clock p.m.
and midnight? And out of all the pairs of times indicated by these
changes, what is the exact time when the minute hand will be nearest to
the point IX?


62.--THE CLUB CLOCK.

One of the big clocks in the Cogitators' Club was found the other night
to have stopped just when, as will be seen in the illustration, the
second hand was exactly midway between the other two hands. One of the
members proposed to some of his friends that they should tell him the
exact time when (if the clock had not stopped) the second hand would
next again have been midway between the minute hand and the hour hand.
Can you find the correct time that it would happen?

[Illustration]


63.--THE STOP-WATCH.

[Illustration]

We have here a stop-watch with three hands. The second hand, which
travels once round the face in a minute, is the one with the little ring
at its end near the centre. Our dial indicates the exact time when its
owner stopped the watch. You will notice that the three hands are nearly
equidistant. The hour and minute hands point to spots that are exactly a
third of the circumference apart, but the second hand is a little too
advanced. An exact equidistance for the three hands is not possible.
Now, we want to know what the time will be when the three hands are next
at exactly the same distances as shown from one another. Can you state
the time?


64.--THE THREE CLOCKS.

On Friday, April 1, 1898, three new clocks were all set going precisely
at the same time--twelve noon. At noon on the following day it was found
that clock A had kept perfect time, that clock B had gained exactly one
minute, and that clock C had lost exactly one minute. Now, supposing
that the clocks B and C had not been regulated, but all three allowed to
go on as they had begun, and that they maintained the same rates of
progress without stopping, on what date and at what time of day would
all three pairs of hands again point at the same moment at twelve
o'clock?


65.--THE RAILWAY STATION CLOCK.

A clock hangs on the wall of a railway station, 71 ft. 9 in. long and 10
ft. 4 in. high. Those are the dimensions of the wall, not of the clock!
While waiting for a train we noticed that the hands of the clock were
pointing in opposite directions, and were parallel to one of the
diagonals of the wall. What was the exact time?


66.--THE VILLAGE SIMPLETON.

A facetious individual who was taking a long walk in the country came
upon a yokel sitting on a stile. As the gentleman was not quite sure of
his road, he thought he would make inquiries of the local inhabitant;
but at the first glance he jumped too hastily to the conclusion that he
had dropped on the village idiot. He therefore decided to test the
fellow's intelligence by first putting to him the simplest question he
could think of, which was, "What day of the week is this, my good man?"
The following is the smart answer that he received:--

"When the day after to-morrow is yesterday, to-day will be as far from
Sunday as to-day was from Sunday when the day before yesterday was
to-morrow."

Can the reader say what day of the week it was? It is pretty evident
that the countryman was not such a fool as he looked. The gentleman went
on his road a puzzled but a wiser man.



LOCOMOTION AND SPEED PUZZLES.

"The race is not to the swift."--_Ecclesiastes_ ix. II.


67.--AVERAGE SPEED.

In a recent motor ride it was found that we had gone at the rate of ten
miles an hour, but we did the return journey over the same route, owing
to the roads being more clear of traffic, at fifteen miles an hour. What
was our average speed? Do not be too hasty in your answer to this simple
little question, or it is pretty certain that you will be wrong.


68.--THE TWO TRAINS.

I put this little question to a stationmaster, and his correct answer
was so prompt that I am convinced there is no necessity to seek talented
railway officials in America or elsewhere.

Two trains start at the same time, one from London to Liverpool, the
other from Liverpool to London. If they arrive at their destinations one
hour and four hours respectively after passing one another, how much
faster is one train running than the other?


69.--THE THREE VILLAGES.

I set out the other day to ride in a motor-car from Acrefield to
Butterford, but by mistake I took the road going _via_ Cheesebury, which
is nearer Acrefield than Butterford, and is twelve miles to the left of
the direct road I should have travelled. After arriving at Butterford I
found that I had gone thirty-five miles. What are the three distances
between these villages, each being a whole number of miles? I may
mention that the three roads are quite straight.


70.--DRAWING HER PENSION.

"Speaking of odd figures," said a gentleman who occupies some post in a
Government office, "one of the queerest characters I know is an old lame
widow who climbs up a hill every week to draw her pension at the village
post office. She crawls up at the rate of a mile and a half an hour and
comes down at the rate of four and a half miles an hour, so that it
takes her just six hours to make the double journey. Can any of you tell
me how far it is from the bottom of the hill to the top?"

[Illustration]


71.--SIR EDWYN DE TUDOR.

In the illustration we have a sketch of Sir Edwyn de Tudor going to
rescue his lady-love, the fair Isabella, who was held a captive by a
neighbouring wicked baron. Sir Edwyn calculated that if he rode fifteen
miles an hour he would arrive at the castle an hour too soon, while if
he rode ten miles an hour he would get there just an hour too late. Now,
it was of the first importance that he should arrive at the exact time
appointed, in order that the rescue that he had planned should be a
success, and the time of the tryst was five o'clock, when the captive
lady would be taking her afternoon tea. The puzzle is to discover
exactly how far Sir Edwyn de Tudor had to ride.


72.--THE HYDROPLANE QUESTION.

The inhabitants of Slocomb-on-Sea were greatly excited over the visit of
a certain flying man. All the town turned out to see the flight of the
wonderful hydroplane, and, of course, Dobson and his family were there.
Master Tommy was in good form, and informed his father that Englishmen
made better airmen than Scotsmen and Irishmen because they are not so
heavy. "How do you make that out?" asked Mr. Dobson. "Well, you see,"
Tommy replied, "it is true that in Ireland there are men of Cork and in
Scotland men of Ayr, which is better still, but in England there are
lightermen." Unfortunately it had to be explained to Mrs. Dobson, and
this took the edge off the thing. The hydroplane flight was from Slocomb
to the neighbouring watering-place Poodleville--five miles distant. But
there was a strong wind, which so helped the airman that he made the
outward journey in the short time of ten minutes, though it took him an
hour to get back to the starting point at Slocomb, with the wind dead
against him. Now, how long would the ten miles have taken him if there
had been a perfect calm? Of course, the hydroplane's engine worked
uniformly throughout.


73.--DONKEY RIDING.

During a visit to the seaside Tommy and Evangeline insisted on having a
donkey race over the mile course on the sands. Mr. Dobson and some of
his friends whom he had met on the beach acted as judges, but, as the
donkeys were familiar acquaintances and declined to part company the
whole way, a dead heat was unavoidable. However, the judges, being
stationed at different points on the course, which was marked off in
quarter-miles, noted the following results:--The first three-quarters
were run in six and three-quarter minutes, the first half-mile took the
same time as the second half, and the third quarter was run in exactly
the same time as the last quarter. From these results Mr. Dobson amused
himself in discovering just how long it took those two donkeys to run
the whole mile. Can you give the answer?


74.--THE BASKET OF POTATOES.

A man had a basket containing fifty potatoes. He proposed to his son, as
a little recreation, that he should place these potatoes on the ground
in a straight line. The distance between the first and second potatoes
was to be one yard, between the second and third three yards, between
the third and fourth five yards, between the fourth and fifth seven
yards, and so on--an increase of two yards for every successive potato
laid down. Then the boy was to pick them up and put them in the basket
one at a time, the basket being placed beside the first potato. How far
would the boy have to travel to accomplish the feat of picking them all
up? We will not consider the journey involved in placing the potatoes,
so that he starts from the basket with them all laid out.


75.--THE PASSENGER'S FARE.

At first sight you would hardly think there was matter for dispute in
the question involved in the following little incident, yet it took the
two persons concerned some little time to come to an agreement. Mr.
Smithers hired a motor-car to take him from Addleford to Clinkerville
and back again for £3. At Bakenham, just midway, he picked up an
acquaintance, Mr. Tompkins, and agreed to take him on to Clinkerville
and bring him back to Bakenham on the return journey. How much should he
have charged the passenger? That is the question. What was a reasonable
fare for Mr. Tompkins?



DIGITAL PUZZLES.

    "Nine worthies were they called."
        DRYDEN: _The Flower and the Leaf._

I give these puzzles, dealing with the nine digits, a class to
themselves, because I have always thought that they deserve more
consideration than they usually receive. Beyond the mere trick of
"casting out nines," very little seems to be generally known of the laws
involved in these problems, and yet an acquaintance with the properties
of the digits often supplies, among other uses, a certain number of
arithmetical checks that are of real value in the saving of labour. Let
me give just one example--the first that occurs to me.

If the reader were required to determine whether or not
15,763,530,163,289 is a square number, how would he proceed? If the
number had ended with a 2, 3, 7, or 8 in the digits place, of course he
would know that it could not be a square, but there is nothing in its
apparent form to prevent its being one. I suspect that in such a case he
would set to work, with a sigh or a groan, at the laborious task of
extracting the square root. Yet if he had given a little attention to
the study of the digital properties of numbers, he would settle the
question in this simple way. The sum of the digits is 59, the sum of
which is 14, the sum of which is 5 (which I call the "digital root"),
and therefore I know that the number cannot be a square, and for this
reason. The digital root of successive square numbers from 1 upwards is
always 1, 4, 7, or 9, and can never be anything else. In fact, the
series, 1, 4, 9, 7, 7, 9, 4, 1, 9, is repeated into infinity. The
analogous series for triangular numbers is 1, 3, 6, 1, 6, 3, 1, 9, 9. So
here we have a similar negative check, for a number cannot be triangular
(that is, (n² + n)/2) if its digital root be 2, 4, 5, 7, or 8.


76.--THE BARREL OF BEER.

A man bought an odd lot of wine in barrels and one barrel containing
beer. These are shown in the illustration, marked with the number of
gallons that each barrel contained. He sold a quantity of the wine to
one man and twice the quantity to another, but kept the beer to himself.
The puzzle is to point out which barrel contains beer. Can you say which
one it is? Of course, the man sold the barrels just as he bought them,
without manipulating in any way the contents.

[Illustration:

               ( 15 Gals )

          (31 Gals)   (19 Gals)

    (20 Gals)   (16 Gals)   (18 Gals)

]


77.--DIGITS AND SQUARES.

[Illustration:

    +---+---+---+
    | 1 | 9 | 2 |
    +---+---+---+
    | 3 | 8 | 4 |
    +---+---+---+
    | 5 | 7 | 6 |
    +---+---+---+

]

It will be seen in the diagram that we have so arranged the nine digits
in a square that the number in the second row is twice that in the first
row, and the number in the bottom row three times that in the top row.
There are three other ways of arranging the digits so as to produce the
same result. Can you find them?


78.--ODD AND EVEN DIGITS.

The odd digits, 1, 3, 5, 7, and 9, add up 25, while the even figures, 2,
4, 6, and 8, only add up 20. Arrange these figures so that the odd ones
and the even ones add up alike. Complex and improper fractions and
recurring decimals are not allowed.


79.--THE LOCKERS PUZZLE.

[Illustration:

            A                   B                   C
    ==================  ==================  ==================
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    |                |  |                |  |                |
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    |                |  |                |  |                |
    ==================  ==================  ==================
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |  | |
    | +--+ +--+ +--+ |  | +--+ +--+ +--+ |  | +--+ +--+ +--+ |
    ------------------  ------------------  ------------------

]

A man had in his office three cupboards, each containing nine lockers,
as shown in the diagram. He told his clerk to place a different
one-figure number on each locker of cupboard A, and to do the same in
the case of B, and of C. As we are here allowed to call nought a digit,
and he was not prohibited from using nought as a number, he clearly had
the option of omitting any one of ten digits from each cupboard.

Now, the employer did not say the lockers were to be numbered in any
numerical order, and he was surprised to find, when the work was done,
that the figures had apparently been mixed up indiscriminately. Calling
upon his clerk for an explanation, the eccentric lad stated that the
notion had occurred to him so to arrange the figures that in each case
they formed a simple addition sum, the two upper rows of figures
producing the sum in the lowest row. But the most surprising point was
this: that he had so arranged them that the addition in A gave the
smallest possible sum, that the addition in C gave the largest possible
sum, and that all the nine digits in the three totals were different.
The puzzle is to show how this could be done. No decimals are allowed
and the nought may not appear in the hundreds place.


80.--THE THREE GROUPS.

There appeared in "Nouvelles Annales de Mathématiques" the following
puzzle as a modification of one of my "Canterbury Puzzles." Arrange the
nine digits in three groups of two, three, and four digits, so that the
first two numbers when multiplied together make the third. Thus, 12 ×
483 = 5,796. I now also propose to include the cases where there are
one, four, and four digits, such as 4 × 1,738 = 6,952. Can you find all
the possible solutions in both cases?


81.--THE NINE COUNTERS.

[Illustration:

    (1)(5)(8)    (7)(9)
       (2)(3)    (4)(6)

]

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4,
5, 6, 7, 8 and 9. I arranged them on the table in two groups, as shown
in the illustration, so as to form two multiplication sums, and found
that both sums gave the same product. You will find that 158 multiplied
by 23 is 3,634, and that 79 multiplied by 46 is also 3,634. Now, the
puzzle I propose is to rearrange the counters so as to get as large a
product as possible. What is the best way of placing them? Remember both
groups must multiply to the same amount, and there must be three
counters multiplied by two in one case, and two multiplied by two
counters in the other, just as at present.



82.--THE TEN COUNTERS.

In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7,
8, 9. The puzzle is, as in the last case, so to arrange the ten counters
that the products of the two multiplications shall be the same, and you
may here have one or more figures in the multiplier, as you choose. The
above is a very easy feat; but it is also required to find the two
arrangements giving pairs of the highest and lowest products possible.
Of course every counter must be used, and the cipher may not be placed
to the left of a row of figures where it would have no effect. Vulgar
fractions or decimals are not allowed.


83.--DIGITAL MULTIPLICATION.

Here is another entertaining problem with the nine digits, the nought
being excluded. Using each figure once, and only once, we can form two
multiplication sums that have the same product, and this may be done in
many ways. For example, 7 × 658 and 14 × 329 contain all the digits
once, and the product in each case is the same--4,606. Now, it will be
seen that the sum of the digits in the product is 16, which is neither
the highest nor the lowest sum so obtainable. Can you find the solution
of the problem that gives the lowest possible sum of digits in the
common product? Also that which gives the highest possible sum?


84.--THE PIERROT'S PUZZLE.

[Illustration]

The Pierrot in the illustration is standing in a posture that represents
the sign of multiplication. He is indicating the peculiar fact that 15
multiplied by 93 produces exactly the same figures (1,395), differently
arranged. The puzzle is to take any four digits you like (all different)
and similarly arrange them so that the number formed on one side of the
Pierrot when multiplied by the number on the other side shall produce
the same figures. There are very few ways of doing it, and I shall give
all the cases possible. Can you find them all? You are allowed to put
two figures on each side of the Pierrot as in the example shown, or to
place a single figure on one side and three figures on the other. If we
only used three digits instead of four, the only possible ways are
these: 3 multiplied by 51 equals 153, and 6 multiplied by 21 equals 126.


85.--THE CAB NUMBERS.

A London policeman one night saw two cabs drive off in opposite
directions under suspicious circumstances. This officer was a
particularly careful and wide-awake man, and he took out his pocket-book
to make an entry of the numbers of the cabs, but discovered that he had
lost his pencil. Luckily, however, he found a small piece of chalk, with
which he marked the two numbers on the gateway of a wharf close by. When
he returned to the same spot on his beat he stood and looked again at
the numbers, and noticed this peculiarity, that all the nine digits (no
nought) were used and that no figure was repeated, but that if he
multiplied the two numbers together they again produced the nine digits,
all once, and once only. When one of the clerks arrived at the wharf in
the early morning, he observed the chalk marks and carefully rubbed them
out. As the policeman could not remember them, certain mathematicians
were then consulted as to whether there was any known method for
discovering all the pairs of numbers that have the peculiarity that the
officer had noticed; but they knew of none. The investigation, however,
was interesting, and the following question out of many was proposed:
What two numbers, containing together all the nine digits, will, when
multiplied together, produce another number (the _highest possible_)
containing also all the nine digits? The nought is not allowed anywhere.


86.--QUEER MULTIPLICATION.

If I multiply 51,249,876 by 3 (thus using all the nine digits once, and
once only), I get 153,749,628 (which again contains all the nine digits
once). Similarly, if I multiply 16,583,742 by 9 the result is
149,253,678, where in each case all the nine digits are used. Now, take
6 as your multiplier and try to arrange the remaining eight digits so as
to produce by multiplication a number containing all nine once, and once
only. You will find it far from easy, but it can be done.


87.--THE NUMBER-CHECKS PUZZLE.

[Illustration]

Where a large number of workmen are employed on a building it is
customary to provide every man with a little disc bearing his number.
These are hung on a board by the men as they arrive, and serve as a
check on punctuality. Now, I once noticed a foreman remove a number of
these checks from his board and place them on a split-ring which he
carried in his pocket. This at once gave me the idea for a good puzzle.
In fact, I will confide to my readers that this is just how ideas for
puzzles arise. You cannot really create an idea: it happens--and you
have to be on the alert to seize it when it does so happen.

It will be seen from the illustration that there are ten of these
checks on a ring, numbered 1 to 9 and 0. The puzzle is to divide them
into three groups without taking any off the ring, so that the first
group multiplied by the second makes the third group. For example, we
can divide them into the three groups, 2--8 9 0 7--1 5 4 6 3, by
bringing the 6 and the 3 round to the 4, but unfortunately the first
two when multiplied together do not make the third. Can you separate
them correctly? Of course you may have as many of the checks as you
like in any group. The puzzle calls for some ingenuity, unless you
have the luck to hit on the answer by chance.


88.--DIGITAL DIVISION.

It is another good puzzle so to arrange the nine digits (the nought
excluded) into two groups so that one group when divided by the other
produces a given number without remainder. For example, 1 3 4 5 8
divided by 6 7 2 9 gives 2. Can the reader find similar arrangements
producing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find the
pairs of smallest possible numbers in each case? Thus, 1 4 6 5 8 divided
by 7 3 2 9 is just as correct for 2 as the other example we have given,
but the numbers are higher.


89.--ADDING THE DIGITS.

If I write the sum of money, £987, 5s. 4½d., and add up the digits,
they sum to 36. No digit has thus been used a second time in the amount
or addition. This is the largest amount possible under the conditions.
Now find the smallest possible amount, pounds, shillings, pence, and
farthings being all represented. You need not use more of the nine
digits than you choose, but no digit may be repeated throughout. The
nought is not allowed.


90.--THE CENTURY PUZZLE.

Can you write 100 in the form of a mixed number, using all the nine
digits once, and only once? The late distinguished French mathematician,
Edouard Lucas, found seven different ways of doing it, and expressed his
doubts as to there being any other ways. As a matter of fact there are
just eleven ways and no more. Here is one of them, 91+5742/638. Nine of
the other ways have similarly two figures in the integral part of the
number, but the eleventh expression has only one figure there. Can the
reader find this last form?


91.--MORE MIXED FRACTIONS.

When I first published my solution to the last puzzle, I was led to
attempt the expression of all numbers in turn up to 100 by a mixed
fraction containing all the nine digits. Here are twelve numbers for the
reader to try his hand at: 13, 14, 15, 16, 18, 20, 27, 36, 40, 69, 72,
94. Use every one of the nine digits once, and only once, in every case.


92.--DIGITAL SQUARE NUMBERS.

Here are the nine digits so arranged that they form four square numbers:
9, 81, 324, 576. Now, can you put them all together so as to form a
single square number--(I) the smallest possible, and (II) the largest
possible?


93.--THE MYSTIC ELEVEN.

Can you find the largest possible number containing any nine of the ten
digits (calling nought a digit) that can be divided by 11 without a
remainder? Can you also find the smallest possible number produced in
the same way that is divisible by 11? Here is an example, where the
digit 5 has been omitted: 896743012. This number contains nine of the
digits and is divisible by 11, but it is neither the largest nor the
smallest number that will work.


94.--THE DIGITAL CENTURY.

1 2 3 4 5 6 7 8 9 = 100.

It is required to place arithmetical signs between the nine figures so
that they shall equal 100. Of course, you must not alter the present
numerical arrangement of the figures. Can you give a correct solution
that employs (1) the fewest possible signs, and (2) the fewest possible
separate strokes or dots of the pen? That is, it is necessary to use as
few signs as possible, and those signs should be of the simplest form.
The signs of addition and multiplication (+ and ×) will thus count as
two strokes, the sign of subtraction (-) as one stroke, the sign of
division (÷) as three, and so on.


95.--THE FOUR SEVENS.

[Illustration]

In the illustration Professor Rackbrane is seen demonstrating one of the
little posers with which he is accustomed to entertain his class. He
believes that by taking his pupils off the beaten tracks he is the
better able to secure their attention, and to induce original and
ingenious methods of thought. He has, it will be seen, just shown how
four 5's may be written with simple arithmetical signs so as to
represent 100. Every juvenile reader will see at a glance that his
example is quite correct. Now, what he wants you to do is this: Arrange
four 7's (neither more nor less) with arithmetical signs so that they
shall represent 100. If he had said we were to use four 9's we might at
once have written 99+9/9, but the four 7's call for rather more
ingenuity. Can you discover the little trick?


96.--THE DICE NUMBERS.

[Illustration]

I have a set of four dice, not marked with spots in the ordinary way,
but with Arabic figures, as shown in the illustration. Each die, of
course, bears the numbers 1 to 6. When put together they will form a
good many, different numbers. As represented they make the number 1246.
Now, if I make all the different four-figure numbers that are possible
with these dice (never putting the same figure more than once in any
number), what will they all add up to? You are allowed to turn the 6
upside down, so as to represent a 9. I do not ask, or expect, the reader
to go to all the labour of writing out the full list of numbers and then
adding them up. Life is not long enough for such wasted energy. Can you
get at the answer in any other way?



VARIOUS ARITHMETICAL AND ALGEBRAICAL PROBLEMS.

    "Variety's the very spice of life,
    That gives it all its flavour."

        COWPER: _The Task._



97.--THE SPOT ON THE TABLE.

A boy, recently home from school, wished to give his father an
exhibition of his precocity. He pushed a large circular table into the
corner of the room, as shown in the illustration, so that it touched
both walls, and he then pointed to a spot of ink on the extreme edge.

[Illustration]

"Here is a little puzzle for you, pater," said the youth. "That spot is
exactly eight inches from one wall and nine inches from the other. Can
you tell me the diameter of the table without measuring it?"

The boy was overheard to tell a friend, "It fairly beat the guv'nor;"
but his father is known to have remarked to a City acquaintance that he
solved the thing in his head in a minute. I often wonder which spoke the
truth.


98.--ACADEMIC COURTESIES.

In a certain mixed school, where a special feature was made of the
inculcation of good manners, they had a curious rule on assembling every
morning. There were twice as many girls as boys. Every girl made a bow
to every other girl, to every boy, and to the teacher. Every boy made a
bow to every other boy, to every girl, and to the teacher. In all there
were nine hundred bows made in that model academy every morning. Now,
can you say exactly how many boys there were in the school? If you are
not very careful, you are likely to get a good deal out in your
calculation.


99.--THE THIRTY-THREE PEARLS.

[Illustration]

"A man I know," said Teddy Nicholson at a certain family party,
"possesses a string of thirty-three pearls. The middle pearl is the
largest and best of all, and the others are so selected and arranged
that, starting from one end, each successive pearl is worth £100 more
than the preceding one, right up to the big pearl. From the other end
the pearls increase in value by £150 up to the large pearl. The whole
string is worth £65,000. What is the value of that large pearl?"

"Pearls and other articles of clothing," said Uncle Walter, when the
price of the precious gem had been discovered, "remind me of Adam and
Eve. Authorities, you may not know, differ as to the number of apples
that were eaten by Adam and Eve. It is the opinion of some that Eve 8
(ate) and Adam 2 (too), a total of 10 only. But certain mathematicians
have figured it out differently, and hold that Eve 8 and Adam a total of
16. Yet the most recent investigators think the above figures entirely
wrong, for if Eve 8 and Adam 8 2, the total must be 90."

"Well," said Harry, "it seems to me that if there were giants in those
days, probably Eve 8 1 and Adam 8 2, which would give a total of 163."

"I am not at all satisfied," said Maud. "It seems to me that if Eve 8 1
and Adam 8 1 2, they together consumed 893."

"I am sure you are all wrong," insisted Mr. Wilson, "for I consider that
Eve 8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of 8,938."

"But, look here," broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2 4
2 oblige Eve, surely the total must have been 82,056!"

At this point Uncle Walter suggested that they might let the matter
rest. He declared it to be clearly what mathematicians call an
indeterminate problem.


100.--THE LABOURER'S PUZZLE.

Professor Rackbrane, during one of his rambles, chanced to come upon a
man digging a deep hole.

"Good morning," he said. "How deep is that hole?"

"Guess," replied the labourer. "My height is exactly five feet ten
inches."

"How much deeper are you going?" said the professor.

"I am going twice as deep," was the answer, "and then my head will be
twice as far below ground as it is now above ground."

Rackbrane now asks if you could tell how deep that hole would be when
finished.


101.--THE TRUSSES OF HAY.

Farmer Tompkins had five trusses of hay, which he told his man Hodge to
weigh before delivering them to a customer. The stupid fellow weighed
them two at a time in all possible ways, and informed his master that
the weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120,
and 121. Now, how was Farmer Tompkins to find out from these figures how
much every one of the five trusses weighed singly? The reader may at
first think that he ought to be told "which pair is which pair," or
something of that sort, but it is quite unnecessary. Can you give the
five correct weights?


102.--MR. GUBBINS IN A FOG.

Mr. Gubbins, a diligent man of business, was much inconvenienced by a
London fog. The electric light happened to be out of order and he had to
manage as best he could with two candles. His clerk assured him that
though both were of the same length one candle would burn for four hours
and the other for five hours. After he had been working some time he put
the candles out as the fog had lifted, and he then noticed that what
remained of one candle was exactly four times the length of what was
left of the other.

When he got home that night Mr. Gubbins, who liked a good puzzle, said
to himself, "Of course it is possible to work out just how long those
two candles were burning to-day. I'll have a shot at it." But he soon
found himself in a worse fog than the atmospheric one. Could you have
assisted him in his dilemma? How long were the candles burning?


103.--PAINTING THE LAMP-POSTS.

Tim Murphy and Pat Donovan were engaged by the local authorities to
paint the lamp-posts in a certain street. Tim, who was an early riser,
arrived first on the job, and had painted three on the south side when
Pat turned up and pointed out that Tim's contract was for the north
side. So Tim started afresh on the north side and Pat continued on the
south. When Pat had finished his side he went across the street and
painted six posts for Tim, and then the job was finished. As there was
an equal number of lamp-posts on each side of the street, the simple
question is: Which man painted the more lamp-posts, and just how many
more?


104.--CATCHING THE THIEF.

"Now, constable," said the defendant's counsel in cross-examination,"
you say that the prisoner was exactly twenty-seven steps ahead of you
when you started to run after him?"

"Yes, sir."

"And you swear that he takes eight steps to your five?"

"That is so."

"Then I ask you, constable, as an intelligent man, to explain how you
ever caught him, if that is the case?"

"Well, you see, I have got a longer stride. In fact, two of my steps are
equal in length to five of the prisoner's. If you work it out, you will
find that the number of steps I required would bring me exactly to the
spot where I captured him."

Here the foreman of the jury asked for a few minutes to figure out the
number of steps the constable must have taken. Can you also say how many
steps the officer needed to catch the thief?


105.--THE PARISH COUNCIL ELECTION.

Here is an easy problem for the novice. At the last election of the
parish council of Tittlebury-in-the-Marsh there were twenty-three
candidates for nine seats. Each voter was qualified to vote for nine of
these candidates or for any less number. One of the electors wants to
know in just how many different ways it was possible for him to vote.


106.--THE MUDDLETOWN ELECTION.

At the last Parliamentary election at Muddletown 5,473 votes were
polled. The Liberal was elected by a majority of 18 over the
Conservative, by 146 over the Independent, and by 575 over the
Socialist. Can you give a simple rule for figuring out how many votes
were polled for each candidate?


107.--THE SUFFRAGISTS' MEETING.

At a recent secret meeting of Suffragists a serious difference of
opinion arose. This led to a split, and a certain number left the
meeting. "I had half a mind to go myself," said the chair-woman, "and if
I had done so, two-thirds of us would have retired." "True," said
another member; "but if I had persuaded my friends Mrs. Wild and
Christine Armstrong to remain we should only have lost half our number."
Can you tell how many were present at the meeting at the start?


108.--THE LEAP-YEAR LADIES.

Last leap-year ladies lost no time in exercising the privilege of making
proposals of marriage. If the figures that reached me from an occult
source are correct, the following represents the state of affairs in
this country.

A number of women proposed once each, of whom one-eighth were widows. In
consequence, a number of men were to be married of whom one-eleventh
were widowers. Of the proposals made to widowers, one-fifth were
declined. All the widows were accepted. Thirty-five forty-fourths of the
widows married bachelors. One thousand two hundred and twenty-one
spinsters were declined by bachelors. The number of spinsters accepted
by bachelors was seven times the number of widows accepted by bachelors.
Those are all the particulars that I was able to obtain. Now, how many
women proposed?


109.--THE GREAT SCRAMBLE.

After dinner, the five boys of a household happened to find a parcel of
sugar-plums. It was quite unexpected loot, and an exciting scramble
ensued, the full details of which I will recount with accuracy, as it
forms an interesting puzzle.

You see, Andrew managed to get possession of just two-thirds of the
parcel of sugar-plums. Bob at once grabbed three-eighths of these, and
Charlie managed to seize three-tenths also. Then young David dashed upon
the scene, and captured all that Andrew had left, except one-seventh,
which Edgar artfully secured for himself by a cunning trick. Now the fun
began in real earnest, for Andrew and Charlie jointly set upon Bob, who
stumbled against the fender and dropped half of all that he had, which
were equally picked up by David and Edgar, who had crawled under a table
and were waiting. Next, Bob sprang on Charlie from a chair, and upset
all the latter's collection on to the floor. Of this prize Andrew got
just a quarter, Bob gathered up one-third, David got two-sevenths, while
Charlie and Edgar divided equally what was left of that stock.

[Illustration]

They were just thinking the fray was over when David suddenly struck out
in two directions at once, upsetting three-quarters of what Bob and
Andrew had last acquired. The two latter, with the greatest difficulty,
recovered five-eighths of it in equal shares, but the three others each
carried off one-fifth of the same. Every sugar-plum was now accounted
for, and they called a truce, and divided equally amongst them the
remainder of the parcel. What is the smallest number of sugar-plums
there could have been at the start, and what proportion did each boy
obtain?


110.--THE ABBOT'S PUZZLE.

The first English puzzlist whose name has come down to us was a
Yorkshireman--no other than Alcuin, Abbot of Canterbury (A.D. 735-804).
Here is a little puzzle from his works, which is at least interesting on
account of its antiquity. "If 100 bushels of corn were distributed among
100 people in such a manner that each man received three bushels, each
woman two, and each child half a bushel, how many men, women, and
children were there?"

Now, there are six different correct answers, if we exclude a case where
there would be no women. But let us say that there were just five times
as many women as men, then what is the correct solution?


111.--REAPING THE CORN.

A farmer had a square cornfield. The corn was all ripe for reaping, and,
as he was short of men, it was arranged that he and his son should share
the work between them. The farmer first cut one rod wide all round the
square, thus leaving a smaller square of standing corn in the middle of
the field. "Now," he said to his son, "I have cut my half of the field,
and you can do your share." The son was not quite satisfied as to the
proposed division of labour, and as the village schoolmaster happened to
be passing, he appealed to that person to decide the matter. He found
the farmer was quite correct, provided there was no dispute as to the
size of the field, and on this point they were agreed. Can you tell the
area of the field, as that ingenious schoolmaster succeeded in doing?


112.--A PUZZLING LEGACY.

A man left a hundred acres of land to be divided among his three
sons--Alfred, Benjamin, and Charles--in the proportion of one-third,
one-fourth, and one-fifth respectively. But Charles died. How was the
land to be divided fairly between Alfred and Benjamin?

113.--THE TORN NUMBER.

[Illustration]

I had the other day in my possession a label bearing the number 3 0 2 5
in large figures. This got accidentally torn in half, so that 3 0 was on
one piece and 2 5 on the other, as shown on the illustration. On looking
at these pieces I began to make a calculation, scarcely conscious of
what I was doing, when I discovered this little peculiarity. If we add
the 3 0 and the 2 5 together and square the sum we get as the result the
complete original number on the label! Thus, 30 added to 25 is 55, and
55 multiplied by 55 is 3025. Curious, is it not? Now, the puzzle is to
find another number, composed of four figures, all different, which may
be divided in the middle and produce the same result.


114.--CURIOUS NUMBERS.

The number 48 has this peculiarity, that if you add 1 to it the result
is a square number (49, the square of 7), and if you add 1 to its half,
you also get a square number (25, the square of 5). Now, there is no
limit to the numbers that have this peculiarity, and it is an
interesting puzzle to find three more of them--the smallest possible
numbers. What are they?


115.--A PRINTER'S ERROR.

In a certain article a printer had to set up the figures 5^4×2^3, which,
of course, means that the fourth power of 5 (625) is to be multiplied by
the cube of 2 (8), the product of which is 5,000. But he printed 5^4×2^3
as 5 4 2 3, which is not correct. Can you place four digits in the
manner shown, so that it will be equally correct if the printer sets it
up aright or makes the same blunder?


116.--THE CONVERTED MISER.

Mr. Jasper Bullyon was one of the very few misers who have ever been
converted to a sense of their duty towards their less fortunate
fellow-men. One eventful night he counted out his accumulated wealth,
and resolved to distribute it amongst the deserving poor.

He found that if he gave away the same number of pounds every day in the
year, he could exactly spread it over a twelvemonth without there being
anything left over; but if he rested on the Sundays, and only gave away
a fixed number of pounds every weekday, there would be one sovereign
left over on New Year's Eve. Now, putting it at the lowest possible,
what was the exact number of pounds that he had to distribute?

Could any question be simpler? A sum of pounds divided by one number of
days leaves no remainder, but divided by another number of days leaves a
sovereign over. That is all; and yet, when you come to tackle this
little question, you will be surprised that it can become so puzzling.


117.--A FENCE PROBLEM.

[Illustration]

The practical usefulness of puzzles is a point that we are liable to
overlook. Yet, as a matter of fact, I have from time to time received
quite a large number of letters from individuals who have found that the
mastering of some little principle upon which a puzzle was built has
proved of considerable value to them in a most unexpected way. Indeed,
it may be accepted as a good maxim that a puzzle is of little real value
unless, as well as being amusing and perplexing, it conceals some
instructive and possibly useful feature. It is, however, very curious
how these little bits of acquired knowledge dovetail into the
occasional requirements of everyday life, and equally curious to what
strange and mysterious uses some of our readers seem to apply them.
What, for example, can be the object of Mr. Wm. Oxley, who writes to me
all the way from Iowa, in wishing to ascertain the dimensions of a field
that he proposes to enclose, containing just as many acres as there
shall be rails in the fence?

The man wishes to fence in a perfectly square field which is to contain
just as many acres as there are rails in the required fence. Each
hurdle, or portion of fence, is seven rails high, and two lengths would
extend one pole (16½ ft.): that is to say, there are fourteen rails
to the pole, lineal measure. Now, what must be the size of the field?


118.--CIRCLING THE SQUARES.

[Illustration]

The puzzle is to place a different number in each of the ten squares so
that the sum of the squares of any two adjacent numbers shall be equal
to the sum of the squares of the two numbers diametrically opposite to
them. The four numbers placed, as examples, must stand as they are. The
square of 16 is 256, and the square of 2 is 4. Add these together, and
the result is 260. Also--the square of 14 is 196, and the square of 8 is
64. These together also make 260. Now, in precisely the same way, B and
C should be equal to G and H (the sum will not necessarily be 260), A
and K to F and E, H and I to C and D, and so on, with any two adjoining
squares in the circle.

All you have to do is to fill in the remaining six numbers. Fractions
are not allowed, and I shall show that no number need contain more than
two figures.


119.--RACKBRANE'S LITTLE LOSS.

Professor Rackbrane was spending an evening with his old friends, Mr.
and Mrs. Potts, and they engaged in some game (he does not say what
game) of cards. The professor lost the first game, which resulted in
doubling the money that both Mr. and Mrs. Potts had laid on the table.
The second game was lost by Mrs. Potts, which doubled the money then
held by her husband and the professor. Curiously enough, the third game
was lost by Mr. Potts, and had the effect of doubling the money then
held by his wife and the professor. It was then found that each person
had exactly the same money, but the professor had lost five shillings in
the course of play. Now, the professor asks, what was the sum of money
with which he sat down at the table? Can you tell him?


120.--THE FARMER AND HIS SHEEP.

[Illustration]

Farmer Longmore had a curious aptitude for arithmetic, and was known in
his district as the "mathematical farmer." The new vicar was not aware
of this fact when, meeting his worthy parishioner one day in the lane,
he asked him in the course of a short conversation, "Now, how many sheep
have you altogether?" He was therefore rather surprised at Longmore's
answer, which was as follows: "You can divide my sheep into two
different parts, so that the difference between the two numbers is the
same as the difference between their squares. Maybe, Mr. Parson, you
will like to work out the little sum for yourself."

Can the reader say just how many sheep the farmer had? Supposing he had
possessed only twenty sheep, and he divided them into the two parts 12
and 8. Now, the difference between their squares, 144 and 64, is 80. So
that will not do, for 4 and 80 are certainly not the same. If you can
find numbers that work out correctly, you will know exactly how many
sheep Farmer Longmore owned.


121.--HEADS OR TAILS.

Crooks, an inveterate gambler, at Goodwood recently said to a friend,
"I'll bet you half the money in my pocket on the toss of a coin--heads I
win, tails I lose." The coin was tossed and the money handed over. He
repeated the offer again and again, each time betting half the money
then in his possession. We are not told how long the game went on, or
how many times the coin was tossed, but this we know, that the number of
times that Crooks lost was exactly equal to the number of times that he
won. Now, did he gain or lose by this little venture?


122.--THE SEE-SAW PUZZLE.

Necessity is, indeed, the mother of invention. I was amused the other
day in watching a boy who wanted to play see-saw and, in his failure to
find another child to share the sport with him, had been driven back
upon the ingenious resort of tying a number of bricks to one end of the
plank to balance his weight at the other.

As a matter of fact, he just balanced against sixteen bricks, when these
were fixed to the short end of plank, but if he fixed them to the long
end of plank he only needed eleven as balance.

Now, what was that boy's weight, if a brick weighs equal to a
three-quarter brick and three-quarters of a pound?


123.--A LEGAL DIFFICULTY.

"A client of mine," said a lawyer, "was on the point of death when his
wife was about to present him with a child. I drew up his will, in which
he settled two-thirds of his estate upon his son (if it should happen to
be a boy) and one-third on the mother. But if the child should be a
girl, then two-thirds of the estate should go to the mother and
one-third to the daughter. As a matter of fact, after his death twins
were born--a boy and a girl. A very nice point then arose. How was the
estate to be equitably divided among the three in the closest possible
accordance with the spirit of the dead man's will?"


124.--A QUESTION OF DEFINITION.

"My property is exactly a mile square," said one landowner to another.

"Curiously enough, mine is a square mile," was the reply.

"Then there is no difference?"

Is this last statement correct?


125.--THE MINERS' HOLIDAY.

Seven coal-miners took a holiday at the seaside during a big strike. Six
of the party spent exactly half a sovereign each, but Bill Harris was
more extravagant. Bill spent three shillings more than the average of
the party. What was the actual amount of Bill's expenditure?


126.--SIMPLE MULTIPLICATION.

If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on the
table in this order:--

    1     4     2     8     5     7

We can demonstrate that in order to multiply by 3 all that is necessary
is to remove the 1 to the other end of the row, and the thing is done.
The answer is 428571. Can you find a number that, when multiplied by 3
and divided by 2, the answer will be the same as if we removed the first
card (which in this case is to be a 3) From the beginning of the row to
the end?


127.--SIMPLE DIVISION.

Sometimes a very simple question in elementary arithmetic will cause a
good deal of perplexity. For example, I want to divide the four numbers,
701, 1,059, 1,417, and 2,312, by the largest number possible that will
leave the same remainder in every case. How am I to set to work Of
course, by a laborious system of trial one can in time discover the
answer, but there is quite a simple method of doing it if you can only
find it.


128.--A PROBLEM IN SQUARES.

We possess three square boards. The surface of the first contains five
square feet more than the second, and the second contains five square
feet more than the third. Can you give exact measurements for the sides
of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7
and also of 13.



129.--THE BATTLE OF HASTINGS.

All historians know that there is a great deal of mystery and
uncertainty concerning the details of the ever-memorable battle on that
fatal day, October 14, 1066. My puzzle deals with a curious passage in
an ancient monkish chronicle that may never receive the attention that
it deserves, and if I am unable to vouch for the authenticity of the
document it will none the less serve to furnish us with a problem that
can hardly fail to interest those of my readers who have arithmetical
predilections. Here is the passage in question.

"The men of Harold stood well together, as their wont was, and formed
sixty and one squares, with a like number of men in every square
thereof, and woe to the hardy Norman who ventured to enter their
redoubts; for a single blow of a Saxon war-hatchet would break his lance
and cut through his coat of mail.... When Harold threw himself into the
fray the Saxons were one mighty square of men, shouting the
battle-cries, 'Ut!' 'Olicrosse!' 'Godemitè!'"

Now, I find that all the contemporary authorities agree that the Saxons
did actually fight in this solid order. For example, in the "Carmen de
Bello Hastingensi," a poem attributed to Guy, Bishop of Amiens, living
at the time of the battle, we are told that "the Saxons stood fixed in a
dense mass," and Henry of Huntingdon records that "they were like unto a
castle, impenetrable to the Normans;" while Robert Wace, a century
after, tells us the same thing. So in this respect my newly-discovered
chronicle may not be greatly in error. But I have reason to believe that
there is something wrong with the actual figures. Let the reader see
what he can make of them.

The number of men would be sixty-one times a square number; but when
Harold himself joined in the fray they were then able to form one large
square. What is the smallest possible number of men there could have
been?

In order to make clear to the reader the simplicity of the question, I
will give the lowest solutions in the case of 60 and 62, the numbers
immediately preceding and following 61. They are 60 × 4² + 1 = 31²,
and 62 × 8² + 1 = 63². That is, 60 squares of 16 men each would be 960
men, and when Harold joined them they would be 961 in number, and so
form a square with 31 men on every side. Similarly in the case of the
figures I have given for 62. Now, find the lowest answer for 61.


130.--THE SCULPTOR'S PROBLEM.

An ancient sculptor was commissioned to supply two statues, each on a
cubical pedestal. It is with these pedestals that we are concerned. They
were of unequal sizes, as will be seen in the illustration, and when the
time arrived for payment a dispute arose as to whether the agreement was
based on lineal or cubical measurement. But as soon as they came to
measure the two pedestals the matter was at once settled, because,
curiously enough, the number of lineal feet was exactly the same as the
number of cubical feet. The puzzle is to find the dimensions for two
pedestals having this peculiarity, in the smallest possible figures. You
see, if the two pedestals, for example, measure respectively 3 ft. and 1
ft. on every side, then the lineal measurement would be 4 ft. and the
cubical contents 28 ft., which are not the same, so these measurements
will not do.

[Illustration]


131.--THE SPANISH MISER.

There once lived in a small town in New Castile a noted miser named Don
Manuel Rodriguez. His love of money was only equalled by a strong
passion for arithmetical problems. These puzzles usually dealt in some
way or other with his accumulated treasure, and were propounded by him
solely in order that he might have the pleasure of solving them himself.
Unfortunately very few of them have survived, and when travelling
through Spain, collecting material for a proposed work on "The Spanish
Onion as a Cause of National Decadence," I only discovered a very few.
One of these concerns the three boxes that appear in the accompanying
authentic portrait.

[Illustration]

Each box contained a different number of golden doubloons. The
difference between the number of doubloons in the upper box and the
number in the middle box was the same as the difference between the
number in the middle box and the number in the bottom box. And if the
contents of any two of the boxes were united they would form a square
number. What is the smallest number of doubloons that there could have
been in any one of the boxes?


132.--THE NINE TREASURE BOXES.

The following puzzle will illustrate the importance on occasions of
being able to fix the minimum and maximum limits of a required number.
This can very frequently be done. For example, it has not yet been
ascertained in how many different ways the knight's tour can be
performed on the chess board; but we know that it is fewer than the
number of combinations of 168 things taken 63 at a time and is greater
than 31,054,144--for the latter is the number of routes of a particular
type. Or, to take a more familiar case, if you ask a man how many coins
he has in his pocket, he may tell you that he has not the slightest
idea. But on further questioning you will get out of him some such
statement as the following: "Yes, I am positive that I have more than
three coins, and equally certain that there are not so many as
twenty-five." Now, the knowledge that a certain number lies between 2
and 12 in my puzzle will enable the solver to find the exact answer;
without that information there would be an infinite number of answers,
from which it would be impossible to select the correct one.

This is another puzzle received from my friend Don Manuel Rodriguez, the
cranky miser of New Castile. On New Year's Eve in 1879 he showed me nine
treasure boxes, and after informing me that every box contained a square
number of golden doubloons, and that the difference between the contents
of A and B was the same as between B and C, D and E, E and F, G and H,
or H and I, he requested me to tell him the number of coins in every one
of the boxes. At first I thought this was impossible, as there would be
an infinite number of different answers, but on consideration I found
that this was not the case. I discovered that while every box contained
coins, the contents of A, B, C increased in weight in alphabetical
order; so did D, E, F; and so did G, H, I; but D or E need not be
heavier than C, nor G or H heavier than F. It was also perfectly certain
that box A could not contain more than a dozen coins at the outside;
there might not be half that number, but I was positive that there were
not more than twelve. With this knowledge I was able to arrive at the
correct answer.

In short, we have to discover nine square numbers such that A, B, C; and
D, E, F; and G, H, I are three groups in arithmetical progression, the
common difference being the same in each group, and A being less than
12. How many doubloons were there in every one of the nine boxes?


133.--THE FIVE BRIGANDS.

The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban,
were counting their spoils after a raid, when it was found that they had
captured altogether exactly 200 doubloons. One of the band pointed out
that if Alfonso had twelve times as much, Benito three times as much,
Carlos the same amount, Diego half as much, and Esteban one-third as
much, they would still have altogether just 200 doubloons. How many
doubloons had each?

There are a good many equally correct answers to this question. Here is
one of them:

    A    6 × 12  = 72
    B   12 ×  3  = 36
    C   17 ×  1  = 17
    D  120 ×  ½  = 60
    E   45 × 1/3 = 15
       ___        ___
       200        200

The puzzle is to discover exactly how many different answers there are,
it being understood that every man had something and that there is to be
no fractional money--only doubloons in every case.

This problem, worded somewhat differently, was propounded by Tartaglia
(died 1559), and he flattered himself that he had found one solution;
but a French mathematician of note (M.A. Labosne), in a recent work,
says that his readers will be astonished when he assures them that there
are 6,639 different correct answers to the question. Is this so? How
many answers are there?


134.--THE BANKER'S PUZZLE.

A banker had a sporting customer who was always anxious to wager on
anything. Hoping to cure him of his bad habit, he proposed as a wager
that the customer would not be able to divide up the contents of a box
containing only sixpences into an exact number of equal piles of
sixpences. The banker was first to put in one or more sixpences (as many
as he liked); then the customer was to put in one or more (but in his
case not more than a pound in value), neither knowing what the other put
in. Lastly, the customer was to transfer from the banker's counter to
the box as many sixpences as the banker desired him to put in. The
puzzle is to find how many sixpences the banker should first put in and
how many he should ask the customer to transfer, so that he may have the
best chance of winning.


135.--THE STONEMASON'S PROBLEM.

A stonemason once had a large number of cubic blocks of stone in his
yard, all of exactly the same size. He had some very fanciful little
ways, and one of his queer notions was to keep these blocks piled in
cubical heaps, no two heaps containing the same number of blocks. He had
discovered for himself (a fact that is well known to mathematicians)
that if he took all the blocks contained in any number of heaps in
regular order, beginning with the single cube, he could always arrange
those on the ground so as to form a perfect square. This will be clear
to the reader, because one block is a square, 1 + 8 = 9 is a square, 1 +
8 + 27 = 36 is a square, 1 + 8 + 27 + 64 = 100 is a square, and so on.
In fact, the sum of any number of consecutive cubes, beginning always
with 1, is in every case a square number.

One day a gentleman entered the mason's yard and offered him a certain
price if he would supply him with a consecutive number of these cubical
heaps which should contain altogether a number of blocks that could be
laid out to form a square, but the buyer insisted on more than three
heaps and _declined to take the single block_ because it contained a
flaw. What was the smallest possible number of blocks of stone that the
mason had to supply?


136.--THE SULTAN'S ARMY.

A certain Sultan wished to send into battle an army that could be formed
into two perfect squares in twelve different ways. What is the smallest
number of men of which that army could be composed? To make it clear to
the novice, I will explain that if there were 130 men, they could be
formed into two squares in only two different ways--81 and 49, or 121
and 9. Of course, all the men must be used on every occasion.


137.--A STUDY IN THRIFT.

Certain numbers are called triangular, because if they are taken to
represent counters or coins they may be laid out on the table so as to
form triangles. The number 1 is always regarded as triangular, just as 1
is a square and a cube number. Place one counter on the table--that is,
the first triangular number. Now place two more counters beneath it, and
you have a triangle of three counters; therefore 3 is triangular. Next
place a row of three more counters, and you have a triangle of six
counters; therefore 6 is triangular. We see that every row of counters
that we add, containing just one more counter than the row above it,
makes a larger triangle.

Now, half the sum of any number and its square is always a triangular
number. Thus half of 2 + 2² = 3; half of 3 + 3² = 6; half of 4 +
4² = 10; half of 5 + 5²= 15; and so on. So if we want to form a
triangle with 8 counters on each side we shall require half of 8 +
8², or 36 counters. This is a pretty little property of numbers.
Before going further, I will here say that if the reader refers to the
"Stonemason's Problem" (No. 135) he will remember that the sum of any
number of consecutive cubes beginning with 1 is always a square, and
these form the series 1², 3², 6², 10², etc. It will now be understood
when I say that one of the keys to the puzzle was the fact that these
are always the squares of triangular numbers--that is, the squares of 1,
3, 6, 10, 15, 21, 28, etc., any of which numbers we have seen will form
a triangle.

Every whole number is either triangular, or the sum of two triangular
numbers or the sum of three triangular numbers. That is, if we take any
number we choose we can always form one, two, or three triangles with
them. The number 1 will obviously, and uniquely, only form one triangle;
some numbers will only form two triangles (as 2, 4, 11, etc.); some
numbers will only form three triangles (as 5, 8, 14, etc.). Then, again,
some numbers will form both one and two triangles (as 6), others both
one and three triangles (as 3 and 10), others both two and three
triangles (as 7 and 9), while some numbers (like 21) will form one, two,
or three triangles, as we desire. Now for a little puzzle in triangular
numbers.

Sandy McAllister, of Aberdeen, practised strict domestic economy, and
was anxious to train his good wife in his own habits of thrift. He told
her last New Year's Eve that when she had saved so many sovereigns that
she could lay them all out on the table so as to form a perfect square,
or a perfect triangle, or two triangles, or three triangles, just as he
might choose to ask he would add five pounds to her treasure. Soon she
went to her husband with a little bag of £36 in sovereigns and claimed
her reward. It will be found that the thirty-six coins will form a
square (with side 6), that they will form a single triangle (with side
8), that they will form two triangles (with sides 5 and 6), and that
they will form three triangles (with sides 3, 5, and 5). In each of the
four cases all the thirty-six coins are used, as required, and Sandy
therefore made his wife the promised present like an honest man.

The Scotsman then undertook to extend his promise for five more years,
so that if next year the increased number of sovereigns that she has
saved can be laid out in the same four different ways she will receive a
second present; if she succeeds in the following year she will get a
third present, and so on until she has earned six presents in all. Now,
how many sovereigns must she put together before she can win the sixth
present?

What you have to do is to find five numbers, the smallest possible,
higher than 36, that can be displayed in the four ways--to form a
square, to form a triangle, to form two triangles, and to form three
triangles. The highest of your five numbers will be your answer.


138.--THE ARTILLERYMEN'S DILEMMA.

[Illustration: [Pyramid of cannon-balls]]


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"All cannon-balls are to be piled in square pyramids," was the order
issued to the regiment. This was done. Then came the further order, "All
pyramids are to contain a square number of balls." Whereupon the trouble
arose. "It can't be done," said the major. "Look at this pyramid, for
example; there are sixteen balls at the base, then nine, then four, then
one at the top, making thirty balls in all. But there must be six more
balls, or five fewer, to make a square number." "It _must_ be done,"
insisted the general. "All you have to do is to put the right number of
balls in your pyramids." "I've got it!" said a lieutenant, the
mathematical genius of the regiment. "Lay the balls out singly." "Bosh!"
exclaimed the general. "You can't _pile_ one ball into a pyramid!" Is it
really possible to obey both orders?


139.--THE DUTCHMEN'S WIVES.

I wonder how many of my readers are acquainted with the puzzle of the
"Dutchmen's Wives"--in which you have to determine the names of three
men's wives, or, rather, which wife belongs to each husband. Some thirty
years ago it was "going the rounds," as something quite new, but I
recently discovered it in the _Ladies' Diary_ for 1739-40, so it was
clearly familiar to the fair sex over one hundred and seventy years ago.
How many of our mothers, wives, sisters, daughters, and aunts could
solve the puzzle to-day? A far greater proportion than then, let us
hope.

Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives,
Gurtrün, Katrün, and Anna, purchase hogs. Each buys as many as he (or
she) gives shillings for one. Each husband pays altogether three guineas
more than his wife. Hendrick buys twenty-three more hogs than Katrün,
and Elas eleven more than Gurtrün. Now, what was the name of each man's
wife?

[Illustration]


140.--FIND ADA'S SURNAME.

This puzzle closely resembles the last one, my remarks on the solution
of which the reader may like to apply in another case. It was recently
submitted to a Sydney evening newspaper that indulges in "intellect
sharpeners," but was rejected with the remark that it is childish and
that they only published problems capable of solution! Five ladies,
accompanied by their daughters, bought cloth at the same shop. Each of
the ten paid as many farthings per foot as she bought feet, and each
mother spent 8s. 5¼d. more than her daughter. Mrs. Robinson spent 6s.
more than Mrs. Evans, who spent about a quarter as much as Mrs. Jones.
Mrs. Smith spent most of all. Mrs. Brown bought 21 yards more than
Bessie--one of the girls. Annie bought 16 yards more than Mary and spent
£3, 0s. 8d. more than Emily. The Christian name of the other girl was
Ada. Now, what was her surname?


141.--SATURDAY MARKETING.

Here is an amusing little case of marketing which, although it deals
with a good many items of money, leads up to a question of a totally
different character. Four married couples went into their village on a
recent Saturday night to do a little marketing. They had to be very
economical, for among them they only possessed forty shilling coins. The
fact is, Ann spent 1s., Mary spent 2s., Jane spent 3s., and Kate spent
4s. The men were rather more extravagant than their wives, for Ned Smith
spent as much as his wife, Tom Brown twice as much as his wife, Bill
Jones three times as much as his wife, and Jack Robinson four times as
much as his wife. On the way home somebody suggested that they should
divide what coin they had left equally among them. This was done, and
the puzzling question is simply this: What was the surname of each
woman? Can you pair off the four couples?



GEOMETRICAL PROBLEMS.

    "God geometrizes continually."

    PLATO.

"There is no study," said Augustus de Morgan, "which presents so simple
a beginning as that of geometry; there is none in which difficulties
grow more rapidly as we proceed." This will be found when the reader
comes to consider the following puzzles, though they are not arranged in
strict order of difficulty. And the fact that they have interested and
given pleasure to man for untold ages is no doubt due in some measure to
the appeal they make to the eye as well as to the brain. Sometimes an
algebraical formula or theorem seems to give pleasure to the
mathematician's eye, but it is probably only an intellectual pleasure.
But there can be no doubt that in the case of certain geometrical
problems, notably dissection or superposition puzzles, the æsthetic
faculty in man contributes to the delight. For example, there are
probably few readers who will examine the various cuttings of the Greek
cross in the following pages without being in some degree stirred by a
sense of beauty. Law and order in Nature are always pleasing to
contemplate, but when they come under the very eye they seem to make a
specially strong appeal. Even the person with no geometrical knowledge
whatever is induced after the inspection of such things to exclaim, "How
very pretty!" In fact, I have known more than one person led on to a
study of geometry by the fascination of cutting-out puzzles. I have,
therefore, thought it well to keep these dissection puzzles distinct
from the geometrical problems on more general lines.



DISSECTION PUZZLES.


    "Take him and cut him out in little stars."

    _Romeo and Juliet_, iii. 2.

Puzzles have infinite variety, but perhaps there is no class more
ancient than dissection, cutting-out, or superposition puzzles. They
were certainly known to the Chinese several thousand years before the
Christian era. And they are just as fascinating to-day as they can have
been at any period of their history. It is supposed by those who have
investigated the matter that the ancient Chinese philosophers used these
puzzles as a sort of kindergarten method of imparting the principles of
geometry. Whether this was so or not, it is certain that all good
dissection puzzles (for the nursery type of jig-saw puzzle, which merely
consists in cutting up a picture into pieces to be put together again,
is not worthy of serious consideration) are really based on geometrical
laws. This statement need not, however, frighten off the novice, for it
means little more than this, that geometry will give us the "reason
why," if we are interested in knowing it, though the solutions may often
be discovered by any intelligent person after the exercise of patience,
ingenuity, and common sagacity.

If we want to cut one plane figure into parts that by readjustment will
form another figure, the first thing is to find a way of doing it at
all, and then to discover how to do it in the fewest possible pieces.
Often a dissection problem is quite easy apart from this limitation of
pieces. At the time of the publication in the _Weekly Dispatch_, in
1902, of a method of cutting an equilateral triangle into four parts
that will form a square (see No. 26, "Canterbury Puzzles"), no
geometrician would have had any difficulty in doing what is required in
five pieces: the whole point of the discovery lay in performing the
little feat in four pieces only.

Mere approximations in the case of these problems are valueless; the
solution must be geometrically exact, or it is not a solution at all.
Fallacies are cropping up now and again, and I shall have occasion to
refer to one or two of these. They are interesting merely as fallacies.
But I want to say something on two little points that are always arising
in cutting-out puzzles--the questions of "hanging by a thread" and
"turning over." These points can best be illustrated by a puzzle that is
frequently to be found in the old books, but invariably with a false
solution. The puzzle is to cut the figure shown in Fig. 1 into three
pieces that will fit together and form a half-square triangle. The
answer that is invariably given is that shown in Figs. 1 and 2. Now, it
is claimed that the four pieces marked C are really only one piece,
because they may be so cut that they are left "hanging together by a
mere thread." But no serious puzzle lover will ever admit this. If the
cut is made so as to leave the four pieces joined in one, then it cannot
result in a perfectly exact solution. If, on the other hand, the
solution is to be exact, then there will be four pieces--or six pieces
in all. It is, therefore, not a solution in three pieces.

[Illustration: Fig. 1]

[Illustration: Fig. 2]

If, however, the reader will look at the solution in Figs. 3 and 4, he
will see that no such fault can be found with it. There is no question
whatever that there are three pieces, and the solution is in this
respect quite satisfactory. But another question arises. It will be
found on inspection that the piece marked F, in Fig. 3, is turned over
in Fig. 4--that is to say, a different side has necessarily to be
presented. If the puzzle were merely to be cut out of cardboard or wood,
there might be no objection to this reversal, but it is quite possible
that the material would not admit of being reversed. There might be a
pattern, a polish, a difference of texture, that prevents it. But it is
generally understood that in dissection puzzles you are allowed to turn
pieces over unless it is distinctly stated that you may not do so. And
very often a puzzle is greatly improved by the added condition, "no
piece may be turned over." I have often made puzzles, too, in which the
diagram has a small repeated pattern, and the pieces have then so to be
cut that not only is there no turning over, but the pattern has to be
matched, which cannot be done if the pieces are turned round, even with
the proper side uppermost.

[Illustration: Fig. 3]

[Illustration: Fig. 4]

Before presenting a varied series of cutting-out puzzles, some very easy
and others difficult, I propose to consider one family alone--those
problems involving what is known as the Greek cross with the square.
This will exhibit a great variety of curious transpositions, and, by
having the solutions as we go along, the reader will be saved the
trouble of perpetually turning to another part of the book, and will
have everything under his eye. It is hoped that in this way the article
may prove somewhat instructive to the novice and interesting to others.


GREEK CROSS PUZZLES.

    "To fret thy soul with crosses."

    SPENSER.

    "But, for my part, it was Greek to me."

    _Julius Cæsar_, i. 2.

Many people are accustomed to consider the cross as a wholly Christian
symbol. This is erroneous: it is of very great antiquity. The ancient
Egyptians employed it as a sacred symbol, and on Greek sculptures we
find representations of a cake (the supposed real origin of our hot
cross buns) bearing a cross. Two such cakes were discovered at
Herculaneum. Cecrops offered to Jupiter Olympus a sacred cake or _boun_
of this kind. The cross and ball, so frequently found on Egyptian
figures, is a circle and the _tau_ cross. The circle signified the
eternal preserver of the world, and the T, named from the Greek letter
_tau_, is the monogram of Thoth, the Egyptian Mercury, meaning wisdom.
This _tau_ cross is also called by Christians the cross of St. Anthony,
and is borne on a badge in the bishop's palace at Exeter. As for the
Greek or mundane cross, the cross with four equal arms, we are told by
competent antiquaries that it was regarded by ancient occultists for
thousands of years as a sign of the dual forces of Nature--the male and
female spirit of everything that was everlasting.

[Illustration: Fig. 5.]

The Greek cross, as shown in Fig. 5, is formed by the assembling
together of five equal squares. We will start with what is known as the
Hindu problem, supposed to be upwards of three thousand years old. It
appears in the seal of Harvard College, and is often given in old works
as symbolical of mathematical science and exactitude. Cut the cross into
five pieces to form a square. Figs. 6 and 7 show how this is done. It
was not until the middle of the nineteenth century that we found that
the cross might be transformed into a square in only four pieces. Figs.
8 and 9 will show how to do it, if we further require the four pieces to
be all of the same size and shape. This Fig. 9 is remarkable because,
according to Dr. Le Plongeon and others, as expounded in a work by
Professor Wilson of the Smithsonian Institute, here we have the great
Swastika, or sign, of "good luck to you "--the most ancient symbol of
the human race of which there is any record. Professor Wilson's work
gives some four hundred illustrations of this curious sign as found in
the Aztec mounds of Mexico, the pyramids of Egypt, the ruins of Troy,
and the ancient lore of India and China. One might almost say there is a
curious affinity between the Greek cross and Swastika! If, however, we
require that the four pieces shall be produced by only two clips of the
scissors (assuming the puzzle is in paper form), then we must cut as in
Fig. 10 to form Fig. 11, the first clip of the scissors being from a
to b. Of course folding the paper, or holding the pieces together
after the first cut, would not in this case be allowed. But there is an
infinite number of different ways of making the cuts to solve the puzzle
in four pieces. To this point I propose to return.

[Illustration: Fig. 6]

[Illustration: Fig. 7]

[Illustration: Fig. 8]

[Illustration: Fig. 9]

[Illustration: Fig. 10]

[Illustration: Fig. 11]

It will be seen that every one of these puzzles has its reverse
puzzle--to cut a square into pieces to form a Greek cross. But as a
square has not so many angles as the cross, it is not always equally
easy to discover the true directions of the cuts. Yet in the case of the
examples given, I will leave the reader to determine their direction for
himself, as they are rather obvious from the diagrams.

Cut a square into five pieces that will form two separate Greek crosses
of _different sizes_. This is quite an easy puzzle. As will be seen in
Fig. 12, we have only to divide our square into 25 little squares and
then cut as shown. The cross A is cut out entire, and the pieces B, C,
D, and E form the larger cross in Fig. 13. The reader may here like to
cut the single piece, B, into four pieces all similar in shape to
itself, and form a cross with them in the manner shown in Fig. 13. I
hardly need give the solution.

[Illustration: FIG. 12.]

[Illustration: FIG. 13.]

Cut a square into five pieces that will form two separate Greek crosses
of exactly the _same size_. This is more difficult. We make the cuts as
in Fig. 14, where the cross A comes out entire and the other four pieces
form the cross in Fig. 15. The direction of the cuts is pretty obvious.
It will be seen that the sides of the square in Fig. 14 are marked off
into six equal parts. The sides of the cross are found by ruling lines
from certain of these points to others.

[Illustration: FIG. 14.]

[Illustration: FIG. 15.]

I will now explain, as I promised, why a Greek cross may be cut into
four pieces in an infinite number of different ways to make a square.
Draw a cross, as in Fig. 16. Then draw on transparent paper the square
shown in Fig. 17, taking care that the distance c to d is exactly
the same as the distance a to b in the cross. Now place the
transparent paper over the cross and slide it about into different
positions, only be very careful always to keep the square at the same
angle to the cross as shown, where a b is parallel to c d. If
you place the point c exactly over a the lines will indicate the
solution (Figs. 10 and 11). If you place c in the very centre of the
dotted square, it will give the solution in Figs. 8 and 9. You will now
see that by sliding the square about so that the point c is always
within the dotted square you may get as many different solutions as you
like; because, since an infinite number of different points may
theoretically be placed within this square, there must be an infinite
number of different solutions. But the point c need not necessarily be
placed within the dotted square. It may be placed, for example, at point
e to give a solution in four pieces. Here the joins at a and f may
be as slender as you like. Yet if you once get over the edge at a or
f you no longer have a solution in four pieces. This proof will be
found both entertaining and instructive. If you do not happen to have
any transparent paper at hand, any thin paper will of course do if you
hold the two sheets against a pane of glass in the window.

[Illustration: FIG. 16.]

[Illustration: FIG. 17.]

It may have been noticed from the solutions of the puzzles that I have
given that the side of the square formed from the cross is always equal
to the distance a to b in Fig. 16. This must necessarily be so, and
I will presently try to make the point quite clear.

We will now go one step further. I have already said that the ideal
solution to a cutting-out puzzle is always that which requires the
fewest possible pieces. We have just seen that two crosses of the same
size may be cut out of a square in five pieces. The reader who
succeeded in solving this perhaps asked himself: "Can it be done in
fewer pieces?" This is just the sort of question that the true puzzle
lover is always asking, and it is the right attitude for him to adopt.
The answer to the question is that the puzzle may be solved in four
pieces--the fewest possible. This, then, is a new puzzle. Cut a square
into four pieces that will form two Greek crosses of the same size.

[Illustration: FIG. 18.]

[Illustration: FIG. 19.]

[Illustration: FIG. 20.]

The solution is very beautiful. If you divide by points the sides of the
square into three equal parts, the directions of the lines in Fig. 18
will be quite obvious. If you cut along these lines, the pieces A and B
will form the cross in Fig. 19 and the pieces C and D the similar cross
in Fig. 20. In this square we have another form of Swastika.

The reader will here appreciate the truth of my remark to the effect
that it is easier to find the directions of the cuts when transforming a
cross to a square than when converting a square into a cross. Thus, in
Figs. 6, 8, and 10 the directions of the cuts are more obvious than in
Fig. 14, where we had first to divide the sides of the square into six
equal parts, and in Fig. 18, where we divide them into three equal
parts. Then, supposing you were required to cut two equal Greek crosses,
each into two pieces, to form a square, a glance at Figs. 19 and 20 will
show how absurdly more easy this is than the reverse puzzle of cutting
the square to make two crosses.

Referring to my remarks on "fallacies," I will now give a little example
of these "solutions" that are not solutions. Some years ago a young
correspondent sent me what he evidently thought was a brilliant new
discovery--the transforming of a square into a Greek cross in four
pieces by cuts all parallel to the sides of the square. I give his
attempt in Figs. 21 and 22, where it will be seen that the four pieces
do not form a symmetrical Greek cross, because the four arms are not
really squares but oblongs. To make it a true Greek cross we should
require the additions that I have indicated with dotted lines. Of course
his solution produces a cross, but it is not the symmetrical Greek
variety required by the conditions of the puzzle. My young friend
thought his attempt was "near enough" to be correct; but if he bought a
penny apple with a sixpence he probably would not have thought it "near
enough" if he had been given only fourpence change. As the reader
advances he will realize the importance of this question of exactitude.

[Illustration: FIG. 21.]

[Illustration: FIG. 22.]

In these cutting-out puzzles it is necessary not only to get the
directions of the cutting lines as correct as possible, but to remember
that these lines have no width. If after cutting up one of the crosses
in a manner indicated in these articles you find that the pieces do not
exactly fit to form a square, you may be certain that the fault is
entirely your own. Either your cross was not exactly drawn, or your cuts
were not made quite in the right directions, or (if you used wood and a
fret-saw) your saw was not sufficiently fine. If you cut out the puzzles
in paper with scissors, or in cardboard with a penknife, no material is
lost; but with a saw, however fine, there is a certain loss. In the case
of most puzzles this slight loss is not sufficient to be appreciable,
if the puzzle is cut out on a large scale, but there have been
instances where I have found it desirable to draw and cut out each part
separately--not from one diagram--in order to produce a perfect result.

[Illustration: FIG. 23.]

[Illustration: FIG. 24.]

Now for another puzzle. If you have cut out the five pieces indicated in
Fig. 14, you will find that these can be put together so as to form the
curious cross shown in Fig. 23. So if I asked you to cut Fig. 24 into
five pieces to form either a square or two equal Greek crosses you would
know how to do it. You would make the cuts as in Fig. 23, and place them
together as in Figs. 14 and 15. But I want something better than that,
and it is this. Cut Fig. 24 into only four pieces that will fit together
and form a square.

[Illustration: FIG. 25.]

[Illustration: FIG. 26.]

The solution to the puzzle is shown in Figs. 25 and 26. The direction of
the cut dividing A and C in the first diagram is very obvious, and the
second cut is made at right angles to it. That the four pieces should
fit together and form a square will surprise the novice, who will do
well to study the puzzle with some care, as it is most instructive.

I will now explain the beautiful rule by which we determine the size of
a square that shall have the same area as a Greek cross, for it is
applicable, and necessary, to the solution of almost every dissection
puzzle that we meet with. It was first discovered by the philosopher
Pythagoras, who died 500 B.C., and is the 47th proposition of Euclid.
The young reader who knows nothing of the elements of geometry will get
some idea of the fascinating character of that science. The triangle ABC
in Fig. 27 is what we call a right-angled triangle, because the side BC
is at right angles to the side AB. Now if we build up a square on each
side of the triangle, the squares on AB and BC will together be exactly
equal to the square on the long side AC, which we call the hypotenuse.
This is proved in the case I have given by subdividing the three squares
into cells of equal dimensions.

[Illustration: FIG. 27.]

[Illustration: FIG. 28.]

It will be seen that 9 added to 16 equals 25, the number of cells in the
large square. If you make triangles with the sides 5, 12 and 13, or with
8, 15 and 17, you will get similar arithmetical proofs, for these are
all "rational" right-angled triangles, but the law is equally true for
all cases. Supposing we cut off the lower arm of a Greek cross and place
it to the left of the upper arm, as in Fig. 28, then the square on EF
added to the square on DE exactly equals a square on DF. Therefore we
know that the square of DF will contain the same area as the cross. This
fact we have proved practically by the solutions of the earlier puzzles
of this series. But whatever length we give to DE and EF, we can never
give the exact length of DF in numbers, because the triangle is not a
"rational" one. But the law is none the less geometrically true.

[Illustration: FIG. 29.]

[Illustration: FIG. 30.]

Now look at Fig. 29, and you will see an elegant method for cutting a
piece of wood of the shape of two squares (of any relative dimensions)
into three pieces that will fit together and form a single square. If
you mark off the distance _ab_ equal to the side _cd_ the directions of
the cuts are very evident. From what we have just been considering, you
will at once see why _bc_ must be the length of the side of the new
square. Make the experiment as often as you like, taking different
relative proportions for the two squares, and you will find the rule
always come true. If you make the two squares of exactly the same size,
you will see that the diagonal of any square is always the side of a
square that is twice the size. All this, which is so simple that anybody
can understand it, is very essential to the solving of cutting-out
puzzles. It is in fact the key to most of them. And it is all so
beautiful that it seems a pity that it should not be familiar to
everybody.

We will now go one step further and deal with the half-square. Take a
square and cut it in half diagonally. Now try to discover how to cut
this triangle into four pieces that will form a Greek cross. The
solution is shown in Figs. 31 and 32. In this case it will be seen that
we divide two of the sides of the triangle into three equal parts and
the long side into four equal parts. Then the direction of the cuts will
be easily found. It is a pretty puzzle, and a little more difficult than
some of the others that I have given. It should be noted again that it
would have been much easier to locate the cuts in the reverse puzzle of
cutting the cross to form a half-square triangle.

[Illustration: FIG. 31.]

[Illustration: FIG. 32.]

[Illustration: FIG. 33.]

[Illustration: FIG. 34.]

Another ideal that the puzzle maker always keeps in mind is to contrive
that there shall, if possible, be only one correct solution. Thus, in
the case of the first puzzle, if we only require that a Greek cross
shall be cut into four pieces to form a square, there is, as I have
shown, an infinite number of different solutions. It makes a better
puzzle to add the condition that all the four pieces shall be of the
same size and shape, because it can then be solved in only one way, as
in Figs. 8 and 9. In this way, too, a puzzle that is too easy to be
interesting may be improved by such an addition. Let us take an example.
We have seen in Fig. 28 that Fig. 33 can be cut into two pieces to form
a Greek cross. I suppose an intelligent child would do it in five
minutes. But suppose we say that the puzzle has to be solved with a
piece of wood that has a bad knot in the position shown in Fig. 33--a
knot that we must not attempt to cut through--then a solution in two
pieces is barred out, and it becomes a more interesting puzzle to solve
it in three pieces. I have shown in Figs. 33 and 34 one way of doing
this, and it will be found entertaining to discover other ways of doing
it. Of course I could bar out all these other ways by introducing more
knots, and so reduce the puzzle to a single solution, but it would then
be overloaded with conditions.

And this brings us to another point in seeking the ideal. Do not
overload your conditions, or you will make your puzzle too complex to be
interesting. The simpler the conditions of a puzzle are, the better. The
solution may be as complex and difficult as you like, or as happens, but
the conditions ought to be easily understood, or people will not attempt
a solution.

If the reader were now asked "to cut a half-square into as few pieces as
possible to form a Greek cross," he would probably produce our solution,
Figs. 31-32, and confidently claim that he had solved the puzzle
correctly. In this way he would be wrong, because it is not now stated
that the square is to be divided diagonally. Although we should always
observe the exact conditions of a puzzle we must not read into it
conditions that are not there. Many puzzles are based entirely on the
tendency that people have to do this.

The very first essential in solving a puzzle is to be sure that you
understand the exact conditions. Now, if you divided your square in half
so as to produce Fig. 35 it is possible to cut it into as few as three
pieces to form a Greek cross. We thus save a piece.

I give another puzzle in Fig. 36. The dotted lines are added merely to
show the correct proportions of the figure--a square of 25 cells with
the four corner cells cut out. The puzzle is to cut this figure into
five pieces that will form a Greek cross (entire) and a square.

[Illustration: FIG. 35.]

[Illustration: FIG. 36.]

The solution to the first of the two puzzles last given--to cut a
rectangle of the shape of a half-square into three pieces that will form
a Greek cross--is shown in Figs. 37 and 38. It will be seen that we
divide the long sides of the oblong into six equal parts and the short
sides into three equal parts, in order to get the points that will
indicate the direction of the cuts. The reader should compare this
solution with some of the previous illustrations. He will see, for
example, that if we continue the cut that divides B and C in the cross,
we get Fig. 15.

[Illustration: FIG. 37.]

[Illustration: FIG. 38.]

The other puzzle, like the one illustrated in Figs. 12 and 13, will show
how useful a little arithmetic may sometimes prove to be in the solution
of dissection puzzles. There are twenty-one of those little square cells
into which our figure is subdivided, from which we have to form both a
square and a Greek cross. Now, as the cross is built up of five squares,
and 5 from 21 leaves 16--a square number--we ought easily to be led to
the solution shown in Fig. 39. It will be seen that the cross is cut out
entire, while the four remaining pieces form the square in Fig. 40.

[Illustration: FIG. 39]

[Illustration: FIG. 40]

Of course a half-square rectangle is the same as a double square, or two
equal squares joined together. Therefore, if you want to solve the
puzzle of cutting a Greek cross into four pieces to form two separate
squares of the same size, all you have to do is to continue the short
cut in Fig. 38 right across the cross, and you will have four pieces of
the same size and shape. Now divide Fig. 37 into two equal squares by a
horizontal cut midway and you will see the four pieces forming the two
squares.

[Illustration: FIG. 41]

Cut a Greek cross into five pieces that will form two separate squares,
one of which shall contain half the area of one of the arms of the
cross. In further illustration of what I have already written, if the
two squares of the same size A B C D and B C F E, in Fig. 41, are cut in
the manner indicated by the dotted lines, the four pieces will form the
large square A G E C. We thus see that the diagonal A C is the side of a
square twice the size of A B C D. It is also clear that half the
diagonal of any square is equal to the side of a square of half the
area. Therefore, if the large square in the diagram is one of the arms
of your cross, the small square is the size of one of the squares
required in the puzzle.

The solution is shown in Figs. 42 and 43. It will be seen that the small
square is cut out whole and the large square composed of the four pieces
B, C, D, and E. After what I have written, the reader will have no
difficulty in seeing that the square A is half the size of one of the
arms of the cross, because the length of the diagonal of the former is
clearly the same as the side of the latter. The thing is now
self-evident. I have thus tried to show that some of these puzzles that
many people are apt to regard as quite wonderful and bewildering, are
really not difficult if only we use a little thought and judgment. In
conclusion of this particular subject I will give four Greek cross
puzzles, with detached solutions.


142.--THE SILK PATCHWORK.

The lady members of the Wilkinson family had made a simple patchwork
quilt, as a small Christmas present, all composed of square pieces of
the same size, as shown in the illustration. It only lacked the four
corner pieces to make it complete. Somebody pointed out to them that if
you unpicked the Greek cross in the middle and then cut the stitches
along the dark joins, the four pieces all of the same size and shape
would fit together and form a square. This the reader knows, from the
solution in Fig. 39, is quite easily done. But George Wilkinson suddenly
suggested to them this poser. He said, "Instead of picking out the cross
entire, and forming the square from four equal pieces, can you cut out a
square entire and four equal pieces that will form a perfect Greek
cross?" The puzzle is, of course, now quite easy.


143.--TWO CROSSES FROM ONE.

Cut a Greek cross into five pieces that will form two such crosses, both
of the same size. The solution of this puzzle is very beautiful.


144.--THE CROSS AND THE TRIANGLE.

Cut a Greek cross into six pieces that will form an equilateral
triangle. This is another hard problem, and I will state here that a
solution is practically impossible without a previous knowledge of my
method of transforming an equilateral triangle into a square (see No.
26, "Canterbury Puzzles").


145.--THE FOLDED CROSS.

Cut out of paper a Greek cross; then so fold it that with a single
straight cut of the scissors the four pieces produced will form a
square.



VARIOUS DISSECTION PUZZLES.


We will now consider a small miscellaneous selection of cutting-out
puzzles, varying in degrees of difficulty.


146.--AN EASY DISSECTION PUZZLE.

First, cut out a piece of paper or cardboard of the shape shown in the
illustration. It will be seen at once that the proportions are simply
those of a square attached to half of another similar square, divided
diagonally. The puzzle is to cut it into four pieces all of precisely
the same size and shape.


147.--AN EASY SQUARE PUZZLE.

If you take a rectangular piece of cardboard, twice as long as it is
broad, and cut it in half diagonally, you will get two of the pieces
shown in the illustration. The puzzle is with five such pieces of equal
size to form a square. One of the pieces may be cut in two, but the
others must be used intact.


148.--THE BUN PUZZLE.

THE three circles represent three buns, and it is simply required to
show how these may be equally divided among four boys. The buns must be
regarded as of equal thickness throughout and of equal thickness to each
other. Of course, they must be cut into as few pieces as possible. To
simplify it I will state the rather surprising fact that only five
pieces are necessary, from which it will be seen that one boy gets his
share in two pieces and the other three receive theirs in a single
piece. I am aware that this statement "gives away" the puzzle, but it
should not destroy its interest to those who like to discover the
"reason why."


149.--THE CHOCOLATE SQUARES.

Here is a slab of chocolate, indented at the dotted lines so that the
twenty squares can be easily separated. Make a copy of the slab in paper
or cardboard and then try to cut it into nine pieces so that they will
form four perfect squares all of exactly the same size.


150.--DISSECTING A MITRE.

The figure that is perplexing the carpenter in the illustration
represents a mitre. It will be seen that its proportions are those of a
square with one quarter removed. The puzzle is to cut it into five
pieces that will fit together and form a perfect square. I show an
attempt, published in America, to perform the feat in four pieces, based
on what is known as the "step principle," but it is a fallacy.

[Illustration]

We are told first to cut oft the pieces 1 and 2 and pack them into the
triangular space marked off by the dotted line, and so form a rectangle.

So far, so good. Now, we are directed to apply the old step principle,
as shown, and, by moving down the piece 4 one step, form the required
square. But, unfortunately, it does _not_ produce a square: only an
oblong. Call the three long sides of the mitre 84 in. each. Then, before
cutting the steps, our rectangle in three pieces will be 84 × 63. The
steps must be 10½ in. in height and 12 in. in breadth. Therefore, by
moving down a step we reduce by 12 in. the side 84 in. and increase by
10½ in. the side 63 in. Hence our final rectangle must be 72 in. × 73½
in., which certainly is not a square! The fact is, the step principle
can only be applied to rectangles with sides of particular relative
lengths. For example, if the shorter side in this case were 61+5/7
(instead of 63), then the step method would apply. For the steps would
then be 10+2/7 in. in height and 12 in. in breadth. Note that 61+5/7 ×
84 = the square of 72. At present no solution has been found in four
pieces, and I do not believe one possible.


151.--THE JOINER'S PROBLEM.

I have often had occasion to remark on the practical utility of puzzles,
arising out of an application to the ordinary affairs of life of the
little tricks and "wrinkles" that we learn while solving recreation
problems.

[Illustration]

The joiner, in the illustration, wants to cut the piece of wood into as
few pieces as possible to form a square table-top, without any waste of
material. How should he go to work? How many pieces would you require?


152.--ANOTHER JOINER'S PROBLEM.

[Illustration]

A joiner had two pieces of wood of the shapes and relative proportions
shown in the diagram. He wished to cut them into as few pieces as
possible so that they could be fitted together, without waste, to form a
perfectly square table-top. How should he have done it? There is no
necessity to give measurements, for if the smaller piece (which is half
a square) be made a little too large or a little too small it will not
affect the method of solution.


153--A CUTTING-OUT PUZZLE.

Here is a little cutting-out poser. I take a strip of paper, measuring
five inches by one inch, and, by cutting it into five pieces, the parts
fit together and form a square, as shown in the illustration. Now, it is
quite an interesting puzzle to discover how we can do this in only four
pieces.

[Illustration]


154.--MRS. HOBSON'S HEARTHRUG.

[Illustration]

Mrs. Hobson's boy had an accident when playing with the fire, and burnt
two of the corners of a pretty hearthrug. The damaged corners have been
cut away, and it now has the appearance and proportions shown in my
diagram. How is Mrs. Hobson to cut the rug into the fewest possible
pieces that will fit together and form a perfectly square rug? It will
be seen that the rug is in the proportions 36 × 27 (it does not matter
whether we say inches or yards), and each piece cut away measured 12 and
6 on the outside.


155.--THE PENTAGON AND SQUARE.

I wonder how many of my readers, amongst those who have not given any
close attention to the elements of geometry, could draw a regular
pentagon, or five-sided figure, if they suddenly required to do so. A
regular hexagon, or six-sided figure, is easy enough, for everybody
knows that all you have to do is to describe a circle and then, taking
the radius as the length of one of the sides, mark off the six points
round the circumference. But a pentagon is quite another matter. So, as
my puzzle has to do with the cutting up of a regular pentagon, it will
perhaps be well if I first show my less experienced readers how this
figure is to be correctly drawn. Describe a circle and draw the two
lines H B and D G, in the diagram, through the centre at right angles.
Now find the point A, midway between C and B. Next place the point of
your compasses at A and with the distance A D describe the arc cutting H
B at E. Then place the point of your compasses at D and with the
distance D E describe the arc cutting the circumference at F. Now, D F
is one of the sides of your pentagon, and you have simply to mark off
the other sides round the circle. Quite simple when you know how, but
otherwise somewhat of a poser.

[Illustration]

Having formed your pentagon, the puzzle is to cut it into the fewest
possible pieces that will fit together and form a perfect square.

[Illustration]


156.--THE DISSECTED TRIANGLE.

A good puzzle is that which the gentleman in the illustration is showing
to his friends. He has simply cut out of paper an equilateral
triangle--that is, a triangle with all its three sides of the same
length. He proposes that it shall be cut into five pieces in such a way
that they will fit together and form either two or three smaller
equilateral triangles, using all the material in each case. Can you
discover how the cuts should be made?

Remember that when you have made your five pieces, you must be able, as
desired, to put them together to form either the single original
triangle or to form two triangles or to form three triangles--all
equilateral.


157.--THE TABLE-TOP AND STOOLS.

I have frequently had occasion to show that the published answers to a
great many of the oldest and most widely known puzzles are either quite
incorrect or capable of improvement. I propose to consider the old poser
of the table-top and stools that most of my readers have probably seen
in some form or another in books compiled for the recreation of
childhood.

The story is told that an economical and ingenious schoolmaster once
wished to convert a circular table-top, for which he had no use, into
seats for two oval stools, each with a hand-hole in the centre. He
instructed the carpenter to make the cuts as in the illustration and
then join the eight pieces together in the manner shown. So impressed
was he with the ingenuity of his performance that he set the puzzle to
his geometry class as a little study in dissection. But the remainder of
the story has never been published, because, so it is said, it was a
characteristic of the principals of academies that they would never
admit that they could err. I get my information from a descendant of the
original boy who had most reason to be interested in the matter.

The clever youth suggested modestly to the master that the hand-holes
were too big, and that a small boy might perhaps fall through them. He
therefore proposed another way of making the cuts that would get over
this objection. For his impertinence he received such severe
chastisement that he became convinced that the larger the hand-hole in
the stools the more comfortable might they be.

[Illustration]

Now what was the method the boy proposed?

Can you show how the circular table-top may be cut into eight pieces
that will fit together and form two oval seats for stools (each of
exactly the same size and shape) and each having similar hand-holes of
smaller dimensions than in the case shown above? Of course, all the wood
must be used.


158.--THE GREAT MONAD.

[Illustration]

Here is a symbol of tremendous antiquity which is worthy of notice. It
is borne on the Korean ensign and merchant flag, and has been adopted as
a trade sign by the Northern Pacific Railroad Company, though probably
few are aware that it is the Great Monad, as shown in the sketch below.
This sign is to the Chinaman what the cross is to the Christian. It is
the sign of Deity and eternity, while the two parts into which the
circle is divided are called the Yin and the Yan--the male and female
forces of nature. A writer on the subject more than three thousand years
ago is reported to have said in reference to it: "The illimitable
produces the great extreme. The great extreme produces the two
principles. The two principles produce the four quarters, and from the
four quarters we develop the quadrature of the eight diagrams of
Feuh-hi." I hope readers will not ask me to explain this, for I have not
the slightest idea what it means. Yet I am persuaded that for ages the
symbol has had occult and probably mathematical meanings for the
esoteric student.

I will introduce the Monad in its elementary form. Here are three easy
questions respecting this great symbol:--

(I.) Which has the greater area, the inner circle containing the Yin and
the Yan, or the outer ring?

(II.) Divide the Yin and the Yan into four pieces of the same size and
shape by one cut.

(III.) Divide the Yin and the Yan into four pieces of the same size, but
different shape, by one straight cut.


159.--THE SQUARE OF VENEER.

The following represents a piece of wood in my possession, 5 in. square.
By markings on the surface it is divided into twenty-five square inches.
I want to discover a way of cutting this piece of wood into the fewest
possible pieces that will fit together and form two perfect squares of
different sizes and of known dimensions. But, unfortunately, at every
one of the sixteen intersections of the cross lines a small nail has
been driven in at some time or other, and my fret-saw will be injured if
it comes in contact with any of these. I have therefore to find a method
of doing the work that will not necessitate my cutting through any of
those sixteen points. How is it to be done? Remember, the exact
dimensions of the two squares must be given.

[Illustration]


160.--THE TWO HORSESHOES.

[Illustration]

Why horseshoes should be considered "lucky" is one of those things
which no man can understand. It is a very old superstition, and John
Aubrey (1626-1700) says, "Most houses at the West End of London have a
horseshoe on the threshold." In Monmouth Street there were seventeen in
1813 and seven so late as 1855. Even Lord Nelson had one nailed to the
mast of the ship _Victory_. To-day we find it more conducive to "good
luck" to see that they are securely nailed on the feet of the horse we
are about to drive.

Nevertheless, so far as the horseshoe, like the Swastika and other
emblems that I have had occasion at times to deal with, has served to
symbolize health, prosperity, and goodwill towards men, we may well
treat it with a certain amount of respectful interest. May there not,
moreover, be some esoteric or lost mathematical mystery concealed in the
form of a horseshoe? I have been looking into this matter, and I wish to
draw my readers' attention to the very remarkable fact that the pair of
horseshoes shown in my illustration are related in a striking and
beautiful manner to the circle, which is the symbol of eternity. I
present this fact in the form of a simple problem, so that it may be
seen how subtly this relation has been concealed for ages and ages. My
readers will, I know, be pleased when they find the key to the mystery.

Cut out the two horseshoes carefully round the outline and then cut them
into four pieces, all different in shape, that will fit together and
form a perfect circle. Each shoe must be cut into two pieces and all the
part of the horse's hoof contained within the outline is to be used and
regarded as part of the area.


161.--THE BETSY ROSS PUZZLE.

A correspondent asked me to supply him with the solution to an old
puzzle that is attributed to a certain Betsy Ross, of Philadelphia, who
showed it to George Washington. It consists in so folding a piece of
paper that with one clip of the scissors a five-pointed star of Freedom
may be produced. Whether the story of the puzzle's origin is a true one
or not I cannot say, but I have a print of the old house in Philadelphia
where the lady is said to have lived, and I believe it still stands
there. But my readers will doubtless be interested in the little poser.

Take a circular piece of paper and so fold it that with one cut of the
scissors you can produce a perfect five-pointed star.


162.--THE CARDBOARD CHAIN.

[Illustration]

Can you cut this chain out of a piece of cardboard without any join
whatever? Every link is solid; without its having been split and
afterwards joined at any place. It is an interesting old puzzle that I
learnt as a child, but I have no knowledge as to its inventor.


163.--THE PAPER BOX.

It may be interesting to introduce here, though it is not strictly a
puzzle, an ingenious method for making a paper box.

Take a square of stout paper and by successive foldings make all the
creases indicated by the dotted lines in the illustration. Then cut away
the eight little triangular pieces that are shaded, and cut through the
paper along the dark lines. The second illustration shows the box half
folded up, and the reader will have no difficulty in effecting its
completion. Before folding up, the reader might cut out the circular
piece indicated in the diagram, for a purpose I will now explain.

This box will be found to serve excellently for the production of vortex
rings. These rings, which were discussed by Von Helmholtz in 1858, are
most interesting, and the box (with the hole cut out) will produce them
to perfection. Fill the box with tobacco smoke by blowing it gently
through the hole. Now, if you hold it horizontally, and softly tap the
side that is opposite to the hole, an immense number of perfect rings
can be produced from one mouthful of smoke. It is best that there should
be no currents of air in the room. People often do not realise that
these rings are formed in the air when no smoke is used. The smoke only
makes them visible. Now, one of these rings, if properly directed on its
course, will travel across the room and put out the flame of a candle,
and this feat is much more striking if you can manage to do it without
the smoke. Of course, with a little practice, the rings may be blown
from the mouth, but the box produces them in much greater perfection,
and no skill whatever is required. Lord Kelvin propounded the theory
that matter may consist of vortex rings in a fluid that fills all space,
and by a development of the hypothesis he was able to explain chemical
combination.

[Illustration:

    ·-----------·-----------·-----------·-----------·
    | ·       ·   ·|||||||·   ·|||||||·   ·       · |
    |   ·           ·|||·       ·|||·       ·   ·   |
    |     ·           ·           ·           ·     |
    |   ·   ·       ·   ·       ·   ·       ·   ·   |
    | ·       ·   ·       ·   ·       ·   ·       · |
    ·           ·           ·           ·           ·
    ||·       ·   ·       ·   ·     /|\   ·       ·||
    ||||·   ·       ·   ·       ·   \|/     ·   ·||||
    ||||||·           ·           ·           ·||||||
    ||||·   ·       ·   ·       ·   ·       ·   ·||||
    ||·       ·   ·       ·   ·       ·   ·       ·||
    ·           ·           ·           ·           ·
    ||·       ·   ·       ·   ·       ·   ·       ·||
    ||||·   ·       ·   ·       ·   ·       ·   ·||||
    ||||||·           ·           ·           ·||||||
    ||||·   ·       ·   ·       ·   ·       ·   ·||||
    ||·       ·   ·       ·   ·       ·   ·       ·||
    ·           ·           ·           ·           ·
    | ·       ·   ·       ·   ·       ·   ·       · |
    |   ·   ·       ·   ·       ·   ·       ·   ·   |
    |     ·           ·           ·           ·     |
    |   ·           ·|||·       ·|||·       ·   ·   |
    | ·       ·   ·|||||||·   ·|||||||·   ·       · |
    ·-----------·-----------·-----------·-----------·

]

[Illustration]


164.--THE POTATO PUZZLE.

Take a circular slice of potato, place it on the table, and see into how
large a number of pieces you can divide it with six cuts of a knife. Of
course you must not readjust the pieces or pile them after a cut. What
is the greatest number of pieces you can make?

[Illustration:

          --------
        /   \ 1/   \
       / \ 2 \/ 3 / \
      /   \  /\  /   \
     / \ 4 \/ 5\/ 6 / \
    |   \  /\  /\  /   |
     \  7\/ 8\/ 9\/10 /
      \  /\  /\  /\  /
       \/11\/12\/13\/
        \  /\  /\  /
         \/14\/15\/
          \  /\  /
           \/16\/
           -----

]

The illustration shows how to make sixteen pieces. This can, of course,
be easily beaten.


165.--THE SEVEN PIGS.

[Illustration]

    +------------------------------+
    |                              |
    |                        P     |
    |                              |
    |          P                   |
    |       P                      |
    |           P                  |
    |              P               |
    |           P                  |
    |     P                        |
    |                              |
    +------------------------------+

Here is a little puzzle that was put to one of the sons of Erin the
other day and perplexed him unduly, for it is really quite easy. It will
be seen from the illustration that he was shown a sketch of a square pen
containing seven pigs. He was asked how he would intersect the pen with
three straight fences so as to enclose every pig in a separate sty. In
other words, all you have to do is to take your pencil and, with three
straight strokes across the square, enclose each pig separately. Nothing
could be simpler.

[Illustration]

The Irishman complained that the pigs would not keep still while he was
putting up the fences. He said that they would all flock together, or
one obstinate beast would go into a corner and flock all by himself. It
was pointed out to him that for the purposes of the puzzle the pigs were
stationary. He answered that Irish pigs are not stationery--they are
pork. Being persuaded to make the attempt, he drew three lines, one of
which cut through a pig. When it was explained that this is not allowed,
he protested that a pig was no use until you cut its throat. "Begorra,
if it's bacon ye want without cutting your pig, it will be all gammon."
We will not do the Irishman the injustice of suggesting that the
miserable pun was intentional. However, he failed to solve the puzzle.
Can you do it?


166.--THE LANDOWNER'S FENCES.

The landowner in the illustration is consulting with his bailiff over a
rather puzzling little question. He has a large plan of one of his
fields, in which there are eleven trees. Now, he wants to divide the
field into just eleven enclosures by means of straight fences, so that
every enclosure shall contain one tree as a shelter for his cattle. How
is he to do it with as few fences as possible? Take your pencil and draw
straight lines across the field until you have marked off the eleven
enclosures (and no more), and then see how many fences you require. Of
course the fences may cross one another.


167.--THE WIZARD'S CATS.

[Illustration]

A wizard placed ten cats inside a magic circle as shown in our
illustration, and hypnotized them so that they should remain stationary
during his pleasure. He then proposed to draw three circles inside the
large one, so that no cat could approach another cat without crossing a
magic circle. Try to draw the three circles so that every cat has its
own enclosure and cannot reach another cat without crossing a line.


168.--THE CHRISTMAS PUDDING.

[Illustration]

"Speaking of Christmas puddings," said the host, as he glanced at the
imposing delicacy at the other end of the table. "I am reminded of the
fact that a friend gave me a new puzzle the other day respecting one.
Here it is," he added, diving into his breast pocket.

"'Problem: To find the contents,' I suppose," said the Eton boy.

"No; the proof of that is in the eating. I will read you the
conditions."

"'Cut the pudding into two parts, each of exactly the same size and
shape, without touching any of the plums. The pudding is to be regarded
as a flat disc, not as a sphere.'"

"Why should you regard a Christmas pudding as a disc? And why should any
reasonable person ever wish to make such an accurate division?" asked
the cynic.

"It is just a puzzle--a problem in dissection." All in turn had a look
at the puzzle, but nobody succeeded in solving it. It is a little
difficult unless you are acquainted with the principle involved in the
making of such puddings, but easy enough when you know how it is done.


169.--A TANGRAM PARADOX.

Many pastimes of great antiquity, such as chess, have so developed and
changed down the centuries that their original inventors would scarcely
recognize them. This is not the case with Tangrams, a recreation that
appears to be at least four thousand years old, that has apparently
never been dormant, and that has not been altered or "improved upon"
since the legendary Chinaman Tan first cut out the seven pieces shown in
Diagram I. If you mark the point B, midway between A and C, on one side
of a square of any size, and D, midway between C and E, on an adjoining
side, the direction of the cuts is too obvious to need further
explanation. Every design in this article is built up from the seven
pieces of blackened cardboard. It will at once be understood that the
possible combinations are infinite.

[Illustration]

The late Mr. Sam Loyd, of New York, who published a small book of very
ingenious designs, possessed the manuscripts of the late Mr. Challenor,
who made a long and close study of Tangrams. This gentleman, it is said,
records that there were originally seven books of Tangrams, compiled in
China two thousand years before the Christian era. These books are so
rare that, after forty years' residence in the country, he only
succeeded in seeing perfect copies of the first and seventh volumes with
fragments of the second. Portions of one of the books, printed in gold
leaf upon parchment, were found in Peking by an English soldier and sold
for three hundred pounds.

A few years ago a little book came into my possession, from the library
of the late Lewis Carroll, entitled _The Fashionable Chinese Puzzle_. It
contains three hundred and twenty-three Tangram designs, mostly
nondescript geometrical figures, to be constructed from the seven
pieces. It was "Published by J. and E. Wallis, 42 Skinner Street, and J.
Wallis, Jun., Marine Library, Sidmouth" (South Devon). There is no date,
but the following note fixes the time of publication pretty closely:
"This ingenious contrivance has for some time past been the favourite
amusement of the ex-Emperor Napoleon, who, being now in a debilitated
state and living very retired, passes many hours a day in thus
exercising his patience and ingenuity." The reader will find, as did the
great exile, that much amusement, not wholly uninstructive, may be
derived from forming the designs of others. He will find many of the
illustrations to this article quite easy to build up, and some rather
difficult. Every picture may thus be regarded as a puzzle.

But it is another pastime altogether to create new and original designs
of a pictorial character, and it is surprising what extraordinary scope
the Tangrams afford for producing pictures of real life--angular and
often grotesque, it is true, but full of character. I give an example of
a recumbent figure (2) that is particularly graceful, and only needs
some slight reduction of its angularities to produce an entirely
satisfactory outline.

As I have referred to the author of _Alice in Wonderland_, I give also
my designs of the March Hare (3) and the Hatter (4). I also give an
attempt at Napoleon (5), and a very excellent Red Indian with his Squaw
by Mr. Loyd (6 and 7). A large number of other designs will be found in
an article by me in _The Strand Magazine_ for November, 1908.

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

On the appearance of this magazine article, the late Sir James Murray,
the eminent philologist, tried, with that amazing industry that
characterized all his work, to trace the word "tangram" to its source.
At length he wrote as follows:--"One of my sons is a professor in the
Anglo-Chinese college at Tientsin. Through him, his colleagues, and his
students, I was able to make inquiries as to the alleged Tan among
Chinese scholars. Our Chinese professor here (Oxford) also took an
interest in the matter and obtained information from the secretary of
the Chinese Legation in London, who is a very eminent representative of
the Chinese literati."

[Illustration: 5]

"The result has been to show that the man Tan, the god Tan, and the
'Book of Tan' are entirely unknown to Chinese literature, history, or
tradition. By most of the learned men the name, or allegation of the
existence, of these had never been heard of. The puzzle is, of course,
well known. It is called in Chinese _ch'i ch'iao t'u_; literally,
'seven-ingenious-plan' or 'ingenious-puzzle figure of seven pieces.' No
name approaching 'tangram,' or even 'tan,' occurs in Chinese, and the
only suggestions for the latter were the Chinese _t'an_, 'to extend'; or
_t'ang_, Cantonese dialect for 'Chinese.' It was suggested that probably
some American or Englishman who knew a little Chinese or Cantonese,
wanting a name for the puzzle, might concoct one out of one of these
words and the European ending 'gram.' I should say the name 'tangram'
was probably invented by an American some little time before 1864 and
after 1847, but I cannot find it in print before the 1864 edition of
Webster. I have therefore had to deal very shortly with the word in the
dictionary, telling what it is applied to and what conjectures or
guesses have been made at the name, and giving a few quotations, one
from your own article, which has enabled me to make more of the subject
than I could otherwise have done."

[Illustration: 6]

[Illustration: 7]

Several correspondents have informed me that they possess, or had
possessed, specimens of the old Chinese books. An American gentleman
writes to me as follows:--"I have in my possession a book made of tissue
paper, printed in black (with a Chinese inscription on the front page),
containing over three hundred designs, which belongs to the box of
'tangrams,' which I also own. The blocks are seven in number, made of
mother-of-pearl, highly polished and finely engraved on either side.
These are contained in a rosewood box 2+1/8 in. square. My great uncle,
----, was one of the first missionaries to visit China. This box and
book, along with quite a collection of other relics, were sent to my
grandfather and descended to myself."

My correspondent kindly supplied me with rubbings of the Tangrams, from
which it is clear that they are cut in the exact proportions that I have
indicated. I reproduce the Chinese inscription (8) for this reason. The
owner of the book informs me that he has submitted it to a number of
Chinamen in the United States and offered as much as a dollar for a
translation. But they all steadfastly refused to read the words,
offering the lame excuse that the inscription is Japanese. Natives of
Japan, however, insist that it is Chinese. Is there something occult and
esoteric about Tangrams, that it is so difficult to lift the veil?
Perhaps this page will come under the eye of some reader acquainted with
the Chinese language, who will supply the required translation, which
may, or may not, throw a little light on this curious question.

[Illustration: 8]

By using several sets of Tangrams at the same time we may construct more
ambitious pictures. I was advised by a friend not to send my picture, "A
Game of Billiards" (9), to the Academy. He assured me that it would not
be accepted because the "judges are so hide-bound by convention."
Perhaps he was right, and it will be more appreciated by
Post-impressionists and Cubists. The players are considering a very
delicate stroke at the top of the table. Of course, the two men, the
table, and the clock are formed from four sets of Tangrams. My second
picture is named "The Orchestra" (10), and it was designed for the
decoration of a large hall of music. Here we have the conductor, the
pianist, the fat little cornet-player, the left-handed player of the
double-bass, whose attitude is life-like, though he does stand at an
unusual distance from his instrument, and the drummer-boy, with his
imposing music-stand. The dog at the back of the pianoforte is not
howling: he is an appreciative listener.

[Illustration: 9]

[Illustration: 10]

One remarkable thing about these Tangram pictures is that they suggest
to the imagination such a lot that is not really there. Who, for
example, can look for a few minutes at Lady Belinda (11) and the Dutch
girl (12) without soon feeling the haughty expression in the one case
and the arch look in the other? Then look again at the stork (13), and
see how it is suggested to the mind that the leg is actually much more
slender than any one of the pieces employed. It is really an optical
illusion. Again, notice in the case of the yacht (14) how, by leaving
that little angular point at the top, a complete mast is suggested. If
you place your Tangrams together on white paper so that they do not
quite touch one another, in some cases the effect is improved by the
white lines; in other cases it is almost destroyed.

[Illustration: 11]

[Illustration: 12]

Finally, I give an example from the many curious paradoxes that one
happens upon in manipulating Tangrams. I show designs of two dignified
individuals (15 and 16) who appear to be exactly alike, except for the
fact that one has a foot and the other has not. Now, both of these
figures are made from the same seven Tangrams. Where does the second man
get his foot from?

[Illustration: 13]

[Illustration: 14]

[Illustration: 15]

[Illustration: 16]



PATCHWORK PUZZLES.

"Of shreds and patches."--_Hamlet_, iii. 4.


170.--THE CUSHION COVERS.

[Illustration]

The above represents a square of brocade. A lady wishes to cut it in
four pieces so that two pieces will form one perfectly square cushion
top, and the remaining two pieces another square cushion top. How is she
to do it? Of course, she can only cut along the lines that divide the
twenty-five squares, and the pattern must "match" properly without any
irregularity whatever in the design of the material. There is only one
way of doing it. Can you find it?


171.--THE BANNER PUZZLE.

[Illustration]

A Lady had a square piece of bunting with two lions on it, of which the
illustration is an exactly reproduced reduction. She wished to cut the
stuff into pieces that would fit together and form two square banners
with a lion on each banner. She discovered that this could be done in as
few as four pieces. How did she manage it? Of course, to cut the British
Lion would be an unpardonable offence, so you must be careful that no
cut passes through any portion of either of them. Ladies are informed
that no allowance whatever has to be made for "turnings," and no part of
the material may be wasted. It is quite a simple little dissection
puzzle if rightly attacked. Remember that the banners have to be perfect
squares, though they need not be both of the same size.


172.--MRS. SMILEY'S CHRISTMAS PRESENT.

Mrs. Smiley's expression of pleasure was sincere when her six
granddaughters sent to her, as a Christmas present, a very pretty
patchwork quilt, which they had made with their own hands. It was
constructed of square pieces of silk material, all of one size, and as
they made a large quilt with fourteen of these little squares on each
side, it is obvious that just 196 pieces had been stitched into it. Now,
the six granddaughters each contributed a part of the work in the form
of a perfect square (all six portions being different in size), but in
order to join them up to form the square quilt it was necessary that the
work of one girl should be unpicked into three separate pieces. Can you
show how the joins might have been made? Of course, no portion can be
turned over.

[Illustration]


173.--MRS. PERKINS'S QUILT.

[Illustration]

It will be seen that in this case the square patchwork quilt is built up
of 169 pieces. The puzzle is to find the smallest possible number of
square portions of which the quilt could be composed and show how they
might be joined together. Or, to put it the reverse way, divide the
quilt into as few square portions as possible by merely cutting the
stitches.


174.--THE SQUARES OF BROCADE.

[Illustration]

I happened to be paying a call at the house of a lady, when I took up
from a table two lovely squares of brocade. They were beautiful
specimens of Eastern workmanship--both of the same design, a delicate
chequered pattern.

"Are they not exquisite?" said my friend. "They were brought to me by a
cousin who has just returned from India. Now, I want you to give me a
little assistance. You see, I have decided to join them together so as
to make one large square cushion-cover. How should I do this so as to
mutilate the material as little as possible? Of course I propose to make
my cuts only along the lines that divide the little chequers."

[Illustration]

I cut the two squares in the manner desired into four pieces that would
fit together and form another larger square, taking care that the
pattern should match properly, and when I had finished I noticed that
two of the pieces were of exactly the same area; that is, each of the
two contained the same number of chequers. Can you show how the cuts
were made in accordance with these conditions?


175--ANOTHER PATCHWORK PUZZLE.

[Illustration]

A lady was presented, by two of her girl friends, with the pretty pieces
of silk patchwork shown in our illustration. It will be seen that both
pieces are made up of squares all of the same size--one 12 × 12 and the
other 5 × 5. She proposes to join them together and make one square
patchwork quilt, 13 × 13, but, of course, she will not cut any of the
material--merely cut the stitches where necessary and join together
again. What perplexes her is this. A friend assures her that there need
be no more than four pieces in all to join up for the new quilt. Could
you show her how this little needlework puzzle is to be solved in so few
pieces?


176.--LINOLEUM CUTTING.

[Illustration]

The diagram herewith represents two separate pieces of linoleum. The
chequered pattern is not repeated at the back, so that the pieces cannot
be turned over. The puzzle is to cut the two squares into four pieces so
that they shall fit together and form one perfect square 10 × 10, so
that the pattern shall properly match, and so that the larger piece
shall have as small a portion as possible cut from it.


177.--ANOTHER LINOLEUM PUZZLE.

[Illustration]

Can you cut this piece of linoleum into four pieces that will fit
together and form a perfect square? Of course the cuts may only be made
along the lines.



VARIOUS GEOMETRICAL PUZZLES.

    "So various are the tastes of men."
        MARK AKENSIDE.


178.--THE CARDBOARD BOX.

This puzzle is not difficult, but it will be found entertaining to
discover the simple rule for its solution. I have a rectangular
cardboard box. The top has an area of 120 square inches, the side 96
square inches, and the end 80 square inches. What are the exact
dimensions of the box?


179.--STEALING THE BELL-ROPES.

Two men broke into a church tower one night to steal the bell-ropes. The
two ropes passed through holes in the wooden ceiling high above them,
and they lost no time in climbing to the top. Then one man drew his
knife and cut the rope above his head, in consequence of which he fell
to the floor and was badly injured. His fellow-thief called out that it
served him right for being such a fool. He said that he should have done
as he was doing, upon which he cut the rope below the place at which he
held on. Then, to his dismay, he found that he was in no better plight,
for, after hanging on as long as his strength lasted, he was compelled
to let go and fall beside his comrade. Here they were both found the
next morning with their limbs broken. How far did they fall? One of the
ropes when they found it was just touching the floor, and when you
pulled the end to the wall, keeping the rope taut, it touched a point
just three inches above the floor, and the wall was four feet from the
rope when it hung at rest. How long was the rope from floor to ceiling?


180.--THE FOUR SONS.

Readers will recognize the diagram as a familiar friend of their youth.
A man possessed a square-shaped estate. He bequeathed to his widow the
quarter of it that is shaded off. The remainder was to be divided
equitably amongst his four sons, so that each should receive land of
exactly the same area and exactly similar in shape. We are shown how
this was done. But the remainder of the story is not so generally known.
In the centre of the estate was a well, indicated by the dark spot, and
Benjamin, Charles, and David complained that the division was not
"equitable," since Alfred had access to this well, while they could not
reach it without trespassing on somebody else's land. The puzzle is to
show how the estate is to be apportioned so that each son shall have
land of the same shape and area, and each have access to the well
without going off his own land.

[Illustration]


181.--THE THREE RAILWAY STATIONS.

As I sat in a railway carriage I noticed at the other end of the
compartment a worthy squire, whom I knew by sight, engaged in
conversation with another passenger, who was evidently a friend of his.

"How far have you to drive to your place from the railway station?"
asked the stranger.

"Well," replied the squire, "if I get out at Appleford, it is just the
same distance as if I go to Bridgefield, another fifteen miles farther
on; and if I changed at Appleford and went thirteen miles from there to
Carterton, it would still be the same distance. You see, I am
equidistant from the three stations, so I get a good choice of trains."

Now I happened to know that Bridgefield is just fourteen miles from
Carterton, so I amused myself in working out the exact distance that the
squire had to drive home whichever station he got out at. What was the
distance?


182.--THE GARDEN PUZZLE.

Professor Rackbrain tells me that he was recently smoking a friendly
pipe under a tree in the garden of a country acquaintance. The garden
was enclosed by four straight walls, and his friend informed him that he
had measured these and found the lengths to be 80, 45, 100, and 63 yards
respectively. "Then," said the professor, "we can calculate the exact
area of the garden." "Impossible," his host replied, "because you can
get an infinite number of different shapes with those four sides." "But
you forget," Rackbrane said, with a twinkle in his eye, "that you told
me once you had planted this tree equidistant from all the four corners
of the garden." Can you work out the garden's area?


183.--DRAWING A SPIRAL.

If you hold the page horizontally and give it a quick rotary motion
while looking at the centre of the spiral, it will appear to revolve.
Perhaps a good many readers are acquainted with this little optical
illusion. But the puzzle is to show how I was able to draw this spiral
with so much exactitude without using anything but a pair of compasses
and the sheet of paper on which the diagram was made. How would you
proceed in such circumstances?

[Illustration]


184.--HOW TO DRAW AN OVAL.

Can you draw a perfect oval on a sheet of paper with one sweep of the
compasses? It is one of the easiest things in the world when you know
how.


185.--ST. GEORGE'S BANNER.

At a celebration of the national festival of St. George's Day I was
contemplating the familiar banner of the patron saint of our country. We
all know the red cross on a white ground, shown in our illustration.
This is the banner of St. George. The banner of St. Andrew (Scotland) is
a white "St. Andrew's Cross" on a blue ground. That of St. Patrick
(Ireland) is a similar cross in red on a white ground. These three are
united in one to form our Union Jack.

Now on looking at St. George's banner it occurred to me that the
following question would make a simple but pretty little puzzle.
Supposing the flag measures four feet by three feet, how wide must the
arm of the cross be if it is required that there shall be used just the
same quantity of red and of white bunting?

[Illustration]


186.--THE CLOTHES LINE PUZZLE.

A boy tied a clothes line from the top of each of two poles to the base
of the other. He then proposed to his father the following question. As
one pole was exactly seven feet above the ground and the other exactly
five feet, what was the height from the ground where the two cords
crossed one another?


187.--THE MILKMAID PUZZLE.

[Illustration]

Here is a little pastoral puzzle that the reader may, at first sight, be
led into supposing is very profound, involving deep calculations. He may
even say that it is quite impossible to give any answer unless we are
told something definite as to the distances. And yet it is really quite
"childlike and bland."

In the corner of a field is seen a milkmaid milking a cow, and on the
other side of the field is the dairy where the extract has to be
deposited. But it has been noticed that the young woman always goes down
to the river with her pail before returning to the dairy. Here the
suspicious reader will perhaps ask why she pays these visits to the
river. I can only reply that it is no business of ours. The alleged milk
is entirely for local consumption.

    "Where are you going to, my pretty maid?"
    "Down to the river, sir," she said.
    "I'll _not_ choose your dairy, my pretty maid."
    "Nobody axed you, sir," she said.

If one had any curiosity in the matter, such an independent spirit would
entirely disarm one. So we will pass from the point of commercial
morality to the subject of the puzzle.

Draw a line from the milking-stool down to the river and thence to the
door of the dairy, which shall indicate the shortest possible route for
the milkmaid. That is all. It is quite easy to indicate the exact spot
on the bank of the river to which she should direct her steps if she
wants as short a walk as possible. Can you find that spot?


188.--THE BALL PROBLEM.

[Illustration]

A stonemason was engaged the other day in cutting out a round ball for
the purpose of some architectural decoration, when a smart schoolboy
came upon the scene.

"Look here," said the mason, "you seem to be a sharp youngster, can you
tell me this? If I placed this ball on the level ground, how many other
balls of the same size could I lay around it (also on the ground) so
that every ball should touch this one?"

The boy at once gave the correct answer, and then put this little
question to the mason:--

"If the surface of that ball contained just as many square feet as its
volume contained cubic feet, what would be the length of its diameter?"

The stonemason could not give an answer. Could you have replied
correctly to the mason's and the boy's questions?


189.--THE YORKSHIRE ESTATES.

[Illustration]

I was on a visit to one of the large towns of Yorkshire. While walking
to the railway station on the day of my departure a man thrust a
hand-bill upon me, and I took this into the railway carriage and read it
at my leisure. It informed me that three Yorkshire neighbouring estates
were to be offered for sale. Each estate was square in shape, and they
joined one another at their corners, just as shown in the diagram.
Estate A contains exactly 370 acres, B contains 116 acres, and C 74
acres.

Now, the little triangular bit of land enclosed by the three square
estates was not offered for sale, and, for no reason in particular, I
became curious as to the area of that piece. How many acres did it
contain?


190.--FARMER WURZEL'S ESTATE.

[Illustration]

I will now present another land problem. The demonstration of the answer
that I shall give will, I think, be found both interesting and easy of
comprehension.

Farmer Wurzel owned the three square fields shown in the annexed plan,
containing respectively 18, 20, and 26 acres. In order to get a
ring-fence round his property he bought the four intervening triangular
fields. The puzzle is to discover what was then the whole area of his
estate.


191.--THE CRESCENT PUZZLE.

[Illustration]

Here is an easy geometrical puzzle. The crescent is formed by two
circles, and C is the centre of the larger circle. The width of the
crescent between B and D is 9 inches, and between E and F 5 inches. What
are the diameters of the two circles?


192.--THE PUZZLE WALL.

[Illustration]

There was a small lake, around which four poor men built their cottages.
Four rich men afterwards built their mansions, as shown in the
illustration, and they wished to have the lake to themselves, so they
instructed a builder to put up the shortest possible wall that would
exclude the cottagers, but give themselves free access to the lake. How
was the wall to be built?


193.--THE SHEEPFOLD.

It is a curious fact that the answers always given to some of the
best-known puzzles that appear in every little book of fireside
recreations that has been published for the last fifty or a hundred
years are either quite unsatisfactory or clearly wrong. Yet nobody ever
seems to detect their faults. Here is an example:--A farmer had a pen
made of fifty hurdles, capable of holding a hundred sheep only.
Supposing he wanted to make it sufficiently large to hold double that
number, how many additional hurdles must he have?


194.--THE GARDEN WALLS.

[Illustration]

A speculative country builder has a circular field, on which he has
erected four cottages, as shown in the illustration. The field is
surrounded by a brick wall, and the owner undertook to put up three
other brick walls, so that the neighbours should not be overlooked by
each other, but the four tenants insist that there shall be no
favouritism, and that each shall have exactly the same length of wall
space for his wall fruit trees. The puzzle is to show how the three
walls may be built so that each tenant shall have the same area of
ground, and precisely the same length of wall.

Of course, each garden must be entirely enclosed by its walls, and it
must be possible to prove that each garden has exactly the same length
of wall. If the puzzle is properly solved no figures are necessary.


195.--LADY BELINDA'S GARDEN.

Lady Belinda is an enthusiastic gardener. In the illustration she is
depicted in the act of worrying out a pleasant little problem which I
will relate. One of her gardens is oblong in shape, enclosed by a high
holly hedge, and she is turning it into a rosary for the cultivation of
some of her choicest roses. She wants to devote exactly half of the area
of the garden to the flowers, in one large bed, and the other half to be
a path going all round it of equal breadth throughout. Such a garden is
shown in the diagram at the foot of the picture. How is she to mark out
the garden under these simple conditions? She has only a tape, the
length of the garden, to do it with, and, as the holly hedge is so thick
and dense, she must make all her measurements inside. Lady Belinda did
not know the exact dimensions of the garden, and, as it was not
necessary for her to know, I also give no dimensions. It is quite a
simple task no matter what the size or proportions of the garden may be.
Yet how many lady gardeners would know just how to proceed? The tape may
be quite plain--that is, it need not be a graduated measure.

[Illustration]


196.--THE TETHERED GOAT.

[Illustration]

Here is a little problem that everybody should know how to solve. The
goat is placed in a half-acre meadow, that is in shape an equilateral
triangle. It is tethered to a post at one corner of the field. What
should be the length of the tether (to the nearest inch) in order that
the goat shall be able to eat just half the grass in the field? It is
assumed that the goat can feed to the end of the tether.


197.--THE COMPASSES PUZZLE.

It is curious how an added condition or restriction will sometimes
convert an absurdly easy puzzle into an interesting and perhaps
difficult one. I remember buying in the street many years ago a little
mechanical puzzle that had a tremendous sale at the time. It consisted
of a medal with holes in it, and the puzzle was to work a ring with a
gap in it from hole to hole until it was finally detached. As I was
walking along the street I very soon acquired the trick of taking off
the ring with one hand while holding the puzzle in my pocket. A friend
to whom I showed the little feat set about accomplishing it himself, and
when I met him some days afterwards he exhibited his proficiency in the
art. But he was a little taken aback when I then took the puzzle from
him and, while simply holding the medal between the finger and thumb of
one hand, by a series of little shakes and jerks caused the ring,
without my even touching it, to fall off upon the floor. The following
little poser will probably prove a rather tough nut for a great many
readers, simply on account of the restricted conditions:--

Show how to find exactly the middle of any straight line by means of the
compasses only. You are not allowed to use any ruler, pencil, or other
article--only the compasses; and no trick or dodge, such as folding the
paper, will be permitted. You must simply use the compasses in the
ordinary legitimate way.


198.--THE EIGHT STICKS.

I have eight sticks, four of them being exactly half the length of the
others. I lay every one of these on the table, so that they enclose
three squares, all of the same size. How do I do it? There must be no
loose ends hanging over.



199.--PAPA'S PUZZLE.

Here is a puzzle by Pappus, who lived at Alexandria about the end of the
third century. It is the fifth proposition in the eighth book of his
_Mathematical Collections_. I give it in the form that I presented it
some years ago under the title "Papa's Puzzle," just to see how many
readers would discover that it was by Pappus himself. "The little maid's
papa has taken two different-sized rectangular pieces of cardboard, and
has clipped off a triangular piece from one of them, so that when it is
suspended by a thread from the point A it hangs with the long side
perfectly horizontal, as shown in the illustration. He has perplexed the
child by asking her to find the point A on the other card, so as to
produce a similar result when cut and suspended by a thread." Of course,
the point must not be found by trial clippings. A curious and pretty
point is involved in this setting of the puzzle. Can the reader discover
it?

[Illustration]


200.--A KITE-FLYING PUZZLE.

While accompanying my friend Professor Highflite during a scientific
kite-flying competition on the South Downs of Sussex I was led into a
little calculation that ought to interest my readers. The Professor was
paying out the wire to which his kite was attached from a winch on which
it had been rolled into a perfectly spherical form. This ball of wire
was just two feet in diameter, and the wire had a diameter of
one-hundredth of an inch. What was the length of the wire?

Now, a simple little question like this that everybody can perfectly
understand will puzzle many people to answer in any way. Let us see
whether, without going into any profound mathematical calculations, we
can get the answer roughly--say, within a mile of what is correct! We
will assume that when the wire is all wound up the ball is perfectly
solid throughout, and that no allowance has to be made for the axle that
passes through it. With that simplification, I wonder how many readers
can state within even a mile of the correct answer the length of that
wire.


201.--HOW TO MAKE CISTERNS.

[Illustration]

Our friend in the illustration has a large sheet of zinc, measuring
(before cutting) eight feet by three feet, and he has cut out square
pieces (all of the same size) from the four corners and now proposes to
fold up the sides, solder the edges, and make a cistern. But the point
that puzzles him is this: Has he cut out those square pieces of the
correct size in order that the cistern may hold the greatest possible
quantity of water? You see, if you cut them very small you get a very
shallow cistern; if you cut them large you get a tall and slender one.
It is all a question of finding a way of cutting put these four square
pieces exactly the right size. How are we to avoid making them too small
or too large?


202.--THE CONE PUZZLE.

[Illustration]

I have a wooden cone, as shown in Fig. 1. How am I to cut out of it the
greatest possible cylinder? It will be seen that I can cut out one that
is long and slender, like Fig. 2, or short and thick, like Fig. 3. But
neither is the largest possible. A child could tell you where to cut, if
he knew the rule. Can you find this simple rule?


203.--CONCERNING WHEELS.

[Illustration]

There are some curious facts concerning the movements of wheels that are
apt to perplex the novice. For example: when a railway train is
travelling from London to Crewe certain parts of the train at any given
moment are actually moving from Crewe towards London. Can you indicate
those parts? It seems absurd that parts of the same train can at any
time travel in opposite directions, but such is the case.

In the accompanying illustration we have two wheels. The lower one is
supposed to be fixed and the upper one running round it in the direction
of the arrows. Now, how many times does the upper wheel turn on its own
axis in making a complete revolution of the other wheel? Do not be in a
hurry with your answer, or you are almost certain to be wrong.
Experiment with two pennies on the table and the correct answer will
surprise you, when you succeed in seeing it.


204.--A NEW MATCH PUZZLE.

[Illustration]

In the illustration eighteen matches are shown arranged so that they
enclose two spaces, one just twice as large as the other. Can you
rearrange them (1) so as to enclose two four-sided spaces, one exactly
three times as large as the other, and (2) so as to enclose two
five-sided spaces, one exactly three times as large as the other? All
the eighteen matches must be fairly used in each case; the two spaces
must be quite detached, and there must be no loose ends or duplicated
matches.


205.--THE SIX SHEEP-PENS.

[Illustration]

Here is a new little puzzle with matches. It will be seen in the
illustration that thirteen matches, representing a farmer's hurdles,
have been so placed that they enclose six sheep-pens all of the same
size. Now, one of these hurdles was stolen, and the farmer wanted still
to enclose six pens of equal size with the remaining twelve. How was he
to do it? All the twelve matches must be fairly used, and there must be
no duplicated matches or loose ends.



POINTS AND LINES PROBLEMS.


"Line upon line, line upon line; here a little and there a
little."--_Isa_. xxviii. 10.

What are known as "Points and Lines" puzzles are found very interesting
by many people. The most familiar example, here given, to plant nine
trees so that they shall form ten straight rows with three trees in
every row, is attributed to Sir Isaac Newton, but the earliest
collection of such puzzles is, I believe, in a rare little book that I
possess--published in 1821--_Rational Amusement for Winter Evenings_, by
John Jackson. The author gives ten examples of "Trees planted in Rows."

These tree-planting puzzles have always been a matter of great
perplexity. They are real "puzzles," in the truest sense of the word,
because nobody has yet succeeded in finding a direct and certain way of
solving them. They demand the exercise of sagacity, ingenuity, and
patience, and what we call "luck" is also sometimes of service. Perhaps
some day a genius will discover the key to the whole mystery. Remember
that the trees must be regarded as mere points, for if we were allowed
to make our trees big enough we might easily "fudge" our diagrams and
get in a few extra straight rows that were more apparent than real.

[Illustration]


206.--THE KING AND THE CASTLES.

There was once, in ancient times, a powerful king, who had eccentric
ideas on the subject of military architecture. He held that there was
great strength and economy in symmetrical forms, and always cited the
example of the bees, who construct their combs in perfect hexagonal
cells, to prove that he had nature to support him. He resolved to build
ten new castles in his country all to be connected by fortified walls,
which should form five lines with four castles in every line. The royal
architect presented his preliminary plan in the form I have shown. But
the monarch pointed out that every castle could be approached from the
outside, and commanded that the plan should be so modified that as many
castles as possible should be free from attack from the outside, and
could only be reached by crossing the fortified walls. The architect
replied that he thought it impossible so to arrange them that even one
castle, which the king proposed to use as a royal residence, could be so
protected, but his majesty soon enlightened him by pointing out how it
might be done. How would you have built the ten castles and
fortifications so as best to fulfil the king's requirements? Remember
that they must form five straight lines with four castles in every line.

[Illustration]


207.--CHERRIES AND PLUMS.

[Illustration]

The illustration is a plan of a cottage as it stands surrounded by an
orchard of fifty-five trees. Ten of these trees are cherries, ten are
plums, and the remainder apples. The cherries are so planted as to form
five straight lines, with four cherry trees in every line. The plum
trees are also planted so as to form five straight lines with four plum
trees in every line. The puzzle is to show which are the ten cherry
trees and which are the ten plums. In order that the cherries and plums
should have the most favourable aspect, as few as possible (under the
conditions) are planted on the north and east sides of the orchard. Of
course in picking out a group of ten trees (cherry or plum, as the case
may be) you ignore all intervening trees. That is to say, four trees may
be in a straight line irrespective of other trees (or the house) being
in between. After the last puzzle this will be quite easy.


208.--A PLANTATION PUZZLE.

[Illustration]

A man had a square plantation of forty-nine trees, but, as will be seen
by the omissions in the illustration, four trees were blown down and
removed. He now wants to cut down all the remainder except ten trees,
which are to be so left that they shall form five straight rows with
four trees in every row. Which are the ten trees that he must leave?


209.--THE TWENTY-ONE TREES.

A gentleman wished to plant twenty-one trees in his park so that they
should form twelve straight rows with five trees in every row. Could you
have supplied him with a pretty symmetrical arrangement that would
satisfy these conditions?


210.--THE TEN COINS.

Place ten pennies on a large sheet of paper or cardboard, as shown in
the diagram, five on each edge. Now remove four of the coins, without
disturbing the others, and replace them on the paper so that the ten
shall form five straight lines with four coins in every line. This in
itself is not difficult, but you should try to discover in how many
different ways the puzzle may be solved, assuming that in every case the
two rows at starting are exactly the same.

[Illustration]


211.--THE TWELVE MINCE-PIES.

It will be seen in our illustration how twelve mince-pies may be placed
on the table so as to form six straight rows with four pies in every
row. The puzzle is to remove only four of them to new positions so that
there shall be _seven_ straight rows with four in every row. Which four
would you remove, and where would you replace them?

[Illustration]


212.--THE BURMESE PLANTATION.

[Illustration]

A short time ago I received an interesting communication from the
British chaplain at Meiktila, Upper Burma, in which my correspondent
informed me that he had found some amusement on board ship on his way
out in trying to solve this little poser.

If he has a plantation of forty-nine trees, planted in the form of a
square as shown in the accompanying illustration, he wishes to know how
he may cut down twenty-seven of the trees so that the twenty-two left
standing shall form as many rows as possible with four trees in every
row.

Of course there may not be more than four trees in any row.


213.--TURKS AND RUSSIANS.

This puzzle is on the lines of the Afridi problem published by me in
_Tit-Bits_ some years ago.

On an open level tract of country a party of Russian infantry, no two of
whom were stationed at the same spot, were suddenly surprised by
thirty-two Turks, who opened fire on the Russians from all directions.
Each of the Turks simultaneously fired a bullet, and each bullet passed
immediately over the heads of three Russian soldiers. As each of these
bullets when fired killed a different man, the puzzle is to discover
what is the smallest possible number of soldiers of which the Russian
party could have consisted and what were the casualties on each side.



MOVING COUNTER PROBLEMS.


    "I cannot do't without counters."

    _Winter's Tale_, iv. 3.

Puzzles of this class, except so far as they occur in connection with
actual games, such as chess, seem to be a comparatively modern
introduction. Mathematicians in recent times, notably Vandermonde and
Reiss, have devoted some attention to them, but they do not appear to
have been considered by the old writers. So far as games with counters
are concerned, perhaps the most ancient and widely known in old times is
"Nine Men's Morris" (known also, as I shall show, under a great many
other names), unless the simpler game, distinctly mentioned in the works
of Ovid (No. 110, "Ovid's Game," in _The Canterbury Puzzles_), from
which "Noughts and Crosses" seems to be derived, is still more ancient.

In France the game is called Marelle, in Poland Siegen Wulf Myll
(She-goat Wolf Mill, or Fight), in Germany and Austria it is called
Muhle (the Mill), in Iceland it goes by the name of Mylla, while the
Bogas (or native bargees) of South America are said to play it, and on
the Amazon it is called Trique, and held to be of Indian origin. In our
own country it has different names in different districts, such as Meg
Merrylegs, Peg Meryll, Nine Peg o'Merryal, Nine-Pin Miracle, Merry Peg,
and Merry Hole. Shakespeare refers to it in "Midsummer Night's Dream"
(Act ii., scene 1):--

    "The nine-men's morris is filled up with mud;
     And the quaint mazes in the wanton green,
     For lack of tread, are undistinguishable."

It was played by the shepherds with stones in holes cut in the turf.
John Clare, the peasant poet of Northamptonshire, in "The Shepherd Boy"
(1835) says:--"Oft we track his haunts .... By nine-peg-morris nicked
upon the green." It is also mentioned by Drayton in his "Polyolbion."

It was found on an old Roman tile discovered during the excavations at
Silchester, and cut upon the steps of the Acropolis at Athens. When
visiting the Christiania Museum a few years ago I was shown the great
Viking ship that was discovered at Gokstad in 1880. On the oak planks
forming the deck of the vessel were found boles and lines marking out
the game, the holes being made to receive pegs. While inspecting the
ancient oak furniture in the Rijks Museum at Amsterdam I became
interested in an old catechumen's settle, and was surprised to find the
game diagram cut in the centre of the seat--quite conveniently for
surreptitious play. It has been discovered cut in the choir stalls of
several of our English cathedrals. In the early eighties it was found
scratched upon a stone built into a wall (probably about the date 1200),
during the restoration of Hargrave church in Northamptonshire. This
stone is now in the Northampton Museum. A similar stone has since been
found at Sempringham, Lincolnshire. It is to be seen on an ancient
tombstone in the Isle of Man, and painted on old Dutch tiles. And in
1901 a stone was dug out of a gravel pit near Oswestry bearing an
undoubted diagram of the game.

The game has been played with different rules at different periods and
places. I give a copy of the board. Sometimes the diagonal lines are
omitted, but this evidently was not intended to affect the play: it
simply meant that the angles alone were thought sufficient to indicate
the points. This is how Strutt, in _Sports and Pastimes_, describes the
game, and it agrees with the way I played it as a boy:--"Two persons,
having each of them nine pieces, or men, lay them down alternately, one
by one, upon the spots; and the business of either party is to prevent
his antagonist from placing three of his pieces so as to form a row of
three, without the intervention of an opponent piece. If a row be
formed, he that made it is at liberty to take up one of his competitor's
pieces from any part he thinks most to his advantage; excepting he has
made a row, which must not be touched if he have another piece upon the
board that is not a component part of that row. When all the pieces are
laid down, they are played backwards and forwards, in any direction that
the lines run, but only can move from one spot to another (next to it)
at one time. He that takes off all his antagonist's pieces is the
conqueror."

[Illustration]


214.--THE SIX FROGS.

[Illustration]

The six educated frogs in the illustration are trained to reverse their
order, so that their numbers shall read 6, 5, 4, 3, 2, 1, with the blank
square in its present position. They can jump to the next square (if
vacant) or leap over one frog to the next square beyond (if vacant),
just as we move in the game of draughts, and can go backwards or
forwards at pleasure. Can you show how they perform their feat in the
fewest possible moves? It is quite easy, so when you have done it add a
seventh frog to the right and try again. Then add more frogs until you
are able to give the shortest solution for any number. For it can always
be done, with that single vacant square, no matter how many frogs there
are.


215.--THE GRASSHOPPER PUZZLE.

It has been suggested that this puzzle was a great favourite among the
young apprentices of the City of London in the sixteenth and seventeenth
centuries. Readers will have noticed the curious brass grasshopper on
the Royal Exchange. This long-lived creature escaped the fires of 1666
and 1838. The grasshopper, after his kind, was the crest of Sir Thomas
Gresham, merchant grocer, who died in 1579, and from this cause it has
been used as a sign by grocers in general. Unfortunately for the legend
as to its origin, the puzzle was only produced by myself so late as the
year 1900. On twelve of the thirteen black discs are placed numbered
counters or grasshoppers. The puzzle is to reverse their order, so that
they shall read, 1, 2, 3, 4, etc., in the opposite direction, with the
vacant disc left in the same position as at present. Move one at a time
in any order, either to the adjoining vacant disc or by jumping over one
grasshopper, like the moves in draughts. The moves or leaps may be made
in either direction that is at any time possible. What are the fewest
possible moves in which it can be done?

[Illustration]


216.--THE EDUCATED FROGS.

[Illustration]

Our six educated frogs have learnt a new and pretty feat. When placed on
glass tumblers, as shown in the illustration, they change sides so that
the three black ones are to the left and the white frogs to the right,
with the unoccupied tumbler at the opposite end--No. 7. They can jump to
the next tumbler (if unoccupied), or over one, or two, frogs to an
unoccupied tumbler. The jumps can be made in either direction, and a
frog may jump over his own or the opposite colour, or both colours. Four
successive specimen jumps will make everything quite plain: 4 to 1, 5 to
4, 3 to 5, 6 to 3. Can you show how they do it in ten jumps?


217.--THE TWICKENHAM PUZZLE.

[Illustration:


          ( I ) ((N))

       ( M )        ((A))

    ( H )              ((T))

      ( E )          ((W))

        ( C )     ((K))
              ( )


]

In the illustration we have eleven discs in a circle. On five of the
discs we place white counters with black letters--as shown--and on five
other discs the black counters with white letters. The bottom disc is
left vacant. Starting thus, it is required to get the counters into
order so that they spell the word "Twickenham" in a clockwise direction,
leaving the vacant disc in the original position. The black counters
move in the direction that a clock-hand revolves, and the white counters
go the opposite way. A counter may jump over one of the opposite colour
if the vacant disc is next beyond. Thus, if your first move is with K,
then C can jump over K. If then K moves towards E, you may next jump W
over C, and so on. The puzzle may be solved in twenty-six moves.
Remember a counter cannot jump over one of its own colour.


218.--THE VICTORIA CROSS PUZZLE.

[Illustration:

    +---------------------+
    |     \... A .../     |
    | (I)  |.......|  (V) |
    |\_____|_______|_____/|
    |......|       |------|
    |.. R .|       |. I ..|
    |......|       |......|
    | _____|_______|_____ |
    |/     |.......|     \|
    | (O)  |.. T ..|  (C) |
    |     /.........\     |
    +---------------------+

]

The puzzle-maker is peculiarly a "snapper-up of unconsidered trifles,"
and his productions are often built up with the slenderest materials.
Trivialities that might entirely escape the observation of others, or,
if they were observed, would be regarded as of no possible moment, often
supply the man who is in quest of posers with a pretty theme or an idea
that he thinks possesses some "basal value."

When seated opposite to a lady in a railway carriage at the time of
Queen Victoria's Diamond Jubilee, my attention was attracted to a brooch
that she was wearing. It was in the form of a Maltese or Victoria Cross,
and bore the letters of the word VICTORIA. The number and arrangement of
the letters immediately gave me the suggestion for the puzzle which I
now present.

The diagram, it will be seen, is composed of nine divisions. The puzzle
is to place eight counters, bearing the letters of the word VICTORIA,
exactly in the manner shown, and then slide one letter at a time from
black to white and white to black alternately, until the word reads
round in the same direction, only with the initial letter V on one of
the black arms of the cross. At no time may two letters be in the same
division. It is required to find the shortest method.

Leaping moves are, of course, not permitted. The first move must
obviously be made with A, I, T, or R. Supposing you move T to the
centre, the next counter played will be O or C, since I or R cannot be
moved. There is something a little remarkable in the solution of this
puzzle which I will explain.


219.--THE LETTER BLOCK PUZZLE.

[Illustration:


    +-----+-----+-----+\
    |     |     |     | |
    |  G  |  E  |  F  | |
    |     |     |     | |
    +-----+-----+-----+\|
    |     |     |     | |
    |  H  |  C  |  B  | |
    |     |     |     | |
    +-----+-----+-----+\|
    |     |\____|     | |
    |  D  ||    |  A  | |
    |     ||    |     | |
    +-----+-----+-----+ |
     \_________________\|

]

Here is a little reminiscence of our old friend the Fifteen Block
Puzzle. Eight wooden blocks are lettered, and are placed in a box, as
shown in the illustration. It will be seen that you can only move one
block at a time to the place vacant for the time being, as no block may
be lifted out of the box. The puzzle is to shift them about until you
get them in the order--

    A   B   C
    D   E   F
    G   H

This you will find by no means difficult if you are allowed as many
moves as you like. But the puzzle is to do it in the fewest possible
moves. I will not say what this smallest number of moves is, because the
reader may like to discover it for himself. In writing down your moves
you will find it necessary to record no more than the letters in the
order that they are shifted. Thus, your first five moves might be C, H,
G, E, F; and this notation can have no possible ambiguity. In practice
you only need eight counters and a simple diagram on a sheet of paper.


220.--A LODGING-HOUSE DIFFICULTY.

[Illustration]

The Dobsons secured apartments at Slocomb-on-Sea. There were six rooms
on the same floor, all communicating, as shown in the diagram. The rooms
they took were numbers 4, 5, and 6, all facing the sea. But a little
difficulty arose. Mr. Dobson insisted that the piano and the bookcase
should change rooms. This was wily, for the Dobsons were not musical,
but they wanted to prevent any one else playing the instrument. Now, the
rooms were very small and the pieces of furniture indicated were very
big, so that no two of these articles could be got into any room at the
same time. How was the exchange to be made with the least possible
labour? Suppose, for example, you first move the wardrobe into No. 2;
then you can move the bookcase to No. 5 and the piano to No. 6, and so
on. It is a fascinating puzzle, but the landlady had reasons for not
appreciating it. Try to solve her difficulty in the fewest possible
removals with counters on a sheet of paper.


221.--THE EIGHT ENGINES.

The diagram represents the engine-yard of a railway company under
eccentric management. The engines are allowed to be stationary only at
the nine points indicated, one of which is at present vacant. It is
required to move the engines, one at a time, from point to point, in
seventeen moves, so that their numbers shall be in numerical order round
the circle, with the central point left vacant. But one of the engines
has had its fire drawn, and therefore cannot move. How is the thing to
be done? And which engine remains stationary throughout?

[Illustration]


222.--A RAILWAY PUZZLE.

[Illustration]

Make a diagram, on a large sheet of paper, like the illustration, and
have three counters marked A, three marked B, and three marked C. It
will be seen that at the intersection of lines there are nine
stopping-places, and a tenth stopping-place is attached to the outer
circle like the tail of a Q. Place the three counters or engines marked
A, the three marked B, and the three marked C at the places indicated.
The puzzle is to move the engines, one at a time, along the lines, from
stopping-place to stopping-place, until you succeed in getting an A, a
B, and a C on each circle, and also A, B, and C on each straight line.
You are required to do this in as few moves as possible. How many moves
do you need?


223.--A RAILWAY MUDDLE.

The plan represents a portion of the line of the London, Clodville, and
Mudford Railway Company. It is a single line with a loop. There is only
room for eight wagons, or seven wagons and an engine, between B and C on
either the left line or the right line of the loop. It happened that two
goods trains (each consisting of an engine and sixteen wagons) got into
the position shown in the illustration. It looked like a hopeless
deadlock, and each engine-driver wanted the other to go back to the next
station and take off nine wagons. But an ingenious stoker undertook to
pass the trains and send them on their respective journeys with their
engines properly in front. He also contrived to reverse the engines the
fewest times possible. Could you have performed the feat? And how many
times would you require to reverse the engines? A "reversal" means a
change of direction, backward or forward. No rope-shunting,
fly-shunting, or other trick is allowed. All the work must be done
legitimately by the two engines. It is a simple but interesting puzzle
if attempted with counters.

[Illustration]


224.--THE MOTOR-GARAGE PUZZLE.

[Illustration]

The difficulties of the proprietor of a motor garage are converted into
a little pastime of a kind that has a peculiar fascination. All you need
is to make a simple plan or diagram on a sheet of paper or cardboard and
number eight counters, 1 to 8. Then a whole family can enter into an
amusing competition to find the best possible solution of the
difficulty.

The illustration represents the plan of a motor garage, with
accommodation for twelve cars. But the premises are so inconveniently
restricted that the proprietor is often caused considerable perplexity.
Suppose, for example, that the eight cars numbered 1 to 8 are in the
positions shown, how are they to be shifted in the quickest possible way
so that 1, 2, 3, and 4 shall change places with 5, 6, 7, and 8--that is,
with the numbers still running from left to right, as at present, but
the top row exchanged with the bottom row? What are the fewest possible
moves?

One car moves at a time, and any distance counts as one move. To prevent
misunderstanding, the stopping-places are marked in squares, and only
one car can be in a square at the same time.


225.--THE TEN PRISONERS.

If prisons had no other use, they might still be preserved for the
special benefit of puzzle-makers. They appear to be an inexhaustible
mine of perplexing ideas. Here is a little poser that will perhaps
interest the reader for a short period. We have in the illustration a
prison of sixteen cells. The locations of the ten prisoners will be
seen. The jailer has queer superstitions about odd and even numbers, and
he wants to rearrange the ten prisoners so that there shall be as many
even rows of men, vertically, horizontally, and diagonally, as
possible. At present it will be seen, as indicated by the arrows, that
there are only twelve such rows of 2 and 4. I will state at once that
the greatest number of such rows that is possible is sixteen. But the
jailer only allows four men to be removed to other cells, and informs me
that, as the man who is seated in the bottom right-hand corner is
infirm, he must not be moved. Now, how are we to get those sixteen rows
of even numbers under such conditions?

[Illustration]


226.--ROUND THE COAST.

[Illustration]

Here is a puzzle that will, I think, be found as amusing as instructive.
We are given a ring of eight circles. Leaving circle 8 blank, we are
required to write in the name of a seven-lettered port in the United
Kingdom in this manner. Touch a blank circle with your pencil, then jump
over two circles in either direction round the ring, and write down the
first letter. Then touch another vacant circle, jump over two circles,
and write down your second letter. Proceed similarly with the other
letters in their proper order until you have completed the word. Thus,
suppose we select "Glasgow," and proceed as follows: 6--1, 7--2, 8--3,
7--4, 8--5, which means that we touch 6, jump over 7 and and write down
"G" on 1; then touch 7, jump over 8 and 1, and write down "l" on 2; and
so on. It will be found that after we have written down the first five
letters--"Glasg"--as above, we cannot go any further. Either there is
something wrong with "Glasgow," or we have not managed our jumps
properly. Can you get to the bottom of the mystery?


227.--CENTRAL SOLITAIRE.

[Illustration]

This ancient puzzle was a great favourite with our grandmothers, and
most of us, I imagine, have on occasions come across a "Solitaire"
board--a round polished board with holes cut in it in a geometrical
pattern, and a glass marble in every hole. Sometimes I have noticed one
on a side table in a suburban front parlour, or found one on a shelf in
a country cottage, or had one brought under my notice at a wayside inn.
Sometimes they are of the form shown above, but it is equally common for
the board to have four more holes, at the points indicated by dots. I
select the simpler form.

Though "Solitaire" boards are still sold at the toy shops, it will be
sufficient if the reader will make an enlarged copy of the above on a
sheet of cardboard or paper, number the "holes," and provide himself
with 33 counters, buttons, or beans. Now place a counter in every hole
except the central one, No. 17, and the puzzle is to take off all the
counters in a series of jumps, except the last counter, which must be
left in that central hole. You are allowed to jump one counter over the
next one to a vacant hole beyond, just as in the game of draughts, and
the counter jumped over is immediately taken off the board. Only
remember every move must be a jump; consequently you will take off a
counter at each move, and thirty-one single jumps will of course remove
all the thirty-one counters. But compound moves are allowed (as in
draughts, again), for so long as one counter continues to jump, the
jumps all count as one move.

Here is the beginning of an imaginary solution which will serve to make
the manner of moving perfectly plain, and show how the solver should
write out his attempts: 5-17, 12-10, 26-12, 24-26 (13-11, 11-25), 9-11
(26-24, 24-10, 10-12), etc., etc. The jumps contained within brackets
count as one move, because they are made with the same counter. Find the
fewest possible moves. Of course, no diagonal jumps are permitted; you
can only jump in the direction of the lines.


228.--THE TEN APPLES.

[Illustration]

The family represented in the illustration are amusing themselves with
this little puzzle, which is not very difficult but quite interesting.
They have, it will be seen, placed sixteen plates on the table in the
form of a square, and put an apple in each of ten plates. They want to
find a way of removing all the apples except one by jumping over one at
a time to the next vacant square, as in draughts; or, better, as in
solitaire, for you are not allowed to make any diagonal moves--only
moves parallel to the sides of the square. It is obvious that as the
apples stand no move can be made, but you are permitted to transfer any
single apple you like to a vacant plate before starting. Then the moves
must be all leaps, taking off the apples leaped over.


229.--THE NINE ALMONDS.

"Here is a little puzzle," said a Parson, "that I have found peculiarly
fascinating. It is so simple, and yet it keeps you interested
indefinitely."

The reverend gentleman took a sheet of paper and divided it off into
twenty-five squares, like a square portion of a chessboard. Then he
placed nine almonds on the central squares, as shown in the
illustration, where we have represented numbered counters for
convenience in giving the solution.

"Now, the puzzle is," continued the Parson, "to remove eight of the
almonds and leave the ninth in the central square. You make the removals
by jumping one almond over another to the vacant square beyond and
taking off the one jumped over--just as in draughts, only here you can
jump in any direction, and not diagonally only. The point is to do the
thing in the fewest possible moves."

The following specimen attempt will make everything clear. Jump 4 over
1, 5 over 9, 3 over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 over 4. But
8 is not left in the central square, as it should be. Remember to remove
those you jump over. Any number of jumps in succession with the same
almond count as one move.

[Illustration]


230.--THE TWELVE PENNIES.

Here is a pretty little puzzle that only requires twelve pennies or
counters. Arrange them in a circle, as shown in the illustration. Now
take up one penny at a time and, passing it over two pennies, place it
on the third penny. Then take up another single penny and do the same
thing, and so on, until, in six such moves, you have the coins in six
pairs in the positions 1, 2, 3, 4, 5, 6. You can move in either
direction round the circle at every play, and it does not matter
whether the two jumped over are separate or a pair. This is quite easy
if you use just a little thought.

[Illustration]


231.--PLATES AND COINS.

Place twelve plates, as shown, on a round table, with a penny or orange
in every plate. Start from any plate you like and, always going in one
direction round the table, take up one penny, pass it over two other
pennies, and place it in the next plate. Go on again; take up another
penny and, having passed it over two pennies, place it in a plate; and
so continue your journey. Six coins only are to be removed, and when
these have been placed there should be two coins in each of six plates
and six plates empty. An important point of the puzzle is to go round
the table as few times as possible. It does not matter whether the two
coins passed over are in one or two plates, nor how many empty plates
you pass a coin over. But you must always go in one direction round the
table and end at the point from which you set out. Your hand, that is to
say, goes steadily forward in one direction, without ever moving
backwards.

[Illustration]


232.--CATCHING THE MICE.

[Illustration]

"Play fair!" said the mice. "You know the rules of the game."

"Yes, I know the rules," said the cat. "I've got to go round and round
the circle, in the direction that you are looking, and eat every
thirteenth mouse, but I must keep the white mouse for a tit-bit at the
finish. Thirteen is an unlucky number, but I will do my best to oblige
you."

"Hurry up, then!" shouted the mice.

"Give a fellow time to think," said the cat. "I don't know which of you
to start at. I must figure it out."

While the cat was working out the puzzle he fell asleep, and, the spell
being thus broken, the mice returned home in safety. At which mouse
should the cat have started the count in order that the white mouse
should be the last eaten?

When the reader has solved that little puzzle, here is a second one for
him. What is the smallest number that the cat can count round and round
the circle, if he must start at the white mouse (calling that "one" in
the count) and still eat the white mouse last of all?

And as a third puzzle try to discover what is the smallest number that
the cat can count round and round if she must start at the white mouse
(calling that "one") and make the white mouse the third eaten.


233.--THE ECCENTRIC CHEESEMONGER.

[Illustration]

The cheesemonger depicted in the illustration is an inveterate puzzle
lover. One of his favourite puzzles is the piling of cheeses in his
warehouse, an amusement that he finds good exercise for the body as well
as for the mind. He places sixteen cheeses on the floor in a straight
row and then makes them into four piles, with four cheeses in every
pile, by always passing a cheese over four others. If you use sixteen
counters and number them in order from 1 to 16, then you may place 1 on
6, 11 on 1, 7 on 4, and so on, until there are four in every pile. It
will be seen that it does not matter whether the four passed over are
standing alone or piled; they count just the same, and you can always
carry a cheese in either direction. There are a great many different
ways of doing it in twelve moves, so it makes a good game of "patience"
to try to solve it so that the four piles shall be left in different
stipulated places. For example, try to leave the piles at the extreme
ends of the row, on Nos. 1, 2, 15 and 16; this is quite easy. Then try
to leave three piles together, on Nos. 13, 14, and 15. Then again play
so that they shall be left on Nos. 3, 5, 12, and 14.


234.--THE EXCHANGE PUZZLE.


Here is a rather entertaining little puzzle with moving counters. You
only need twelve counters--six of one colour, marked A, C, E, G, I, and
K, and the other six marked B, D, F, H, J, and L. You first place them
on the diagram, as shown in the illustration, and the puzzle is to get
them into regular alphabetical order, as follows:--

    A   B   C   D
    E   F   G   H
    I   J   K   L

The moves are made by exchanges of opposite colours standing on the same
line. Thus, G and J may exchange places, or F and A, but you cannot
exchange G and C, or F and D, because in one case they are both white
and in the other case both black. Can you bring about the required
arrangement in seventeen exchanges?

[Illustration]

It cannot be done in fewer moves. The puzzle is really much easier than
it looks, if properly attacked.


235.--TORPEDO PRACTICE.

[Illustration]

If a fleet of sixteen men-of-war were lying at anchor and surrounded by
the enemy, how many ships might be sunk if every torpedo, projected in a
straight line, passed under three vessels and sank the fourth? In the
diagram we have arranged the fleet in square formation, where it will be
seen that as many as seven ships may be sunk (those in the top row and
first column) by firing the torpedoes indicated by arrows. Anchoring the
fleet as we like, to what extent can we increase this number? Remember
that each successive ship is sunk before another torpedo is launched,
and that every torpedo proceeds in a different direction; otherwise, by
placing the ships in a straight line, we might sink as many as thirteen!
It is an interesting little study in naval warfare, and eminently
practical--provided the enemy will allow you to arrange his fleet for
your convenience and promise to lie still and do nothing!


236.--THE HAT PUZZLE.

Ten hats were hung on pegs as shown in the illustration--five silk hats
and five felt "bowlers," alternately silk and felt. The two pegs at the
end of the row were empty.

[Illustration]

The puzzle is to remove two contiguous hats to the vacant pegs, then two
other adjoining hats to the pegs now unoccupied, and so on until five
pairs have been moved and the hats again hang in an unbroken row, but
with all the silk ones together and all the felt hats together.

Remember, the two hats removed must always be contiguous ones, and you
must take one in each hand and place them on their new pegs without
reversing their relative position. You are not allowed to cross your
hands, nor to hang up one at a time.

Can you solve this old puzzle, which I give as introductory to the next?
Try it with counters of two colours or with coins, and remember that the
two empty pegs must be left at one end of the row.


237.--BOYS AND GIRLS.

If you mark off ten divisions on a sheet of paper to represent the
chairs, and use eight numbered counters for the children, you will have
a fascinating pastime. Let the odd numbers represent boys and even
numbers girls, or you can use counters of two colours, or coins.

The puzzle is to remove two children who are occupying adjoining chairs
and place them in two empty chairs, _making them first change sides_;
then remove a second pair of children from adjoining chairs and place
them in the two now vacant, making them change sides; and so on, until
all the boys are together and all the girls together, with the two
vacant chairs at one end as at present. To solve the puzzle you must do
this in five moves. The two children must always be taken from chairs
that are next to one another; and remember the important point of making
the two children change sides, as this latter is the distinctive feature
of the puzzle. By "change sides" I simply mean that if, for example, you
first move 1 and 2 to the vacant chairs, then the first (the outside)
chair will be occupied by 2 and the second one by 1.

[Illustration]


238.--ARRANGING THE JAMPOTS.

I happened to see a little girl sorting out some jam in a cupboard for
her mother. She was putting each different kind of preserve apart on the
shelves. I noticed that she took a pot of damson in one hand and a pot
of gooseberry in the other and made them change places; then she changed
a strawberry with a raspberry, and so on. It was interesting to observe
what a lot of unnecessary trouble she gave herself by making more
interchanges than there was any need for, and I thought it would work
into a good puzzle.

It will be seen in the illustration that little Dorothy has to
manipulate twenty-four large jampots in as many pigeon-holes. She wants
to get them in correct numerical order--that is, 1, 2, 3, 4, 5, 6 on the
top shelf, 7, 8, 9, 10, 11, 12 on the next shelf, and so on. Now, if she
always takes one pot in the right hand and another in the left and makes
them change places, how many of these interchanges will be necessary to
get all the jampots in proper order? She would naturally first change
the 1 and the 3, then the 2 and the 3, when she would have the first
three pots in their places. How would you advise her to go on then?
Place some numbered counters on a sheet of paper divided into squares
for the pigeon-holes, and you will find it an amusing puzzle.

[Illustration]



UNICURSAL AND ROUTE PROBLEMS.

    "I see them on their winding way."
                           REGINALD HEBER.

It is reasonable to suppose that from the earliest ages one man has
asked another such questions as these: "Which is the nearest way home?"
"Which is the easiest or pleasantest way?" "How can we find a way that
will enable us to dodge the mastodon and the plesiosaurus?" "How can we
get there without ever crossing the track of the enemy?" All these are
elementary route problems, and they can be turned into good puzzles by
the introduction of some conditions that complicate matters. A variety
of such complications will be found in the following examples. I have
also included some enumerations of more or less difficulty. These afford
excellent practice for the reasoning faculties, and enable one to
generalize in the case of symmetrical forms in a manner that is most
instructive.


239.--A JUVENILE PUZZLE.

For years I have been perpetually consulted by my juvenile friends about
this little puzzle. Most children seem to know it, and yet, curiously
enough, they are invariably unacquainted with the answer. The question
they always ask is, "Do, please, tell me whether it is really possible."
I believe Houdin the conjurer used to be very fond of giving it to his
child friends, but I cannot say whether he invented the little puzzle or
not. No doubt a large number of my readers will be glad to have the
mystery of the solution cleared up, so I make no apology for introducing
this old "teaser."

The puzzle is to draw with three strokes of the pencil the diagram that
the little girl is exhibiting in the illustration. Of course, you must
not remove your pencil from the paper during a stroke or go over the
same line a second time. You will find that you can get in a good deal
of the figure with one continuous stroke, but it will always appear as
if four strokes are necessary.

[Illustration]

Another form of the puzzle is to draw the diagram on a slate and then
rub it out in three rubs.


240.--THE UNION JACK.

[Illustration]

The illustration is a rough sketch somewhat resembling the British flag,
the Union Jack. It is not possible to draw the whole of it without
lifting the pencil from the paper or going over the same line twice. The
puzzle is to find out just _how much_ of the drawing it is possible to
make without lifting your pencil or going twice over the same line. Take
your pencil and see what is the best you can do.


241.--THE DISSECTED CIRCLE.

How many continuous strokes, without lifting your pencil from the paper,
do you require to draw the design shown in our illustration? Directly
you change the direction of your pencil it begins a new stroke. You may
go over the same line more than once if you like. It requires just a
little care, or you may find yourself beaten by one stroke.

[Illustration]


242.--THE TUBE INSPECTOR'S PUZZLE.

The man in our illustration is in a little dilemma. He has just been
appointed inspector of a certain system of tube railways, and it is his
duty to inspect regularly, within a stated period, all the company's
seventeen lines connecting twelve stations, as shown on the big poster
plan that he is contemplating. Now he wants to arrange his route so that
it shall take him over all the lines with as little travelling as
possible. He may begin where he likes and end where he likes. What is
his shortest route?

Could anything be simpler? But the reader will soon find that, however
he decides to proceed, the inspector must go over some of the lines more
than once. In other words, if we say that the stations are a mile apart,
he will have to travel more than seventeen miles to inspect every line.
There is the little difficulty. How far is he compelled to travel, and
which route do you recommend?

[Illustration]


243.--VISITING THE TOWNS.

[Illustration]

A traveller, starting from town No. 1, wishes to visit every one of the
towns once, and once only, going only by roads indicated by straight
lines. How many different routes are there from which he can select? Of
course, he must end his journey at No. 1, from which he started, and
must take no notice of cross roads, but go straight from town to town.
This is an absurdly easy puzzle, if you go the right way to work.


244.--THE FIFTEEN TURNINGS.

Here is another queer travelling puzzle, the solution of which calls for
ingenuity. In this case the traveller starts from the black town and
wishes to go as far as possible while making only fifteen turnings and
never going along the same road twice. The towns are supposed to be a
mile apart. Supposing, for example, that he went straight to A, then
straight to B, then to C, D, E, and F, you will then find that he has
travelled thirty-seven miles in five turnings. Now, how far can he go in
fifteen turnings?

[Illustration]


245.--THE FLY ON THE OCTAHEDRON.

"Look here," said the professor to his colleague, "I have been watching
that fly on the octahedron, and it confines its walks entirely to the
edges. What can be its reason for avoiding the sides?"

"Perhaps it is trying to solve some route problem," suggested the other.
"Supposing it to start from the top point, how many different routes are
there by which it may walk over all the edges, without ever going twice
along the same edge in any route?"

[Illustration]

The problem was a harder one than they expected, and after working at it
during leisure moments for several days their results did not agree--in
fact, they were both wrong. If the reader is surprised at their failure,
let him attempt the little puzzle himself. I will just explain that the
octahedron is one of the five regular, or Platonic, bodies, and is
contained under eight equal and equilateral triangles. If you cut out
the two pieces of cardboard of the shape shown in the margin of the
illustration, cut half through along the dotted lines and then bend them
and put them together, you will have a perfect octahedron. In any route
over all the edges it will be found that the fly must end at the point
of departure at the top.


246.--THE ICOSAHEDRON PUZZLE.

The icosahedron is another of the five regular, or Platonic, bodies
having all their sides, angles, and planes similar and equal. It is
bounded by twenty similar equilateral triangles. If you cut out a piece
of cardboard of the form shown in the smaller diagram, and cut half
through along the dotted lines, it will fold up and form a perfect
icosahedron.

Now, a Platonic body does not mean a heavenly body; but it will suit the
purpose of our puzzle if we suppose there to be a habitable planet of
this shape. We will also suppose that, owing to a superfluity of water,
the only dry land is along the edges, and that the inhabitants have no
knowledge of navigation. If every one of those edges is 10,000 miles
long and a solitary traveller is placed at the North Pole (the highest
point shown), how far will he have to travel before he will have visited
every habitable part of the planet--that is, have traversed every one of
the edges?

[Illustration]


247.--INSPECTING A MINE.

The diagram is supposed to represent the passages or galleries in a
mine. We will assume that every passage, A to B, B to C, C to H, H to I,
and so on, is one furlong in length. It will be seen that there are
thirty-one of these passages. Now, an official has to inspect all of
them, and he descends by the shaft to the point A. How far must he
travel, and what route do you recommend? The reader may at first say,
"As there are thirty-one passages, each a furlong in length, he will
have to travel just thirty-one furlongs." But this is assuming that he
need never go along a passage more than once, which is not the case.
Take your pencil and try to find the shortest route. You will soon
discover that there is room for considerable judgment. In fact, it is a
perplexing puzzle.

[Illustration]


248.--THE CYCLISTS' TOUR.

Two cyclists were consulting a road map in preparation for a little tour
together. The circles represent towns, and all the good roads are
represented by lines. They are starting from the town with a star, and
must complete their tour at E. But before arriving there they want to
visit every other town once, and only once. That is the difficulty. Mr.
Spicer said, "I am certain we can find a way of doing it;" but Mr. Maggs
replied, "No way, I'm sure." Now, which of them was correct? Take your
pencil and see if you can find any way of doing it. Of course you must
keep to the roads indicated.

[Illustration]


249.--THE SAILOR'S PUZZLE.

The sailor depicted in the illustration stated that he had since his
boyhood been engaged in trading with a small vessel among some twenty
little islands in the Pacific. He supplied the rough chart of which I
have given a copy, and explained that the lines from island to island
represented the only routes that he ever adopted. He always started from
island A at the beginning of the season, and then visited every island
once, and once only, finishing up his tour at the starting-point A. But
he always put off his visit to C as long as possible, for trade reasons
that I need not enter into. The puzzle is to discover his exact route,
and this can be done with certainty. Take your pencil and, starting at
A, try to trace it out. If you write down the islands in the order in
which you visit them--thus, for example, A, I, O, L, G, etc.--you can at
once see if you have visited an island twice or omitted any. Of course,
the crossings of the lines must be ignored--that is, you must continue
your route direct, and you are not allowed to switch off at a crossing
and proceed in another direction. There is no trick of this kind in the
puzzle. The sailor knew the best route. Can you find it?

[Illustration]


250.--THE GRAND TOUR.

One of the everyday puzzles of life is the working out of routes. If you
are taking a holiday on your bicycle, or a motor tour, there always
arises the question of how you are to make the best of your time and
other resources. You have determined to get as far as some particular
place, to include visits to such-and-such a town, to try to see
something of special interest elsewhere, and perhaps to try to look up
an old friend at a spot that will not take you much out of your way.
Then you have to plan your route so as to avoid bad roads, uninteresting
country, and, if possible, the necessity of a return by the same way
that you went. With a map before you, the interesting puzzle is attacked
and solved. I will present a little poser based on these lines.

I give a rough map of a country--it is not necessary to say what
particular country--the circles representing towns and the dotted lines
the railways connecting them. Now there lived in the town marked A a man
who was born there, and during the whole of his life had never once left
his native place. From his youth upwards he had been very industrious,
sticking incessantly to his trade, and had no desire whatever to roam
abroad. However, on attaining his fiftieth birthday he decided to see
something of his country, and especially to pay a visit to a very old
friend living at the town marked Z. What he proposed was this: that he
would start from his home, enter every town once and only once, and
finish his journey at Z. As he made up his mind to perform this grand
tour by rail only, he found it rather a puzzle to work out his route,
but he at length succeeded in doing so. How did he manage it? Do not
forget that every town has to be visited once, and not more than once.

[Illustration]


251.--WATER, GAS, AND ELECTRICITY.

There are some half-dozen puzzles, as old as the hills, that are
perpetually cropping up, and there is hardly a month in the year that
does not bring inquiries as to their solution. Occasionally one of
these, that one had thought was an extinct volcano, bursts into eruption
in a surprising manner. I have received an extraordinary number of
letters respecting the ancient puzzle that I have called "Water, Gas,
and Electricity." It is much older than electric lighting, or even gas,
but the new dress brings it up to date. The puzzle is to lay on water,
gas, and electricity, from W, G, and E, to each of the three houses, A,
B, and C, without any pipe crossing another. Take your pencil and draw
lines showing how this should be done. You will soon find yourself
landed in difficulties.

[Illustration]


252.--A PUZZLE FOR MOTORISTS.

[Illustration]

Eight motorists drove to church one morning. Their respective houses
and churches, together with the only roads available (the dotted lines),
are shown. One went from his house A to his church A, another from his
house B to his church B, another from C to C, and so on, but it was
afterwards found that no driver ever crossed the track of another car.
Take your pencil and try to trace out their various routes.


253.--A BANK HOLIDAY PUZZLE.

Two friends were spending their bank holiday on a cycling trip. Stopping
for a rest at a village inn, they consulted a route map, which is
represented in our illustration in an exceedingly simplified form, for
the puzzle is interesting enough without all the original complexities.
They started from the town in the top left-hand corner marked A. It will
be seen that there are one hundred and twenty such towns, all connected
by straight roads. Now they discovered that there are exactly 1,365
different routes by which they may reach their destination, always
travelling either due south or due east. The puzzle is to discover which
town is their destination.

[Illustration]

Of course, if you find that there are more than 1,365 different routes
to a town it cannot be the right one.


254.--THE MOTOR-CAR TOUR.

[Illustration]

In the above diagram the circles represent towns and the lines good
roads. In just how many different ways can a motorist, starting from
London (marked with an L), make a tour of all these towns, visiting
every town once, and only once, on a tour, and always coming back to
London on the last ride? The exact reverse of any route is not counted
as different.


255.--THE LEVEL PUZZLE.

[Illustration]

This is a simple counting puzzle. In how many different ways can you
spell out the word LEVEL by placing the point of your pencil on an L and
then passing along the lines from letter to letter. You may go in any
direction, backwards or forwards. Of course you are not allowed to miss
letters--that is to say, if you come to a letter you must use it.


256.--THE DIAMOND PUZZLE.

IN how many different ways may the word DIAMOND be read in the
arrangement shown? You may start wherever you like at a D and go up or
down, backwards or forwards, in and out, in any direction you like, so
long as you always pass from one letter to another that adjoins it. How
many ways are there?

[Illustration]


257.--THE DEIFIED PUZZLE.

In how many different ways may the word DEIFIED be read in this
arrangement under the same conditions as in the last puzzle, with the
addition that you can use any letters twice in the same reading?

[Illustration]


258.--THE VOTERS' PUZZLE.

[Illustration]

Here we have, perhaps, the most interesting form of the puzzle. In how
many different ways can you read the political injunction, "RISE TO
VOTE, SIR," under the same conditions as before? In this case every
reading of the palindrome requires the use of the central V as the
middle letter.


259.--HANNAH'S PUZZLE.

A man was in love with a young lady whose Christian name was Hannah.
When he asked her to be his wife she wrote down the letters of her name
in this manner:--

    H  H  H  H  H  H
    H  A  A  A  A  H
    H  A  N  N  A  H
    H  A  N  N  A  H
    H  A  A  A  A  H
    H  H  H  H  H  H

and promised that she would be his if he could tell her correctly in how
many different ways it was possible to spell out her name, always
passing from one letter to another that was adjacent. Diagonal steps are
here allowed. Whether she did this merely to tease him or to test his
cleverness is not recorded, but it is satisfactory to know that he
succeeded. Would you have been equally successful? Take your pencil and
try. You may start from any of the H's and go backwards or forwards and
in any direction, so long as all the letters in a spelling are adjoining
one another. How many ways are there, no two exactly alike?


260.--THE HONEYCOMB PUZZLE.

[Illustration]

Here is a little puzzle with the simplest possible conditions. Place the
point of your pencil on a letter in one of the cells of the honeycomb,
and trace out a very familiar proverb by passing always from a cell to
one that is contiguous to it. If you take the right route you will have
visited every cell once, and only once. The puzzle is much easier than
it looks.


261.--THE MONK AND THE BRIDGES.

In this case I give a rough plan of a river with an island and five
bridges. On one side of the river is a monastery, and on the other side
is seen a monk in the foreground. Now, the monk has decided that he will
cross every bridge once, and only once, on his return to the monastery.
This is, of course, quite easy to do, but on the way he thought to
himself, "I wonder how many different routes there are from which I
might have selected." Could you have told him? That is the puzzle. Take
your pencil and trace out a route that will take you once over all the
five bridges. Then trace out a second route, then a third, and see if
you can count all the variations. You will find that the difficulty is
twofold: you have to avoid dropping routes on the one hand and counting
the same routes more than once on the other.

[Illustration]



COMBINATION AND GROUP PROBLEMS.

    "A combination and a form indeed."
                         _Hamlet_, iii. 4.

Various puzzles in this class might be termed problems in the "geometry
of situation," but their solution really depends on the theory of
combinations which, in its turn, is derived directly from the theory of
permutations. It has seemed convenient to include here certain group
puzzles and enumerations that might, perhaps, with equal reason have
been placed elsewhere; but readers are again asked not to be too
critical about the classification, which is very difficult and
arbitrary. As I have included my problem of "The Round Table" (No. 273),
perhaps a few remarks on another well-known problem of the same class,
known by the French as La Problême des Ménages, may be interesting. If
n married ladies are seated at a round table in any determined order,
in how many different ways may their n husbands be placed so that
every man is between two ladies but never next to his own wife?

This difficult problem was first solved by Laisant, and the method shown
in the following table is due to Moreau:--

     4        0        2
     5        3       13
     6       13       80
     7       83      579
     8      592     4738
     9     4821    43387
    10    43979   439792

The first column shows the number of married couples. The numbers in the
second column are obtained in this way: 5 × 3 + 0 - 2 = 13; 6 × 13 + 3 +
2 = 83; 7 × 83 + 13 - 2 = 592; 8 × 592 + 83 + 2 = 4821; and so on. Find
all the numbers, except 2, in the table, and the method will be evident.
It will be noted that the 2 is subtracted when the first number (the
number of couples) is odd, and added when that number is even. The
numbers in the third column are obtained thus: 13 - 0 = 13; 83 - 3 = 80;
592 - 13 = 579; 4821 - 83 = 4738; and so on. The numbers in this last
column give the required solutions. Thus, four husbands may be seated in
two ways, five husbands may be placed in thirteen ways, and six husbands
in eighty ways.

The following method, by Lucas, will show the remarkable way in which
chessboard analysis may be applied to the solution of a circular problem
of this kind. Divide a square into thirty-six cells, six by six, and
strike out all the cells in the long diagonal from the bottom left-hand
corner to the top right-hand corner, also the five cells in the diagonal
next above it and the cell in the bottom right-hand corner. The answer
for six couples will be the same as the number of ways in which you can
place six rooks (not using the cancelled cells) so that no rook shall
ever attack another rook. It will be found that the six rooks may be
placed in eighty different ways, which agrees with the above table.


262.--THOSE FIFTEEN SHEEP.

A certain cyclopædia has the following curious problem, I am told:
"Place fifteen sheep in four pens so that there shall be the same number
of sheep in each pen." No answer whatever is vouchsafed, so I thought I
would investigate the matter. I saw that in dealing with apples or
bricks the thing would appear to be quite impossible, since four times
any number must be an even number, while fifteen is an odd number. I
thought, therefore, that there must be some quality peculiar to the
sheep that was not generally known. So I decided to interview some
farmers on the subject. The first one pointed out that if we put one pen
inside another, like the rings of a target, and placed all sheep in the
smallest pen, it would be all right. But I objected to this, because you
admittedly place all the sheep in one pen, not in four pens. The second
man said that if I placed four sheep in each of three pens and three
sheep in the last pen (that is fifteen sheep in all), and one of the
ewes in the last pen had a lamb during the night, there would be the
same number in each pen in the morning. This also failed to satisfy me.

[Illustration]

The third farmer said, "I've got four hurdle pens down in one of my
fields, and a small flock of wethers, so if you will just step down with
me I will show you how it is done." The illustration depicts my friend
as he is about to demonstrate the matter to me. His lucid explanation
was evidently that which was in the mind of the writer of the article in
the cyclopædia. What was it? Can you place those fifteen sheep?


263.--KING ARTHUR'S KNIGHTS.

King Arthur sat at the Round Table on three successive evenings with his
knights--Beleobus, Caradoc, Driam, Eric, Floll, and Galahad--but on no
occasion did any person have as his neighbour one who had before sat
next to him. On the first evening they sat in alphabetical order round
the table. But afterwards King Arthur arranged the two next sittings so
that he might have Beleobus as near to him as possible and Galahad as
far away from him as could be managed. How did he seat the knights to
the best advantage, remembering that rule that no knight may have the
same neighbour twice?


264.--THE CITY LUNCHEONS.

Twelve men connected with a large firm in the City of London sit down to
luncheon together every day in the same room. The tables are small ones
that only accommodate two persons at the same time. Can you show how
these twelve men may lunch together on eleven days in pairs, so that no
two of them shall ever sit twice together? We will represent the men by
the first twelve letters of the alphabet, and suppose the first day's
pairing to be as follows--

    (A B) (C D) (E F) (G H) (I J) (K L).

Then give any pairing you like for the next day, say--

    (A C) (B D) (E G) (F H) (I K) (J L),

and so on, until you have completed your eleven lines, with no pair ever
occurring twice. There are a good many different arrangements possible.
Try to find one of them.


265.--A PUZZLE FOR CARD-PLAYERS.

Twelve members of a club arranged to play bridge together on eleven
evenings, but no player was ever to have the same partner more than
once, or the same opponent more than twice. Can you draw up a scheme
showing how they may all sit down at three tables every evening? Call
the twelve players by the first twelve letters of the alphabet and try
to group them.


266.--A TENNIS TOURNAMENT.

Four married couples played a "mixed double" tennis tournament, a man
and a lady always playing against a man and a lady. But no person ever
played with or against any other person more than once. Can you show how
they all could have played together in the two courts on three
successive days? This is a little puzzle of a quite practical kind, and
it is just perplexing enough to be interesting.


267.--THE WRONG HATS.

"One of the most perplexing things I have come across lately," said Mr.
Wilson, "is this. Eight men had been dining not wisely but too well at a
certain London restaurant. They were the last to leave, but not one man
was in a condition to identify his own hat. Now, considering that they
took their hats at random, what are the chances that every man took a
hat that did not belong to him?"

"The first thing," said Mr. Waterson, "is to see in how many different
ways the eight hats could be taken."

"That is quite easy," Mr. Stubbs explained. "Multiply together the
numbers, 1, 2, 3, 4, 5, 6, 7, and 8. Let me see--half a minute--yes;
there are 40,320 different ways."

"Now all you've got to do is to see in how many of these cases no man
has his own hat," said Mr. Waterson.

"Thank you, I'm not taking any," said Mr. Packhurst. "I don't envy the
man who attempts the task of writing out all those forty-thousand-odd
cases and then picking out the ones he wants."

They all agreed that life is not long enough for that sort of amusement;
and as nobody saw any other way of getting at the answer, the matter was
postponed indefinitely. Can you solve the puzzle?


268.--THE PEAL OF BELLS.

A correspondent, who is apparently much interested in campanology, asks
me how he is to construct what he calls a "true and correct" peal for
four bells. He says that every possible permutation of the four bells
must be rung once, and once only. He adds that no bell must move more
than one place at a time, that no bell must make more than two
successive strokes in either the first or the last place, and that the
last change must be able to pass into the first. These fantastic
conditions will be found to be observed in the little peal for three
bells, as follows:--

    1   2   3
    2   1   3
    2   3   1
    3   2   1
    3   1   2
    1   3   2

How are we to give him a correct solution for his four bells?


269.--THREE MEN IN A BOAT.

A certain generous London manufacturer gives his workmen every year a
week's holiday at the seaside at his own expense. One year fifteen of
his men paid a visit to Herne Bay. On the morning of their departure
from London they were addressed by their employer, who expressed the
hope that they would have a very pleasant time.

"I have been given to understand," he added, "that some of you fellows
are very fond of rowing, so I propose on this occasion to provide you
with this recreation, and at the same time give you an amusing little
puzzle to solve. During the seven days that you are at Herne Bay every
one of you will go out every day at the same time for a row, but there
must always be three men in a boat and no more. No two men may ever go
out in a boat together more than once, and no man is allowed to go out
twice in the same boat. If you can manage to do this, and use as few
different boats as possible, you may charge the firm with the expense."

One of the men tells me that the experience he has gained in such
matters soon enabled him to work out the answer to the entire
satisfaction of themselves and their employer. But the amusing part of
the thing is that they never really solved the little mystery. I find
their method to have been quite incorrect, and I think it will amuse my
readers to discover how the men should have been placed in the boats. As
their names happen to have been Andrews, Baker, Carter, Danby, Edwards,
Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, and
Onslow, we can call them by their initials and write out the five groups
for each of the seven days in the following simple way:

                 1     2     3     4     5
    First Day: (ABC) (DEF) (GHI) (JKL) (MNO).

The men within each pair of brackets are here seen to be in the same
boat, and therefore A can never go out with B or with C again, and C can
never go out again with B. The same applies to the other four boats. The
figures show the number on the boat, so that A, B, or C, for example,
can never go out in boat No. 1 again.


270.--THE GLASS BALLS.

A number of clever marksmen were staying at a country house, and the
host, to provide a little amusement, suspended strings of glass balls,
as shown in the illustration, to be fired at. After they had all put
their skill to a sufficient test, somebody asked the following question:
"What is the total number of different ways in which these sixteen balls
may be broken, if we must always break the lowest ball that remains on
any string?" Thus, one way would be to break all the four balls on each
string in succession, taking the strings from left to right. Another
would be to break all the fourth balls on the four strings first, then
break the three remaining on the first string, then take the balls on
the three other strings alternately from right to left, and so on. There
is such a vast number of different ways (since every little variation of
order makes a different way) that one is apt to be at first impressed by
the great difficulty of the problem. Yet it is really quite simple when
once you have hit on the proper method of attacking it. How many
different ways are there?

[Illustration]


271.--FIFTEEN LETTER PUZZLE.

    ALE FOE HOD BGN
    CAB HEN JOG KFM
    HAG GEM MOB BFH
    FAN KIN JEK DFL
    JAM HIM GCL LJH
    AID JIB FCJ NJD
    OAK FIG HCK MLN
    BED OIL MCD BLK
    ICE CON DGK

The above is the solution of a puzzle I gave in _Tit-bits_ in the summer
of 1896. It was required to take the letters, A, B, C, D, E, F, G, H, I,
J, K, L, M, N, and O, and with them form thirty-five groups of three
letters so that the combinations should include the greatest number
possible of common English words. No two letters may appear together in
a group more than once. Thus, A and L having been together in ALE, must
never be found together again; nor may A appear again in a group with E,
nor L with E. These conditions will be found complied with in the above
solution, and the number of words formed is twenty-one. Many persons
have since tried hard to beat this number, but so far have not
succeeded.

More than thirty-five combinations of the fifteen letters cannot be
formed within the conditions. Theoretically, there cannot possibly be
more than twenty-three words formed, because only this number of
combinations is possible with a vowel or vowels in each. And as no
English word can be formed from three of the given vowels (A, E, I, and
O), we must reduce the number of possible words to twenty-two. This is
correct theoretically, but practically that twenty-second word cannot be
got in. If JEK, shown above, were a word it would be all right; but it
is not, and no amount of juggling with the other letters has resulted in
a better answer than the one shown. I should, say that proper nouns and
abbreviations, such as Joe, Jim, Alf, Hal, Flo, Ike, etc., are
disallowed.

Now, the present puzzle is a variation of the above. It is simply this:
Instead of using the fifteen letters given, the reader is allowed to
select any fifteen different letters of the alphabet that he may prefer.
Then construct thirty-five groups in accordance with the conditions, and
show as many good English words as possible.


272.--THE NINE SCHOOLBOYS.

This is a new and interesting companion puzzle to the "Fifteen
Schoolgirls" (see solution of No. 269), and even in the simplest
possible form in which I present it there are unquestionable
difficulties. Nine schoolboys walk out in triplets on the six week days
so that no boy ever walks _side by side_ with any other boy more than
once. How would you arrange them?

If we represent them by the first nine letters of the alphabet, they
might be grouped on the first day as follows:--

    A   B   C
    D   E   F
    G   H   I

Then A can never walk again side by side with B, or B with C, or D with
E, and so on. But A can, of course, walk side by side with C. It is here
not a question of being together in the same triplet, but of walking
side by side in a triplet. Under these conditions they can walk out on
six days; under the "Schoolgirls" conditions they can only walk on four
days.


273.--THE ROUND TABLE.

Seat the same n persons at a round table on

    (n - 1)(n - 2)
    --------------
           2

occasions so that no person shall ever have the same two neighbours
twice. This is, of course, equivalent to saying that every person must
sit once, and once only, between every possible pair.


274.--THE MOUSE-TRAP PUZZLE.

[Illustration


         6 20 2 19
      13           21
     7               5
    3                18
    17                8
    15               11
     14             16
       1           9
          10 4 12

]

This is a modern version, with a difference, of an old puzzle of the
same name. Number twenty-one cards, 1, 2, 3, etc., up to 21, and place
them in a circle in the particular order shown in the illustration.
These cards represent mice. You start from any card, calling that card
"one," and count, "one, two, three," etc., in a clockwise direction, and
when your count agrees with the number on the card, you have made a
"catch," and you remove the card. Then start at the next card, calling
that "one," and try again to make another "catch." And so on. Supposing
you start at 18, calling that card "one," your first "catch" will be 19.
Remove 19 and your next "catch" is 10. Remove 10 and your next "catch"
is 1. Remove the 1, and if you count up to 21 (you must never go
beyond), you cannot make another "catch." Now, the ideal is to "catch"
all the twenty-one mice, but this is not here possible, and if it were
it would merely require twenty-one different trials, at the most, to
succeed. But the reader may make any two cards change places before he
begins. Thus, you can change the 6 with the 2, or the 7 with the 11, or
any other pair. This can be done in several ways so as to enable you to
"catch" all the twenty-one mice, if you then start at the right place.
You may never pass over a "catch"; you must always remove the card and
start afresh.


275.--THE SIXTEEN SHEEP.

[Illustration:

    +========================+
    ||    |     |     |     ||
    || 0  |  0  |  0  |  0  ||
    +-----+-----+-----+------+
    ||    |     |     |     ||
    || 0  |  0  |  0  |  0  ||
    +========================+
    ||    ||    |     ||    ||
    || 0  || 0  |  0  || 0  ||
    +-----+=====+=====+------+
    ||    |     ||    |     ||
    || 0  |  0  || 0  |  0  ||
    +========================+

]

Here is a new puzzle with matches and counters or coins. In the
illustration the matches represent hurdles and the counters sheep. The
sixteen hurdles on the outside, and the sheep, must be regarded as
immovable; the puzzle has to do entirely with the nine hurdles on the
inside. It will be seen that at present these nine hurdles enclose four
groups of 8, 3, 3, and 2 sheep. The farmer requires to readjust some of
the hurdles so as to enclose 6, 6, and 4 sheep. Can you do it by only
replacing two hurdles? When you have succeeded, then try to do it by
replacing three hurdles; then four, five, six, and seven in succession.
Of course, the hurdles must be legitimately laid on the dotted lines,
and no such tricks are allowed as leaving unconnected ends of hurdles,
or two hurdles placed side by side, or merely making hurdles change
places. In fact, the conditions are so simple that any farm labourer
will understand it directly.


276.--THE EIGHT VILLAS.

In one of the outlying suburbs of London a man had a square plot of
ground on which he decided to build eight villas, as shown in the
illustration, with a common recreation ground in the middle. After the
houses were completed, and all or some of them let, he discovered that
the number of occupants in the three houses forming a side of the square
was in every case nine. He did not state how the occupants were
distributed, but I have shown by the numbers on the sides of the houses
one way in which it might have happened. The puzzle is to discover the
total number of ways in which all or any of the houses might be
occupied, so that there should be nine persons on each side. In order
that there may be no misunderstanding, I will explain that although B is
what we call a reflection of A, these would count as two different
arrangements, while C, if it is turned round, will give four
arrangements; and if turned round in front of a mirror, four other
arrangements. All eight must be counted.


[Illustration:

     /\      /\      /\
    |2 |    |5 |    |2 |

     /\              /\
    |5 |            |5 |

     /\      /\      /\
    |2 |    |5 |    |2 |

    +---+---+---+    +---+---+---+    +---+---+---+
    | 1 | 6 | 2 |    | 2 | 6 | 1 |    | 1 | 6 | 2 |
    +---+---+---+    +---+---+---+    +---+---+---+
    | 6 |   | 6 |    | 6 |   | 6 |    | 4 |   | 4 |
    +---+---+---+    +---+---+---+    +---+---+---+
    | 2 | 6 | 1 |    | 1 | 6 | 2 |    | 4 | 2 | 3 |
    +---+---+---+    +---+---+---+    +---+---+---+
          A                B                C

]


277.--COUNTER CROSSES.

All that we need for this puzzle is nine counters, numbered 1, 2, 3, 4,
5, 6, 7, 8, and 9. It will be seen that in the illustration A these are
arranged so as to form a Greek cross, while in the case of B they form a
Latin cross. In both cases the reader will find that the sum of the
numbers in the upright of the cross is the same as the sum of the
numbers in the horizontal arm. It is quite easy to hit on such an
arrangement by trial, but the problem is to discover in exactly how many
different ways it may be done in each case. Remember that reversals and
reflections do not count as different. That is to say, if you turn this
page round you get four arrangements of the Greek cross, and if you turn
it round again in front of a mirror you will get four more. But these
eight are all regarded as one and the same. Now, how many different ways
are there in each case?

[Illustration:

            (1)                        (2)

            (2)                (4) (5) (1) (6) (7)

    (3) (4) (9) (5) (6)                (3)

            (7)                        (8)

    A       (8)                B       (9)

]


278.--A DORMITORY PUZZLE.

In a certain convent there were eight large dormitories on one floor,
approached by a spiral staircase in the centre, as shown in our plan. On
an inspection one Monday by the abbess it was found that the south
aspect was so much preferred that six times as many nuns slept on the
south side as on each of the other three sides. She objected to this
overcrowding, and ordered that it should be reduced. On Tuesday she
found that five times as many slept on the south side as on each of the
other sides. Again she complained. On Wednesday she found four times as
many on the south side, on Thursday three times as many, and on Friday
twice as many. Urging the nuns to further efforts, she was pleased to
find on Saturday that an equal number slept on each of the four sides of
the house. What is the smallest number of nuns there could have been,
and how might they have arranged themselves on each of the six nights?
No room may ever be unoccupied.

[Illustration

    +---+---+---+
    |   |   |   |
    |   |   |   |
    |   |   |   |
    +---+---+---+
    |   |\|/|   |
    |   |-*-|   |
    |   |/|\|   |
    +---+---+---+
    |   |   |   |
    |   |   |   |
    |   |   |   |
    +---+---+---+

]

279.--THE BARRELS OF BALSAM.

A merchant of Bagdad had ten barrels of precious balsam for sale. They
were numbered, and were arranged in two rows, one on top of the other,
as shown in the picture. The smaller the number on the barrel, the
greater was its value. So that the best quality was numbered "1" and the
worst numbered "10," and all the other numbers of graduating values.
Now, the rule of Ahmed Assan, the merchant, was that he never put a
barrel either beneath or to the right of one of less value. The
arrangement shown is, of course, the simplest way of complying with this
condition. But there are many other ways--such, for example, as this:--

    1  2  5  7  8
    3  4  6  9 10

Here, again, no barrel has a smaller number than itself on its right or
beneath it. The puzzle is to discover in how many different ways the
merchant of Bagdad might have arranged his barrels in the two rows
without breaking his rule. Can you count the number of ways?


280.--BUILDING THE TETRAHEDRON.

I possess a tetrahedron, or triangular pyramid, formed of six sticks
glued together, as shown in the illustration. Can you count correctly
the number of different ways in which these six sticks might have been
stuck together so as to form the pyramid?

Some friends worked at it together one evening, each person providing
himself with six lucifer matches to aid his thoughts; but it was found
that no two results were the same. You see, if we remove one of the
sticks and turn it round the other way, that will be a different
pyramid. If we make two of the sticks change places the result will
again be different. But remember that every pyramid may be made to stand
on either of its four sides without being a different one. How many ways
are there altogether?

[Illustration]


281.--PAINTING A PYRAMID.

This puzzle concerns the painting of the four sides of a tetrahedron, or
triangular pyramid. If you cut out a piece of cardboard of the
triangular shape shown in Fig. 1, and then cut half through along the
dotted lines, it will fold up and form a perfect triangular pyramid. And
I would first remind my readers that the primary colours of the solar
spectrum are seven--violet, indigo, blue, green, yellow, orange, and
red. When I was a child I was taught to remember these by the ungainly
word formed by the initials of the colours, "Vibgyor."

[Illustration]

In how many different ways may the triangular pyramid be coloured, using
in every case one, two, three, or four colours of the solar spectrum? Of
course a side can only receive a single colour, and no side can be left
uncoloured. But there is one point that I must make quite clear. The
four sides are not to be regarded as individually distinct. That is to
say, if you paint your pyramid as shown in Fig. 2 (where the bottom side
is green and the other side that is out of view is yellow), and then
paint another in the order shown in Fig. 3, these are really both the
same and count as one way. For if you tilt over No. 2 to the right it
will so fall as to represent No. 3. The avoidance of repetitions of this
kind is the real puzzle of the thing. If a coloured pyramid cannot be
placed so that it exactly resembles in its colours and their relative
order another pyramid, then they are different. Remember that one way
would be to colour all the four sides red, another to colour two sides
green, and the remaining sides yellow and blue; and so on.


282.--THE ANTIQUARY'S CHAIN.

An antiquary possessed a number of curious old links, which he took to a
blacksmith, and told him to join together to form one straight piece of
chain, with the sole condition that the two circular links were not to
be together. The following illustration shows the appearance of the
chain and the form of each link. Now, supposing the owner should
separate the links again, and then take them to another smith and repeat
his former instructions exactly, what are the chances against the links
being put together exactly as they were by the first man? Remember that
every successive link can be joined on to another in one of two ways,
just as you can put a ring on your finger in two ways, or link your
forefingers and thumbs in two ways.

[Illustration]


283.--THE FIFTEEN DOMINOES.

In this case we do not use the complete set of twenty-eight dominoes to
be found in the ordinary box. We dispense with all those dominoes that
have a five or a six on them and limit ourselves to the fifteen that
remain, where the double-four is the highest.

In how many different ways may the fifteen dominoes be arranged in a
straight line in accordance with the simple rule of the game that a
number must always be placed against a similar number--that is, a four
against a four, a blank against a blank, and so on? Left to right and
right to left of the same arrangement are to be counted as two different
ways.


384.--THE CROSS TARGET.

        +-+-+
        |*|*|
        +-+-+
        |*|*|
    +-+-+-+-+-+-+
    | | | |*| | |
    +-+-+-+-+-+-+
    | | |*| |*| |
    +-+-+-+-+-+-+
        | |*|
        +-+-+
        | | |
        +-+-+

In the illustration we have a somewhat curious target designed by an
eccentric sharpshooter. His idea was that in order to score you must hit
four circles in as many shots so that those four shots shall form a
square. It will be seen by the results recorded on the target that two
attempts have been successful. The first man hit the four circles at the
top of the cross, and thus formed his square. The second man intended to
hit the four in the bottom arm, but his second shot, on the left, went
too high. This compelled him to complete his four in a different way
than he intended. It will thus be seen that though it is immaterial
which circle you hit at the first shot, the second shot may commit you
to a definite procedure if you are to get your square. Now, the puzzle
is to say in just how many different ways it is possible to form a
square on the target with four shots.


285.--THE FOUR POSTAGE STAMPS.

    +---+----+----+----+
    | 1 |  2 |  3 |  4 |
    +---+----+----+----+
    | 5 |  6 |  7 |  8 |
    +---+----+----+----+
    | 9 | 10 | 11 | 12 |
    +---+----+----+----+

"It is as easy as counting," is an expression one sometimes hears. But
mere counting may be puzzling at times. Take the following simple
example. Suppose you have just bought twelve postage stamps, in this
form--three by four--and a friend asks you to oblige him with four
stamps, all joined together--no stamp hanging on by a mere corner. In
how many different ways is it possible for you to tear off those four
stamps? You see, you can give him 1, 2, 3, 4, or 2, 3, 6, 7, or 1, 2, 3,
6, or 1, 2, 3, 7, or 2, 3, 4, 8, and so on. Can you count the number of
different ways in which those four stamps might be delivered? There are
not many more than fifty ways, so it is not a big count. Can you get the
exact number?


286.--PAINTING THE DIE.

In how many different ways may the numbers on a single die be marked,
with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4
must be on opposite sides? It is a simple enough question, and yet it
will puzzle a good many people.


287.--AN ACROSTIC PUZZLE.

In the making or solving of double acrostics, has it ever occurred to
you to consider the variety and limitation of the pair of initial and
final letters available for cross words? You may have to find a word
beginning with A and ending with B, or A and C, or A and D, and so on.
Some combinations are obviously impossible--such, for example, as those
with Q at the end. But let us assume that a good English word can be
found for every case. Then how many possible pairs of letters are
available?



CHESSBOARD PROBLEMS.

    "You and I will goe to the chesse."

    GREENE'S _Groatsworth of Wit._


During a heavy gale a chimney-pot was hurled through the air, and
crashed upon the pavement just in front of a pedestrian. He quite calmly
said, "I have no use for it: I do not smoke." Some readers, when they
happen to see a puzzle represented on a chessboard with chess pieces,
are apt to make the equally inconsequent remark, "I have no use for it:
I do not play chess." This is largely a result of the common, but
erroneous, notion that the ordinary chess puzzle with which we are
familiar in the press (dignified, for some reason, with the name
"problem") has a vital connection with the game of chess itself. But
there is no condition in the game that you shall checkmate your opponent
in two moves, in three moves, or in four moves, while the majority of
the positions given in these puzzles are such that one player would have
so great a superiority in pieces that the other would have resigned
before the situations were reached. And the solving of them helps you
but little, and that quite indirectly, in playing the game, it being
well known that, as a rule, the best "chess problemists" are indifferent
players, and _vice versa_. Occasionally a man will be found strong on
both subjects, but he is the exception to the rule.

Yet the simple chequered board and the characteristic moves of the
pieces lend themselves in a very remarkable manner to the devising of
the most entertaining puzzles. There is room for such infinite variety
that the true puzzle lover cannot afford to neglect them. It was with a
view to securing the interest of readers who are frightened off by the
mere presentation of a chessboard that so many puzzles of this class
were originally published by me in various fanciful dresses. Some of
these posers I still retain in their disguised form; others I have
translated into terms of the chessboard. In the majority of cases the
reader will not need any knowledge whatever of chess, but I have thought
it best to assume throughout that he is acquainted with the terminology,
the moves, and the notation of the game.

I first deal with a few questions affecting the chessboard itself; then
with certain statical puzzles relating to the Rook, the Bishop, the
Queen, and the Knight in turn; then dynamical puzzles with the pieces in
the same order; and, finally, with some miscellaneous puzzles on the
chessboard. It is hoped that the formulæ and tables given at the end of
the statical puzzles will be of interest, as they are, for the most
part, published for the first time.



THE CHESSBOARD.

    "Good company's a chessboard."
             BYRON'S _Don Juan_, xiii. 89.

A chessboard is essentially a square plane divided into sixty-four
smaller squares by straight lines at right angles. Originally it was not
chequered (that is, made with its rows and columns alternately black and
white, or of any other two colours), and this improvement was introduced
merely to help the eye in actual play. The utility of the chequers is
unquestionable. For example, it facilitates the operation of the
bishops, enabling us to see at the merest glance that our king or pawns
on black squares are not open to attack from an opponent's bishop
running on the white diagonals. Yet the chequering of the board is not
essential to the game of chess. Also, when we are propounding puzzles on
the chessboard, it is often well to remember that additional interest
may result from "generalizing" for boards containing any number of
squares, or from limiting ourselves to some particular chequered
arrangement, not necessarily a square. We will give a few puzzles
dealing with chequered boards in this general way.


288.--CHEQUERED BOARD DIVISIONS.

I recently asked myself the question: In how many different ways may a
chessboard be divided into two parts of the same size and shape by cuts
along the lines dividing the squares? The problem soon proved to be both
fascinating and bristling with difficulties. I present it in a
simplified form, taking a board of smaller dimensions.

[Illustration:

    +---+---*---+---+   +---+---+---*---+   +---+---+---*---+
    |   |   H   |   |   |   |   |   H   |   |   |   |   H   |
    +---+---*---+---+   +---+---*===*---+   +---*===*---*---+
    |   |   H   |   |   |   |   H   |   |   |   H   H   H   |
    +---+---*---+---+   +---+---*---+---+   +---*---*---*---+
    |   |   H   |   |   |   |   H   |   |   |   H   H   H   |
    +---+---*---+---+   +---*===*---+---+   +---*---*===*---+
    |   |   H   |   |   |   H   |   |   |   |   H   |   |   |
    +---+---*---+---+   +---*---+---+---+   +---*---+---+---+

                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+
                    |   |   |   |   |   |   |
                    +---+---+---+---+---+---+

    +---+---*---+---+   +---+---+---*---+   +---+---+---*---+
    |   |   H   |   |   |   |   |   H   |   |   |   |   H   |
    +---*===*---+---+   +---*===*===*---+   +---+---*===*---+
    |   H   |   |   |   |   H   |   |   |   |   |   H   |   |
    +---*===*===*---+   +---*===*===*---+   +---+---*---+---+
    |   |   |   H   |   |   |   |   H   |   |   |   H   |   |
    +---+---*===*---+   +---*===*===*---+   +---*===*---+---+
    |   |   H   |   |   |   H   |   |   |   |   H   |   |   |
    +---+---*---+---+   +---*---+---+---+   +---*---+---+---+

]

It is obvious that a board of four squares can only be so divided in one
way--by a straight cut down the centre--because we shall not count
reversals and reflections as different. In the case of a board of
sixteen squares--four by four--there are just six different ways. I have
given all these in the diagram, and the reader will not find any others.
Now, take the larger board of thirty-six squares, and try to discover in
how many ways it may be cut into two parts of the same size and shape.


289.--LIONS AND CROWNS.

The young lady in the illustration is confronted with a little
cutting-out difficulty in which the reader may be glad to assist her.
She wishes, for some reason that she has not communicated to me, to cut
that square piece of valuable material into four parts, all of exactly
the same size and shape, but it is important that every piece shall
contain a lion and a crown. As she insists that the cuts can only be
made along the lines dividing the squares, she is considerably perplexed
to find out how it is to be done. Can you show her the way? There is
only one possible method of cutting the stuff.

[Illustration:

    +-+-+-+-+-+-+
    | | | | | | |
    +-+-+-+-+-+-+
    | |L|L|L| | |
    +-+-+-+-+-+-+
    | | |C|C| | |
    +-+-+-+-+-+-+
    | | |C|C| | |
    +-+-+-+-+-+-+
    |L| | | | | |
    +-+-+-+-+-+-+
    | | | | | | |
    +-+-+-+-+-+-+

]


290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

We will here consider the question of those boards that contain an odd
number of squares. We will suppose that the central square is first cut
out, so as to leave an even number of squares for division. Now, it is
obvious that a square three by three can only be divided in one way, as
shown in Fig. 1. It will be seen that the pieces A and B are of the same
size and shape, and that any other way of cutting would only produce the
same shaped pieces, so remember that these variations are not counted as
different ways. The puzzle I propose is to cut the board five by five
(Fig. 2) into two pieces of the same size and shape in as many different
ways as possible. I have shown in the illustration one way of doing it.
How many different ways are there altogether? A piece which when turned
over resembles another piece is not considered to be of a different
shape.

[Illustration:

    +---*---+---+
    |   H   |   |
    +---*===*---+
    |   HHHHH   |
    +---*===*---+
    |   |   H   |
    +---+---*---+

Fig 1]

[Illustration:

    +---+---+---+---+---+
    |   |   |   |   |   |
    *===*===*===*---+---+
    |   |   |   H   |   |
    +---+---*===*---+---+
    |   |   HHHHH   |   |
    +---+---*===*---+---+
    |   |   H   |   |   |
    +---+---*===*===*===*
    |   H   |   |   |   |
    +---*---+---+---+---+

Fig 2]


291.--THE GRAND LAMA'S PROBLEM.

Once upon a time there was a Grand Lama who had a chessboard made of
pure gold, magnificently engraved, and, of course, of great value. Every
year a tournament was held at Lhassa among the priests, and whenever any
one beat the Grand Lama it was considered a great honour, and his name
was inscribed on the back of the board, and a costly jewel set in the
particular square on which the checkmate had been given. After this
sovereign pontiff had been defeated on four occasions he died--possibly
of chagrin.

[Illustration:

    +---+---+---+---+---+---+---+---+
    | * |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   | * |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   | * |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   | * |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+

]

Now the new Grand Lama was an inferior chess-player, and preferred other
forms of innocent amusement, such as cutting off people's heads. So he
discouraged chess as a degrading game, that did not improve either the
mind or the morals, and abolished the tournament summarily. Then he sent
for the four priests who had had the effrontery to play better than a
Grand Lama, and addressed them as follows: "Miserable and heathenish
men, calling yourselves priests! Know ye not that to lay claim to a
capacity to do anything better than my predecessor is a capital offence?
Take that chessboard and, before day dawns upon the torture chamber, cut
it into four equal parts of the same shape, each containing sixteen
perfect squares, with one of the gems in each part! If in this you fail,
then shall other sports be devised for your special delectation. Go!"
The four priests succeeded in their apparently hopeless task. Can you
show how the board may be divided into four equal parts, each of
exactly the same shape, by cuts along the lines dividing the squares,
each part to contain one of the gems?


292.--THE ABBOT'S WINDOW.

[Illustration]

Once upon a time the Lord Abbot of St. Edmondsbury, in consequence of
"devotions too strong for his head," fell sick and was unable to leave
his bed. As he lay awake, tossing his head restlessly from side to side,
the attentive monks noticed that something was disturbing his mind; but
nobody dared ask what it might be, for the abbot was of a stern
disposition, and never would brook inquisitiveness. Suddenly he called
for Father John, and that venerable monk was soon at the bedside.

"Father John," said the Abbot, "dost thou know that I came into this
wicked world on a Christmas Even?"

The monk nodded assent.

"And have I not often told thee that, having been born on Christmas
Even, I have no love for the things that are odd? Look there!"

The Abbot pointed to the large dormitory window, of which I give a
sketch. The monk looked, and was perplexed.

"Dost thou not see that the sixty-four lights add up an even number
vertically and horizontally, but that all the _diagonal_ lines, except
fourteen are of a number that is odd? Why is this?"

"Of a truth, my Lord Abbot, it is of the very nature of things, and
cannot be changed."

"Nay, but it _shall_ be changed. I command thee that certain of the
lights be closed this day, so that every line shall have an even number
of lights. See thou that this be done without delay, lest the cellars be
locked up for a month and other grievous troubles befall thee."

Father John was at his wits' end, but after consultation with one who
was learned in strange mysteries, a way was found to satisfy the whim of
the Lord Abbot. Which lights were blocked up, so that those which
remained added up an even number in every line horizontally, vertically,
and diagonally, while the least possible obstruction of light was
caused?


293.--THE CHINESE CHESSBOARD.

Into how large a number of different pieces may the chessboard be cut
(by cuts along the lines only), no two pieces being exactly alike?
Remember that the arrangement of black and white constitutes a
difference. Thus, a single black square will be different from a single
white square, a row of three containing two white squares will differ
from a row of three containing two black, and so on. If two pieces
cannot be placed on the table so as to be exactly alike, they count as
different. And as the back of the board is plain, the pieces cannot be
turned over.


294.--THE CHESSBOARD SENTENCE.

[Illustration]

I once set myself the amusing task of so dissecting an ordinary
chessboard into letters of the alphabet that they would form a complete
sentence. It will be seen from the illustration that the pieces
assembled give the sentence, "CUT THY LIFE," with the stops between. The
ideal sentence would, of course, have only one full stop, but that I did
not succeed in obtaining.

The sentence is an appeal to the transgressor to cut himself adrift from
the evil life he is living. Can you fit these pieces together to form a
perfect chessboard?



STATICAL CHESS PUZZLES.

    "They also serve who only stand and wait."
                                MILTON.


295.--THE EIGHT ROOKS.


[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    | R | R | R | R | R | R | R | R |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+

FIG. 1.]

[Illustration:

    +---+---+---+---+---+---+---+---+
    | R |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   | R |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   | R |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   | R |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   | R |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   | R |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   | R |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   | R |
    +---+---+---+---+---+---+---+---+

FIG. 2.]

It will be seen in the first diagram that every square on the board is
either occupied or attacked by a rook, and that every rook is "guarded"
(if they were alternately black and white rooks we should say
"attacked") by another rook. Placing the eight rooks on any row or file
obviously will have the same effect. In diagram 2 every square is again
either occupied or attacked, but in this case every rook is unguarded.
Now, in how many different ways can you so place the eight rooks on the
board that every square shall be occupied or attacked and no rook ever
guarded by another? I do not wish to go into the question of reversals
and reflections on this occasion, so that placing the rooks on the other
diagonal will count as different, and similarly with other repetitions
obtained by turning the board round.


296.--THE FOUR LIONS.

The puzzle is to find in how many different ways the four lions may be
placed so that there shall never be more than one lion in any row or
column. Mere reversals and reflections will not count as different.
Thus, regarding the example given, if we place the lions in the other
diagonal, it will be considered the same arrangement. For if you hold
the second arrangement in front of a mirror or give it a quarter turn,
you merely get the first arrangement. It is a simple little puzzle, but
requires a certain amount of careful consideration.

[Illustration

    +---+---+---+---+
    | L |   |   |   |
    +---+---+---+---+
    |   | L |   |   |
    +---+---+---+---+
    |   |   | L |   |
    +---+---+---+---+
    |   |   |   | L |
    +---+---+---+---+

]


297.--BISHOPS--UNGUARDED.

Place as few bishops as possible on an ordinary chessboard so that every
square of the board shall be either occupied or attacked. It will be
seen that the rook has more scope than the bishop: for wherever you
place the former, it will always attack fourteen other squares; whereas
the latter will attack seven, nine, eleven, or thirteen squares,
according to the position of the diagonal on which it is placed. And it
is well here to state that when we speak of "diagonals" in connection
with the chessboard, we do not limit ourselves to the two long diagonals
from corner to corner, but include all the shorter lines that are
parallel to these. To prevent misunderstanding on future occasions, it
will be well for the reader to note carefully this fact.


298.--BISHOPS--GUARDED.

Now, how many bishops are necessary in order that every square shall be
either occupied or attacked, and every bishop guarded by another bishop?
And how may they be placed?


299.--BISHOPS IN CONVOCATION.

[Illustration:

    +---+---+---+---+---+---+---+---+
    | B | B | B | B | B | B | B | B |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   | B | B | B | B | B | B |   |
    +---+---+---+---+---+---+---+---+

]

The greatest number of bishops that can be placed at the same time on
the chessboard, without any bishop attacking another, is fourteen. I
show, in diagram, the simplest way of doing this. In fact, on a square
chequered board of any number of squares the greatest number of bishops
that can be placed without attack is always two less than twice the
number of squares on the side. It is an interesting puzzle to discover
in just how many different ways the fourteen bishops may be so placed
without mutual attack. I shall give an exceedingly simple rule for
determining the number of ways for a square chequered board of any
number of squares.


300.--THE EIGHT QUEENS.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   |   |   ..Q |   |   |   |
    +---+---+---+...+---+---+---+---+
    |   |   ..Q..   |   |   |   |   |
    +---+...+---+---+---+---+---+---+
    | Q..   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   | Q |   |
    +---+---+---+---+---+---+---+---+
    |   | Q |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   ..Q |
    +---+---+---+---+---+---+...+---+
    |   |   |   |   |   ..Q..   |   |
    +---+---+---+---+...+---+---+---+
    |   |   |   | Q..   |   |   |   |
    +---+---+---+---+---+---+---+---+

]

The queen is by far the strongest piece on the chessboard. If you place
her on one of the four squares in the centre of the board, she attacks
no fewer than twenty-seven other squares; and if you try to hide her in
a corner, she still attacks twenty-one squares. Eight queens may be
placed on the board so that no queen attacks another, and it is an old
puzzle (first proposed by Nauck in 1850, and it has quite a little
literature of its own) to discover in just how many different ways this
may be done. I show one way in the diagram, and there are in all twelve
of these fundamentally different ways. These twelve produce ninety-two
ways if we regard reversals and reflections as different. The diagram is
in a way a symmetrical arrangement. If you turn the page upside down, it
will reproduce itself exactly; but if you look at it with one of the
other sides at the bottom, you get another way that is not identical.
Then if you reflect these two ways in a mirror you get two more ways.
Now, all the other eleven solutions are non-symmetrical, and therefore
each of them may be presented in eight ways by these reversals and
reflections. It will thus be seen why the twelve fundamentally different
solutions produce only ninety-two arrangements, as I have said, and not
ninety-six, as would happen if all twelve were non-symmetrical. It is
well to have a clear understanding on the matter of reversals and
reflections when dealing with puzzles on the chessboard.

Can the reader place the eight queens on the board so that no queen
shall attack another and so that no three queens shall be in a straight
line in any oblique direction? Another glance at the diagram will show
that this arrangement will not answer the conditions, for in the two
directions indicated by the dotted lines there are three queens in a
straight line. There is only one of the twelve fundamental ways that
will solve the puzzle. Can you find it?


301.--THE EIGHT STARS.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |///|   |   |   |   |   |   |///|
    +---+---+---+---+---+---+---+---+
    |   |///|   |   |   |   |///| * |
    +---+---+---+---+---+---+---+---+
    |   |   |///|   |   |///|   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |///|///|   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |///|///|   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |///|   |   |///|   |   |
    +---+---+---+---+---+---+---+---+
    |   |///|   |   |   |   |///|   |
    +---+---+---+---+---+---+---+---+
    |///|   |   |   |   |   |   |///|
    +---+---+---+---+---+---+---+---+

]

The puzzle in this case is to place eight stars in the diagram so that
no star shall be in line with another star horizontally, vertically, or
diagonally. One star is already placed, and that must not be moved, so
there are only seven for the reader now to place. But you must not place
a star on any one of the shaded squares. There is only one way of
solving this little puzzle.


302.--A PROBLEM IN MOSAICS.

The art of producing pictures or designs by means of joining together
pieces of hard substances, either naturally or artificially coloured, is
of very great antiquity. It was certainly known in the time of the
Pharaohs, and we find a reference in the Book of Esther to "a pavement
of red, and blue, and white, and black marble." Some of this ancient
work that has come down to us, especially some of the Roman mosaics,
would seem to show clearly, even where design is not at first evident,
that much thought was bestowed upon apparently disorderly arrangements.
Where, for example, the work has been produced with a very limited
number of colours, there are evidences of great ingenuity in preventing
the same tints coming in close proximity. Lady readers who are familiar
with the construction of patchwork quilts will know how desirable it is
sometimes, when they are limited in the choice of material, to prevent
pieces of the same stuff coming too near together. Now, this puzzle will
apply equally to patchwork quilts or tesselated pavements.

It will be seen from the diagram how a square piece of flooring may be
paved with sixty-two square tiles of the eight colours violet, red,
yellow, green, orange, purple, white, and blue (indicated by the initial
letters), so that no tile is in line with a similarly coloured tile,
vertically, horizontally, or diagonally. Sixty-four such tiles could not
possibly be placed under these conditions, but the two shaded squares
happen to be occupied by iron ventilators.

[Illustration:

    +---+---+---+---+---+---+---+---+
    | V | R | Y | G | O | P | W | B |
    +---+---+---+---+---+---+---+---+
    | W | B | O | P | Y | G | V | R |
    +---+---*===*---+---*===*---+---+
    | G | P H W H V | B H R H Y | O |
    +---+---*===*---+---*===*---+---+
    | R | Y | B | O | G | V | P | W |
    +---+---+---+---+---+---+---+---+
    | B | G | R | Y | P | W | O | V |
    +---+---+---+---+---+---+---+---+
    | O | V | P | W | R | Y | B | G |
    +---+---+---+---+---+---+---+---+
    | P | W | G | B | V | O | R | Y |
    +---+---+---+---+---+---+---+---+
    |///| O | V | R | W | B | G |///|
    +---+---+---+---+---+---+---+---+

]

The puzzle is this. These two ventilators have to be removed to the
positions indicated by the darkly bordered tiles, and two tiles placed
in those bottom corner squares. Can you readjust the thirty-two tiles so
that no two of the same colour shall still be in line?


303.--UNDER THE VEIL.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   | V | E | I | L |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   | I | L | V | E |   |   |
    +---+---+---+---+---+---+---+---+
    | I | V |   |   |   |   | L | E |
    +---+---+---+---+---+---+---+---+
    | L | E |   |   |   |   | I | V |
    +---+---+---+---+---+---+---+---+
    | V | I |   |   |   |   | E | L |
    +---+---+---+---+---+---+---+---+
    | E | L |   |   |   |   | V | I |
    +---+---+---+---+---+---+---+---+
    |   |   | E | V | L | I |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   | L | I | E | V |   |   |
    +---+---+---+---+---+---+---+---+

]

If the reader will examine the above diagram, he will see that I have so
placed eight V's, eight E's, eight I's, and eight L's in the diagram
that no letter is in line with a similar one horizontally, vertically,
or diagonally. Thus, no V is in line with another V, no E with another
E, and so on. There are a great many different ways of arranging the
letters under this condition. The puzzle is to find an arrangement that
produces the greatest possible number of four-letter words, reading
upwards and downwards, backwards and forwards, or diagonally. All
repetitions count as different words, and the five variations that may
be used are: VEIL, VILE, LEVI, LIVE, and EVIL.

This will be made perfectly clear when I say that the above arrangement
scores eight, because the top and bottom row both give VEIL; the second
and seventh columns both give VEIL; and the two diagonals, starting from
the L in the 5th row and E in the 8th row, both give LIVE and EVIL.
There are therefore eight different readings of the words in all.

This difficult word puzzle is given as an example of the use of
chessboard analysis in solving such things. Only a person who is
familiar with the "Eight Queens" problem could hope to solve it.


304.--BACHET'S SQUARE.

One of the oldest card puzzles is by Claude Caspar Bachet de Méziriac,
first published, I believe, in the 1624 edition of his work. Rearrange
the sixteen court cards (including the aces) in a square so that in no
row of four cards, horizontal, vertical, or diagonal, shall be found two
cards of the same suit or the same value. This in itself is easy enough,
but a point of the puzzle is to find in how many different ways this may
be done. The eminent French mathematician A. Labosne, in his modern
edition of Bachet, gives the answer incorrectly. And yet the puzzle is
really quite easy. Any arrangement produces seven more by turning the
square round and reflecting it in a mirror. These are counted as
different by Bachet.

Note "row of four cards," so that the only diagonals we have here to
consider are the two long ones.


305.--THE THIRTY-SIX LETTER-BLOCKS.

[Illustration]

The illustration represents a box containing thirty-six letter-blocks.
The puzzle is to rearrange these blocks so that no A shall be in a line
vertically, horizontally, or diagonally with another A, no B with
another B, no C with another C, and so on. You will find it impossible
to get all the letters into the box under these conditions, but the
point is to place as many as possible. Of course no letters other than
those shown may be used.


306.--THE CROWDED CHESSBOARD.

[Illustration]

The puzzle is to rearrange the fifty-one pieces on the chessboard so
that no queen shall attack another queen, no rook attack another rook,
no bishop attack another bishop, and no knight attack another knight. No
notice is to be taken of the intervention of pieces of another type from
that under consideration--that is, two queens will be considered to
attack one another although there may be, say, a rook, a bishop, and a
knight between them. And so with the rooks and bishops. It is not
difficult to dispose of each type of piece separately; the difficulty
comes in when you have to find room for all the arrangements on the
board simultaneously.


307.--THE COLOURED COUNTERS.

[Illustration]

The diagram represents twenty-five coloured counters, Red, Blue, Yellow,
Orange, and Green (indicated by their initials), and there are five of
each colour, numbered 1, 2, 3, 4, and 5. The problem is so to place them
in a square that neither colour nor number shall be found repeated in
any one of the five rows, five columns, and two diagonals. Can you so
rearrange them?


308.--THE GENTLE ART OF STAMP-LICKING.

The Insurance Act is a most prolific source of entertaining puzzles,
particularly entertaining if you happen to be among the exempt. One's
initiation into the gentle art of stamp-licking suggests the following
little poser: If you have a card divided into sixteen spaces (4 × 4),
and are provided with plenty of stamps of the values 1d., 2d., 3d., 4d.,
and 5d., what is the greatest value that you can stick on the card if
the Chancellor of the Exchequer forbids you to place any stamp in a
straight line (that is, horizontally, vertically, or diagonally) with
another stamp of similar value? Of course, only one stamp can be affixed
in a space. The reader will probably find, when he sees the solution,
that, like the stamps themselves, he is licked He will most likely be
twopence short of the maximum. A friend asked the Post Office how it was
to be done; but they sent him to the Customs and Excise officer, who
sent him to the Insurance Commissioners, who sent him to an approved
society, who profanely sent him--but no matter.


309.--THE FORTY-NINE COUNTERS.

[Illustration]

Can you rearrange the above forty-nine counters in a square so that no
letter, and also no number, shall be in line with a similar one,
vertically, horizontally, or diagonally? Here I, of course, mean in the
lines parallel with the diagonals, in the chessboard sense.


310.--THE THREE SHEEP.

[Illustration]

A farmer had three sheep and an arrangement of sixteen pens, divided off
by hurdles in the manner indicated in the illustration. In how many
different ways could he place those sheep, each in a separate pen, so
that every pen should be either occupied or in line (horizontally,
vertically, or diagonally) with at least one sheep? I have given one
arrangement that fulfils the conditions. How many others can you find?
Mere reversals and reflections must not be counted as different. The
reader may regard the sheep as queens. The problem is then to place the
three queens so that every square shall be either occupied or attacked
by at least one queen--in the maximum number of different ways.


311.--THE FIVE DOGS PUZZLE.

In 1863, C.F. de Jaenisch first discussed the "Five Queens Puzzle"--to
place five queens on the chessboard so that every square shall be
attacked or occupied--which was propounded by his friend, a "Mr. de R."
Jaenisch showed that if no queen may attack another there are ninety-one
different ways of placing the five queens, reversals and reflections not
counting as different. If the queens may attack one another, I have
recorded hundreds of ways, but it is not practicable to enumerate them
exactly.

[Illustration]

The illustration is supposed to represent an arrangement of sixty-four
kennels. It will be seen that five kennels each contain a dog, and on
further examination it will be seen that every one of the sixty-four
kennels is in a straight line with at least one dog--either
horizontally, vertically, or diagonally. Take any kennel you like, and
you will find that you can draw a straight line to a dog in one or other
of the three ways mentioned. The puzzle is to replace the five dogs and
discover in just how many different ways they may be placed in five
kennels _in a straight row_, so that every kennel shall always be in
line with at least one dog. Reversals and reflections are here counted
as different.


312.--THE FIVE CRESCENTS OF BYZANTIUM.

When Philip of Macedon, the father of Alexander the Great, found himself
confronted with great difficulties in the siege of Byzantium, he set his
men to undermine the walls. His desires, however, miscarried, for no
sooner had the operations been begun than a crescent moon suddenly
appeared in the heavens and discovered his plans to his adversaries. The
Byzantines were naturally elated, and in order to show their gratitude
they erected a statue to Diana, and the crescent became thenceforward a
symbol of the state. In the temple that contained the statue was a
square pavement composed of sixty-four large and costly tiles. These
were all plain, with the exception of five, which bore the symbol of the
crescent. These five were for occult reasons so placed that every tile
should be watched over by (that is, in a straight line, vertically,
horizontally, or diagonally with) at least one of the crescents. The
arrangement adopted by the Byzantine architect was as follows:--

[Illustration]

Now, to cover up one of these five crescents was a capital offence, the
death being something very painful and lingering. But on a certain
occasion of festivity it was necessary to lay down on this pavement a
square carpet of the largest dimensions possible, and I have shown in
the illustration by dark shading the largest dimensions that would be
available.

The puzzle is to show how the architect, if he had foreseen this
question of the carpet, might have so arranged his five crescent tiles
in accordance with the required conditions, and yet have allowed for the
largest possible square carpet to be laid down without any one of the
five crescent tiles being covered, or any portion of them.


313.--QUEENS AND BISHOP PUZZLE.

It will be seen that every square of the board is either occupied or
attacked. The puzzle is to substitute a bishop for the rook on the same
square, and then place the four queens on other squares so that every
square shall again be either occupied or attacked.

[Illustration]


314.--THE SOUTHERN CROSS.

[Illustration]

In the above illustration we have five Planets and eighty-one Fixed
Stars, five of the latter being hidden by the Planets. It will be found
that every Star, with the exception of the ten that have a black spot in
their centres, is in a straight line, vertically, horizontally, or
diagonally, with at least one of the Planets. The puzzle is so to
rearrange the Planets that all the Stars shall be in line with one or
more of them.

In rearranging the Planets, each of the five may be moved once in a
straight line, in either of the three directions mentioned. They will,
of course, obscure five other Stars in place of those at present
covered.


315.--THE HAT-PEG PUZZLE.

Here is a five-queen puzzle that I gave in a fanciful dress in 1897. As
the queens were there represented as hats on sixty-four pegs, I will
keep to the title, "The Hat-Peg Puzzle." It will be seen that every
square is occupied or attacked. The puzzle is to remove one queen to a
different square so that still every square is occupied or attacked,
then move a second queen under a similar condition, then a third queen,
and finally a fourth queen. After the fourth move every square must be
attacked or occupied, but no queen must then attack another. Of course,
the moves need not be "queen moves;" you can move a queen to any part of
the board.

[Illustration]


316.--THE AMAZONS.

[Illustration]

This puzzle is based on one by Captain Turton. Remove three of the
queens to other squares so that there shall be eleven squares on the
board that are not attacked. The removal of the three queens need not be
by "queen moves." You may take them up and place them anywhere. There is
only one solution.


317.--A PUZZLE WITH PAWNS.

Place two pawns in the middle of the chessboard, one at Q 4 and the
other at K 5. Now, place the remaining fourteen pawns (sixteen in all)
so that no three shall be in a straight line in any possible direction.

Note that I purposely do not say queens, because by the words "any
possible direction" I go beyond attacks on diagonals. The pawns must be
regarded as mere points in space--at the centres of the squares. See
dotted lines in the case of No. 300, "The Eight Queens."


318.--LION-HUNTING.

[Illustration]

My friend Captain Potham Hall, the renowned hunter of big game, says
there is nothing more exhilarating than a brush with a herd--a pack--a
team--a flock--a swarm (it has taken me a full quarter of an hour to
recall the right word, but I have it at last)--a _pride_ of lions. Why a
number of lions are called a "pride," a number of whales a "school," and
a number of foxes a "skulk" are mysteries of philology into which I will
not enter.

Well, the captain says that if a spirited lion crosses your path in the
desert it becomes lively, for the lion has generally been looking for
the man just as much as the man has sought the king of the forest. And
yet when they meet they always quarrel and fight it out. A little
contemplation of this unfortunate and long-standing feud between two
estimable families has led me to figure out a few calculations as to the
probability of the man and the lion crossing one another's path in the
jungle. In all these cases one has to start on certain more or less
arbitrary assumptions. That is why in the above illustration I have
thought it necessary to represent the paths in the desert with such
rigid regularity. Though the captain assures me that the tracks of the
lions usually run much in this way, I have doubts.

The puzzle is simply to find out in how many different ways the man and
the lion may be placed on two different spots that are not on the same
path. By "paths" it must be understood that I only refer to the ruled
lines. Thus, with the exception of the four corner spots, each combatant
is always on two paths and no more. It will be seen that there is a lot
of scope for evading one another in the desert, which is just what one
has always understood.


319.--THE KNIGHT-GUARDS.

[Illustration]

The knight is the irresponsible low comedian of the chessboard. "He is a
very uncertain, sneaking, and demoralizing rascal," says an American
writer. "He can only move two squares, but makes up in the quality of
his locomotion for its quantity, for he can spring one square sideways
and one forward simultaneously, like a cat; can stand on one leg in the
middle of the board and jump to any one of eight squares he chooses; can
get on one side of a fence and blackguard three or four men on the
other; has an objectionable way of inserting himself in safe places
where he can scare the king and compel him to move, and then gobble a
queen. For pure cussedness the knight has no equal, and when you chase
him out of one hole he skips into another." Attempts have been made over
and over again to obtain a short, simple, and exact definition of the
move of the knight--without success. It really consists in moving one
square like a rook, and then another square like a bishop--the two
operations being done in one leap, so that it does not matter whether
the first square passed over is occupied by another piece or not. It is,
in fact, the only leaping move in chess. But difficult as it is to
define, a child can learn it by inspection in a few minutes.

I have shown in the diagram how twelve knights (the fewest possible that
will perform the feat) may be placed on the chessboard so that every
square is either occupied or attacked by a knight. Examine every square
in turn, and you will find that this is so. Now, the puzzle in this case
is to discover what is the smallest possible number of knights that is
required in order that every square shall be either occupied or
attacked, and every knight protected by another knight. And how would
you arrange them? It will be found that of the twelve shown in the
diagram only four are thus protected by being a knight's move from
another knight.


THE GUARDED CHESSBOARD.

On an ordinary chessboard, 8 by 8, every square can be guarded--that is,
either occupied or attacked--by 5 queens, the fewest possible. There are
exactly 91 fundamentally different arrangements in which no queen
attacks another queen. If every queen must attack (or be protected by)
another queen, there are at fewest 41 arrangements, and I have recorded
some 150 ways in which some of the queens are attacked and some not, but
this last case is very difficult to enumerate exactly.

On an ordinary chessboard every square can be guarded by 8 rooks (the
fewest possible) in 40,320 ways, if no rook may attack another rook, but
it is not known how many of these are fundamentally different. (See
solution to No. 295, "The Eight Rooks.") I have not enumerated the ways
in which every rook shall be protected by another rook.

On an ordinary chessboard every square can be guarded by 8 bishops (the
fewest possible), if no bishop may attack another bishop. Ten bishops
are necessary if every bishop is to be protected. (See Nos. 297 and 298,
"Bishops unguarded" and "Bishops guarded.")

On an ordinary chessboard every square can be guarded by 12 knights if
all but 4 are unprotected. But if every knight must be protected, 14 are
necessary. (See No. 319, "The Knight-Guards.")

Dealing with the queen on n² boards generally, where n is less
than 8, the following results will be of interest:--


1 queen guards 2² board in 1 fundamental way.

1 queen guards 3² board in 1 fundamental way.

2 queens guard 4² board in 3 fundamental ways (protected).

3 queens guard 4² board in 2 fundamental ways (not protected).

3 queens guard 5² board in 37 fundamental ways (protected).

3 queens guard 5² board in 2 fundamental ways (not protected).

3 queens guard 6² board in 1 fundamental way (protected).

4 queens guard 6² board in 17 fundamental ways (not protected).

4 queens guard 7² board in 5 fundamental ways (protected).

4 queens guard 7² board in 1 fundamental way (not protected).


NON-ATTACKING CHESSBOARD ARRANGEMENTS.

We know that n queens may always be placed on a square board of n²
squares (if n be greater than 3) without any queen attacking another
queen. But no general formula for enumerating the number of different
ways in which it may be done has yet been discovered; probably it is
undiscoverable. The known results are as follows:--

Where n = 4 there is 1 fundamental solution and 2 in all.

Where n = 5 there are 2 fundamental solutions and 10 in all.

Where n = 6 there is 1 fundamental solution and 4 in all.

Where n = 7 there are 6 fundamental solutions and 40 in all.

Where n = 8 there are 12 fundamental solutions and 92 in all.

Where n = 9 there are 46 fundamental solutions.

Where n = 10 there are 92 fundamental solutions.

Where n = 11 there are 341 fundamental solutions.

Obviously n rooks may be placed without attack on an n² board in n!
ways, but how many of these are fundamentally different I have only
worked out in the four cases where n equals 2, 3, 4, and 5. The answers
here are respectively 1, 2, 7, and 23. (See No. 296, "The Four Lions.")

We can place 2n-2 bishops on an n² board in 2^{n} ways. (See No. 299,
"Bishops in Convocation.") For boards containing 2, 3, 4, 5, 6, 7, 8
squares, on a side there are respectively 1, 2, 3, 6, 10, 20, 36
fundamentally different arrangements. Where n is odd there are
2^{½(n-1)} such arrangements, each giving 4 by reversals and
reflections, and 2^{n-3} - 2^{½(n-3)} giving 8. Where n is even there
are 2^{½(n-2)}, each giving 4 by reversals and reflections, and 2^{n-3}
- 2^{½(n-4)}, each giving 8.

We can place ½(n²+1) knights on an n² board without attack, when n
is odd, in 1 fundamental way; and ½n² knights on an n² board, when
n is even, in 1 fundamental way. In the first case we place all the
knights on the same colour as the central square; in the second case we
place them all on black, or all on white, squares.


THE TWO PIECES PROBLEM.

On a board of n² squares, two queens, two rooks, two bishops, or two
knights can always be placed, irrespective of attack or not, in ½(n^{4}
- n²) ways. The following formulæ will show in how many of these ways
the two pieces may be placed with attack and without:--

               With Attack.          Without Attack.

    2 Queens      5n³ - 6n² + n         3n^{4} - 10n³ + 9n² - 2n
               -------------------    ------------------------------
                        3                           6

    2 Rooks          n³ - n²                n^{4} - 2n³ + n²
                                          ----------------------
                                                    2

    2 Bishops   4n³ - 6n² + 2n       3n^{4} - 4n³ + 3n² - 2n
               --------------------   -----------------------------
                        6                           6

    2 Knights    4n² - 12n + 8            n^{4} - 9n² + 24n
                                          --------------------
                                                    2

(See No. 318, " Lion Hunting.")



DYNAMICAL CHESS PUZZLES.

    "Push on--keep moving."
    THOS. MORTON: _Cure for the Heartache_.


320.--THE ROOK'S TOUR.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   | R |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+

]


The puzzle is to move the single rook over the whole board, so that it
shall visit every square of the board once, and only once, and end its
tour on the square from which it starts. You have to do this in as few
moves as possible, and unless you are very careful you will take just
one move too many. Of course, a square is regarded equally as "visited"
whether you merely pass over it or make it a stopping-place, and we will
not quibble over the point whether the original square is actually
visited twice. We will assume that it is not.


321.--THE ROOK'S JOURNEY.

This puzzle I call "The Rook's Journey," because the word "tour"
(derived from a turner's wheel) implies that we return to the point from
which we set out, and we do not do this in the present case. We should
not be satisfied with a personally conducted holiday tour that ended by
leaving us, say, in the middle of the Sahara. The rook here makes
twenty-one moves, in the course of which journey it visits every square
of the board once and only once, stopping at the square marked 10 at the
end of its tenth move, and ending at the square marked 21. Two
consecutive moves cannot be made in the same direction--that is to say,
you must make a turn after every move.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   | R |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+
    |   | 21|   | 10|   |   |   |   |
    +---+---+---+---+---+---+---+---+

]


322.--THE LANGUISHING MAIDEN.

[Illustration:

        --+-----+-----+-----+-----+-----+-----+-----+
          |     |     |     |     |     |     |     |
    |  Kt                                           |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                      M        |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A wicked baron in the good old days imprisoned an innocent maiden in one
of the deepest dungeons beneath the castle moat. It will be seen from
our illustration that there were sixty-three cells in the dungeon, all
connected by open doors, and the maiden was chained in the cell in which
she is shown. Now, a valiant knight, who loved the damsel, succeeded in
rescuing her from the enemy. Having gained an entrance to the dungeon at
the point where he is seen, he succeeded in reaching the maiden after
entering every cell once and only once. Take your pencil and try to
trace out such a route. When you have succeeded, then try to discover a
route in twenty-two straight paths through the cells. It can be done in
this number without entering any cell a second time.


323.--A DUNGEON PUZZLE.

[Illustration:

    +-----+-----+-----+-----+-----+-----+-----+-----+
    |     |     |     |     |     |     |     |     |
    |  .............     .......     .............  |
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    |  .......     .......     .......     .......  |
    |     |  .  |     |     |     |     |  .  |     |
    +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+
    |     |  .  |     |     |     |     |  .  |     |
    |  .......     .......     .......     .......  |
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    |  .............     .......     .     .......  |
    |     |     |     |     |     |  .  |  .  |     |
    +-- --+-- --+-- --+-- --+-- --+--.--+--.--+-- --+
    |     |     |     |     |     |  .  |  .  |     |
    |  .............     .......     .     .......  |
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    |  .......     .......     .......     .......  |
    |     |  .  |     |     |     |     |  .  |     |
    +-- --+--.--+-- --+-- --+-- --+-- --+--.--+-- --+
    |     |  .  |     |     |     |     |  .  |     |
    |  .......     .......     .......     .......  |
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    +--.--+-- --+--.--+--.--+--.--+--.--+-- --+--.--+
    |  .  |     |  .  |  .  |  .  |  .  |     |  .  |
    |  .............     .     P     .............  |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

A French prisoner, for his sins (or other people's), was confined in an
underground dungeon containing sixty-four cells, all communicating with
open doorways, as shown in our illustration. In order to reduce the
tedium of his restricted life, he set himself various puzzles, and this
is one of them. Starting from the cell in which he is shown, how could
he visit every cell once, and only once, and make as many turnings as
possible? His first attempt is shown by the dotted track. It will be
found that there are as many as fifty-five straight lines in his path,
but after many attempts he improved upon this. Can you get more than
fifty-five? You may end your path in any cell you like. Try the puzzle
with a pencil on chessboard diagrams, or you may regard them as rooks'
moves on a board.


324.--THE LION AND THE MAN.

In a public place in Rome there once stood a prison divided into
sixty-four cells, all open to the sky and all communicating with one
another, as shown in the illustration. The sports that here took place
were watched from a high tower. The favourite game was to place a
Christian in one corner cell and a lion in the diagonally opposite
corner and then leave them with all the inner doors open. The consequent
effect was sometimes most laughable. On one occasion the man was given a
sword. He was no coward, and was as anxious to find the lion as the lion
undoubtedly was to find him.

[Illustration:

    +-----+-----+-----+-----+-----+-----+-----+-----+
    |     |     |     |     |     |     |     |     |
    |                                            L  |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |                                               |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+
    |     |     |     |     |     |     |     |     |
    |  C                                            |
    |     |     |     |     |     |     |     |     |
    +-- --+-- --+-- --+-- --+-- --+-- --+-- --+-- --+

]

The man visited every cell once and only once in the fewest possible
straight lines until he reached the lion's cell. The lion, curiously
enough, also visited every cell once and only once in the fewest
possible straight lines until he finally reached the man's cell. They
started together and went at the same speed; yet, although they
occasionally got glimpses of one another, they never once met. The
puzzle is to show the route that each happened to take.


325.--AN EPISCOPAL VISITATION.

The white squares on the chessboard represent the parishes of a diocese.
Place the bishop on any square you like, and so contrive that (using the
ordinary bishop's move of chess) he shall visit every one of his
parishes in the fewest possible moves. Of course, all the parishes
passed through on any move are regarded as "visited." You can visit any
squares more than once, but you are not allowed to move twice between
the same two adjoining squares. What are the fewest possible moves? The
bishop need not end his visitation at the parish from which he first set
out.


326.--A NEW COUNTER PUZZLE.

Here is a new puzzle with moving counters, or coins, that at first
glance looks as if it must be absurdly simple. But it will be found
quite a little perplexity. I give it in this place for a reason that I
will explain when we come to the next puzzle. Copy the simple diagram,
enlarged, on a sheet of paper; then place two white counters on the
points 1 and 2, and two red counters on 9 and 10, The puzzle is to make
the red and white change places. You may move the counters one at a time
in any order you like, along the lines from point to point, with the
only restriction that a red and a white counter may never stand at once
on the same straight line. Thus the first move can only be from 1 or 2
to 3, or from 9 or 10 to 7.

[Illustration:

      4   8
     / \ / \
    2   6   10
     \ / \ /
      3   7
     / \ / \
    1   5   9

]


327.--A NEW BISHOP'S PUZZLE.

[Illustration:

    +---+---+---+---+
    | b | b | b | b |
    +---+---+---+---+
    |   |   |   |   |
    +---+---+---+---+
    |   |   |   |   |
    +---+---+---+---+
    | B | B | B | B |
    +---+---+---+---+

]

This is quite a fascinating little puzzle. Place eight bishops (four
black and four white) on the reduced chessboard, as shown in the
illustration. The problem is to make the black bishops change places
with the white ones, no bishop ever attacking another of the opposite
colour. They must move alternately--first a white, then a black, then a
white, and so on. When you have succeeded in doing it at all, try to
find the fewest possible moves.

If you leave out the bishops standing on black squares, and only play on
the white squares, you will discover my last puzzle turned on its side.


328.--THE QUEEN'S TOUR.

The puzzle of making a complete tour of the chessboard with the queen in
the fewest possible moves (in which squares may be visited more than
once) was first given by the late Sam Loyd in his _Chess Strategy_. But
the solution shown below is the one he gave in _American Chess-Nuts_ in
1868. I have recorded at least six different solutions in the minimum
number of moves--fourteen--but this one is the best of all, for reasons
I will explain.

[Illustration:

    +---+---+---+---+---+---+---+---+
    |   |   |   |   |   |   |   |   |
    | ............................. |
    | . |   |   |   |   |   |   | . |
    +-.-+---+---+---+---+---+---+-.-+
    | . |   |   |   |   |   |   | . |
    | . | ..........................|
    | . |  .|   |   |   |   |   | . |
    +-.-+---.---+---+---+---+---+..-+
    | . |   |.  |   |   |   |   . . |
    | . | ................. |  .| . |
    | . |  .|  .|   |   |.  | . | . |
    +-.-+---.---.---+---.---+.--+-.-+
    | . |   |.  |.  |  .|   .   | . |
    | . | . | . | . | . |  .| . | . |
    | . | ..|  .|  .|.  | . |.. | . |
    +-.-+-.-.---.---.---+.--.-.-+-.-+
    | . | . |.  |. .|.  .  .| . | . |
    | . | . | . | . | ..| . | . | . |
    | . | . |  .|. .| ..|.  | . | . |
    +-.-+-.-+---.---..--.---+-.-+-.-+
    | . | . |  .|.  .. .|.  | . | . |
    | . | . | . | ..| . | . | . | . |
    | . | . |.  | ..|. .|  .| . | . |
    +-.-+-.-.---+.--.---.---.-.-+-.-+
    | . | ..|   .  .|.  |.  |.. | . |
    | . | . |  .| . | . | . | . | . |
    | . |.. | . |.  |  .|  .| ..| . |
    +-.-.-.-+.--.---+---.---.-.-.-.-+
    | ..| . .  .|   |   |.  |.. |.. |
    | . | ..| ............. | . | . |
    |   | . |   |   |   |   |   |   |
    +---+---+---+---+---+---+---+---+

]

If you will look at the lettered square you will understand that there
are only ten really differently placed squares on a chessboard--those
enclosed by a dark line--all the others are mere reversals or
reflections. For example, every A is a corner square, and every J a
central square. Consequently, as the solution shown has a turning-point
at the enclosed D square, we can obtain a solution starting from and
ending at any square marked D--by just turning the board about. Now,
this scheme will give you a tour starting from any A, B, C, D, E, F, or
H, while no other route that I know can be adapted to more than five
different starting-points. There is no Queen's Tour in fourteen moves
(remember a tour must be re-entrant) that may start from a G, I, or J.
But we can have a non-re-entrant path over the whole board in fourteen
moves, starting from any given square. Hence the following puzzle:--

[Illustration:

    +---+---+---+---*---+---+---+---+
    | A | B | C | G " G | C | B | A |
    *===*---+---+---*---+---+---+---+
    | B " D | E | H " H | E | D | B |
    +---*===*---+---*---+---+---+---+
    | C | E " F | I " I | F | E | C |
    +---+---*===*---*---+---+---+---+
    | G | H | I " J " J | I | H | G |
    +---+---+---*===*---+---+---+---+
    | G | H | I | J | J | I | H | G |
    +---+---+---+---+---+---+---+---+
    | C | E | F | I | I | F | E | C |
    +---+---+---+---+---+---+---+---+
    | B | D | E | H | H | E | D | B |
    +---+---+---+---+---+---+---+---+
    | A | B | C | G | G | C | B | A |
    +---+---+---+---+---+---+---+---+

]

Start from the J in the enclosed part of the lettered diagram and visit
every square of the board in fourteen moves, ending wherever you like.


329.--THE STAR PUZZLE.

[Illustration:

    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | ¤ | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | ¤ | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+
    | * | * | * | * | * | * | * | * |
    +---+---+---+---+---+---+---+---+

]

Put the point of your pencil on one of the white stars and (without ever
lifting your pencil from the paper) strike out all the stars in fourteen
continuous straight strokes, ending at the second white star. Your
straight strokes may be in any direction you like, only every turning
must be made on a star. There is no objection to striking out any star
more than once.

In this case, where both your starting and ending squares are fixed
inconveniently, you cannot obtain a solution by breaking a Queen's Tour,
or in any other way by queen moves alone. But you are allowed to use
oblique straight lines--such as from the upper white star direct to a
corner star.


330.--THE YACHT RACE.

Now then, ye land-lubbers, hoist your baby-jib-topsails, break out your
spinnakers, ease off your balloon sheets, and get your head-sails set!

Our race consists in starting from the point at which the yacht is lying
in the illustration and touching every one of the sixty-four buoys in
fourteen straight courses, returning in the final tack to the buoy from
which we start. The seventh course must finish at the buoy from which a
flag is flying.

This puzzle will call for a lot of skilful seamanship on account of the
sharp angles at which it will occasionally be necessary to tack. The
point of a lead pencil and a good nautical eye are all the outfit that
we require.

[Illustration]

This is difficult, because of the condition as to the flag-buoy, and
because it is a re-entrant tour. But again we are allowed those oblique
lines.


331.--THE SCIENTIFIC SKATER.

[Illustration]

It will be seen that this skater has marked on the ice sixty-four points
or stars, and he proposes to start _from his present position_ near the
corner and enter every one of the points in fourteen straight lines. How
will he do it? Of course there is no objection to his passing over any
point more than once, but his last straight stroke must bring him back
to the position from which he started.

It is merely a matter of taking your pencil and starting from the spot
on which the skater's foot is at present resting, and striking out all
the stars in fourteen continuous straight lines, returning to the point
from which you set out.


332.--THE FORTY-NINE STARS.

[Illustration]

The puzzle in this case is simply to take your pencil and, starting from
one black star, strike out all the stars in twelve straight strokes,
ending at the other black star. It will be seen that the attempt shown
in the illustration requires fifteen strokes. Can you do it in twelve?
Every turning must be made on a star, and the lines must be parallel to
the sides and diagonals of the square, as shown. In this case we are
dealing with a chessboard of reduced dimensions, but only queen moves
(without going outside the boundary as in the last case) are required.


333.--THE QUEEN'S JOURNEY.

[Illustration]

Place the queen on her own square, as shown in the illustration, and
then try to discover the greatest distance that she can travel over the
board in five queen's moves without passing over any square a second
time. Mark the queen's path on the board, and note carefully also that
she must never cross her own track. It seems simple enough, but the
reader may find that he has tripped.


334.--ST. GEORGE AND THE DRAGON.

[Illustration]

Here is a little puzzle on a reduced chessboard of forty-nine squares.
St. George wishes to kill the dragon. Killing dragons was a well-known
pastime of his, and, being a knight, it was only natural that he should
desire to perform the feat in a series of knight's moves. Can you show
how, starting from that central square, he may visit once, and only
once, every square of the board in a chain of chess knight's moves, and
end by capturing the dragon on his last move? Of course a variety of
different ways are open to him, so try to discover a route that forms
some pretty design when you have marked each successive leap by a
straight line from square to square.


335.--FARMER LAWRENCE'S CORNFIELDS.

One of the most beautiful districts within easy distance of London for a
summer ramble is that part of Buckinghamshire known as the Valley of the
Chess--at least, it was a few years ago, before it was discovered by the
speculative builder. At the beginning of the present century there
lived, not far from Latimers, a worthy but eccentric farmer named
Lawrence. One of his queer notions was that every person who lived near
the banks of the river Chess ought to be in some way acquainted with the
noble game of the same name, and in order to impress this fact on his
men and his neighbours he adopted at times strange terminology. For
example, when one of his ewes presented him with a lamb, he would say
that it had "queened a pawn"; when he put up a new barn against the
highway, he called it "castling on the king's side"; and when he sent a
man with a gun to keep his neighbour's birds off his fields, he spoke of
it as "attacking his opponent's rooks." Everybody in the neighbourhood
used to be amused at Farmer Lawrence's little jokes, and one boy (the
wag of the village) who got his ears pulled by the old gentleman for
stealing his "chestnuts" went so far as to call him "a silly old
chess-protector!"

One year he had a large square field divided into forty-nine square
plots, as shown in the illustration. The white squares were sown with
wheat and the black squares with barley. When the harvest time came
round he gave orders that his men were first to cut the corn in the
patch marked 1, and that each successive cutting should be exactly a
knight's move from the last one, the thirteenth cutting being in the
patch marked 13, the twenty-fifth in the patch marked 25, the
thirty-seventh in the one marked 37, and the last, or forty-ninth
cutting, in the patch marked 49. This was too much for poor Hodge, and
each day Farmer Lawrence had to go down to the field and show which
piece had to be operated upon. But the problem will perhaps present no
difficulty to my readers.

[Illustration]


336.--THE GREYHOUND PUZZLE.

In this puzzle the twenty kennels do not communicate with one another by
doors, but are divided off by a low wall. The solitary occupant is the
greyhound which lives in the kennel in the top left-hand corner. When he
is allowed his liberty he has to obtain it by visiting every kennel once
and only once in a series of knight's moves, ending at the bottom
right-hand corner, which is open to the world. The lines in the above
diagram show one solution. The puzzle is to discover in how many
different ways the greyhound may thus make his exit from his corner
kennel.

[Illustration]


337.--THE FOUR KANGAROOS.

[Illustration]

In introducing a little Commonwealth problem, I must first explain that
the diagram represents the sixty-four fields, all properly fenced off
from one another, of an Australian settlement, though I need hardly say
that our kith and kin "down under" always _do_ set out their land in
this methodical and exact manner. It will be seen that in every one of
the four corners is a kangaroo. Why kangaroos have a marked preference
for corner plots has never been satisfactorily explained, and it would
be out of place to discuss the point here. I should also add that
kangaroos, as is well known, always leap in what we call "knight's
moves." In fact, chess players would probably have adopted the better
term "kangaroo's move" had not chess been invented before kangaroos.

The puzzle is simply this. One morning each kangaroo went for his
morning hop, and in sixteen consecutive knight's leaps visited just
fifteen different fields and jumped back to his corner. No field was
visited by more than one of the kangaroos. The diagram shows how they
arranged matters. What you are asked to do is to show how they might
have performed the feat without any kangaroo ever crossing the
horizontal line in the middle of the square that divides the board into
two equal parts.


338.--THE BOARD IN COMPARTMENTS.

[Illustration]

We cannot divide the ordinary chessboard into four equal square
compartments, and describe a complete tour, or even path, in each
compartment. But we may divide it into four compartments, as in the
illustration, two containing each twenty squares, and the other two each
twelve squares, and so obtain an interesting puzzle. You are asked to
describe a complete re-entrant tour on this board, starting where you
like, but visiting every square in each successive compartment before
passing into another one, and making the final leap back to the square
from which the knight set out. It is not difficult, but will be found
very entertaining and not uninstructive.

Whether a re-entrant "tour" or a complete knight's "path" is possible or
not on a rectangular board of given dimensions depends not only on its
dimensions, but also on its shape. A tour is obviously not possible on a
board containing an odd number of cells, such as 5 by 5 or 7 by 7, for
this reason: Every successive leap of the knight must be from a white
square to a black and a black to a white alternately. But if there be an
odd number of cells or squares there must be one more square of one
colour than of the other, therefore the path must begin from a square of
the colour that is in excess, and end on a similar colour, and as a
knight's move from one colour to a similar colour is impossible the
path cannot be re-entrant. But a perfect tour may be made on a
rectangular board of any dimensions provided the number of squares be
even, and that the number of squares on one side be not less than 6 and
on the other not less than 5. In other words, the smallest rectangular
board on which a re-entrant tour is possible is one that is 6 by 5.

A complete knight's path (not re-entrant) over all the squares of a
board is never possible if there be only two squares on one side; nor is
it possible on a square board of smaller dimensions than 5 by 5. So that
on a board 4 by 4 we can neither describe a knight's tour nor a complete
knight's path; we must leave one square unvisited. Yet on a board 4 by 3
(containing four squares fewer) a complete path may be described in
sixteen different ways. It may interest the reader to discover all
these. Every path that starts from and ends at different squares is here
counted as a different solution, and even reverse routes are called
different.


339.--THE FOUR KNIGHTS' TOURS.

[Illustration]

I will repeat that if a chessboard be cut into four equal parts, as
indicated by the dark lines in the illustration, it is not possible to
perform a knight's tour, either re-entrant or not, on one of the parts.
The best re-entrant attempt is shown, in which each knight has to
trespass twice on other parts. The puzzle is to cut the board
differently into four parts, each of the same size and shape, so that a
re-entrant knight's tour may be made on each part. Cuts along the dotted
lines will not do, as the four central squares of the board would be
either detached or hanging on by a mere thread.


340.--THE CUBIC KNIGHT'S TOUR.

Some few years ago I happened to read somewhere that Abnit Vandermonde,
a clever mathematician, who was born in 1736 and died in 1793, had
devoted a good deal of study to the question of knight's tours. Beyond
what may be gathered from a few fragmentary references, I am not aware
of the exact nature or results of his investigations, but one thing
attracted my attention, and that was the statement that he had proposed
the question of a tour of the knight over the six surfaces of a cube,
each surface being a chessboard. Whether he obtained a solution or not I
do not know, but I have never seen one published. So I at once set to
work to master this interesting problem. Perhaps the reader may like to
attempt it.


341.--THE FOUR FROGS.

[Illustration]

In the illustration we have eight toadstools, with white frogs on 1 and
3 and black frogs on 6 and 8. The puzzle is to move one frog at a time,
in any order, along one of the straight lines from toadstool to
toadstool, until they have exchanged places, the white frogs being left
on 6 and 8 and the black ones on 1 and 3. If you use four counters on a
simple diagram, you will find this quite easy, but it is a little more
puzzling to do it in only seven plays, any number of successive moves by
one frog counting as one play. Of course, more than one frog cannot be
on a toadstool at the same time.


342.--THE MANDARIN'S PUZZLE.

The following puzzle has an added interest from the circumstance that a
correct solution of it secured for a certain young Chinaman the hand of
his charming bride. The wealthiest mandarin within a radius of a hundred
miles of Peking was Hi-Chum-Chop, and his beautiful daughter, Peeky-Bo,
had innumerable admirers. One of her most ardent lovers was Winky-Hi,
and when he asked the old mandarin for his consent to their marriage,
Hi-Chum-Chop presented him with the following puzzle and promised his
consent if the youth brought him the correct answer within a week.
Winky-Hi, following a habit which obtains among certain solvers to this
day, gave it to all his friends, and when he had compared their
solutions he handed in the best one as his own. Luckily it was quite
right. The mandarin thereupon fulfilled his promise. The fatted pup was
killed for the wedding feast, and when Hi-Chum-Chop passed Winky-Hi the
liver wing all present knew that it was a token of eternal goodwill, in
accordance with Chinese custom from time immemorial.

The mandarin had a table divided into twenty-five squares, as shown in
the diagram. On each of twenty-four of these squares was placed a
numbered counter, just as I have indicated. The puzzle is to get the
counters in numerical order by moving them one at a time in what we call
"knight's moves." Counter 1 should be where 16 is, 2 where 11 is, 4
where 13 now is, and so on. It will be seen that all the counters on
shaded squares are in their proper positions. Of course, two counters
may never be on a square at the same time. Can you perform the feat in
the fewest possible moves?

[Illustration]

In order to make the manner of moving perfectly clear I will point out
that the first knight's move can only be made by 1 or by 2 or by 10.
Supposing 1 moves, then the next move must be by 23, 4, 8, or 21. As
there is never more than one square vacant, the order in which the
counters move may be written out as follows: 1--21--14--18--22, etc. A
rough diagram should be made on a larger scale for practice, and
numbered counters or pieces of cardboard used.


343.--EXERCISE FOR PRISONERS.

The following is the plan of the north wing of a certain gaol, showing
the sixteen cells all communicating by open doorways. Fifteen prisoners
were numbered and arranged in the cells as shown. They were allowed to
change their cells as much as they liked, but if two prisoners were ever
in the same cell together there was a severe punishment promised them.

[Illustration]

Now, in order to reduce their growing obesity, and to combine physical
exercise with mental recreation, the prisoners decided, on the
suggestion of one of their number who was interested in knight's tours,
to try to form themselves into a perfect knight's path without breaking
the prison regulations, and leaving the bottom right-hand corner cell
vacant, as originally. The joke of the matter is that the arrangement at
which they arrived was as follows:--

     8    3   12    1
    11   14    9    6
     4    7    2   13
    15   10    5

The warders failed to detect the important fact that the men could not
possibly get into this position without two of them having been at some
time in the same cell together. Make the attempt with counters on a
ruled diagram, and you will find that this is so. Otherwise the solution
is correct enough, each member being, as required, a knight's move from
the preceding number, and the original corner cell vacant.

The puzzle is to start with the men placed as in the illustration and
show how it might have been done in the fewest moves, while giving a
complete rest to as many prisoners as possible.

As there is never more than one vacant cell for a man to enter, it is
only necessary to write down the numbers of the men in the order in
which they move. It is clear that very few men can be left throughout in
their cells undisturbed, but I will leave the solver to discover just
how many, as this is a very essential part of the puzzle.


344.--THE KENNEL PUZZLE.

[Illustration]

A man has twenty-five dog kennels all communicating with each other by
doorways, as shown in the illustration. He wishes to arrange his twenty
dogs so that they shall form a knight's string from dog No. 1 to dog No.
20, the bottom row of five kennels to be left empty, as at present. This
is to be done by moving one dog at a time into a vacant kennel. The dogs
are well trained to obedience, and may be trusted to remain in the
kennels in which they are placed, except that if two are placed in the
same kennel together they will fight it out to the death. How is the
puzzle to be solved in the fewest possible moves without two dogs ever
being together?


345.--THE TWO PAWNS.

[Illustration]

Here is a neat little puzzle in counting. In how many different ways may
the two pawns advance to the eighth square? You may move them in any
order you like to form a different sequence. For example, you may move
the Q R P (one or two squares) first, or the K R P first, or one pawn as
far as you like before touching the other. Any sequence is permissible,
only in this puzzle as soon as a pawn reaches the eighth square it is
dead, and remains there unconverted. Can you count the number of
different sequences? At first it will strike you as being very
difficult, but I will show that it is really quite simple when properly
attacked.



VARIOUS CHESS PUZZLES.

    "Chesse-play is a good and wittie exercise of
    the minde for some kinde of men."
              Burton's _Anatomy of Melancholy_.

346.--SETTING THE BOARD.

I have a single chessboard and a single set of chessmen. In how many
different ways may the men be correctly set up for the beginning of a
game? I find that most people slip at a particular point in making the
calculation.


347.--COUNTING THE RECTANGLES.

Can you say correctly just how many squares and other rectangles the
chessboard contains? In other words, in how great a number of different
ways is it possible to indicate a square or other rectangle enclosed by
lines that separate the squares of the board?


348.--THE ROOKERY.

[Illustration]

The White rooks cannot move outside the little square in which they are
enclosed except on the final move, in giving checkmate. The puzzle is
how to checkmate Black in the fewest possible moves with No. 8 rook, the
other rooks being left in numerical order round the sides of their
square with the break between 1 and 7.


349.--STALEMATE.

Some years ago the puzzle was proposed to construct an imaginary game of
chess, in which White shall be stalemated in the fewest possible moves
with all the thirty-two pieces on the board. Can you build up such a
position in fewer than twenty moves?


350.--THE FORSAKEN KING.

[Illustration]

Set up the position shown in the diagram. Then the condition of the
puzzle is--White to play and checkmate in six moves. Notwithstanding the
complexities, I will show how the manner of play may be condensed into
quite a few lines, merely stating here that the first two moves of White
cannot be varied.


351.--THE CRUSADER.

The following is a prize puzzle propounded by me some years ago. Produce
a game of chess which, after sixteen moves, shall leave White with all
his sixteen men on their original squares and Black in possession of his
king alone (not necessarily on his own square). White is then to _force_
mate in three moves.


352.--IMMOVABLE PAWNS.

Starting from the ordinary arrangement of the pieces as for a game, what
is the smallest possible number of moves necessary in order to arrive at
the following position? The moves for both sides must, of course, be
played strictly in accordance with the rules of the game, though the
result will necessarily be a very weird kind of chess.

[Illustration]


353.--THIRTY-SIX MATES.

[Illustration]

Place the remaining eight White pieces in such a position that White
shall have the choice of thirty-six different mates on the move. Every
move that checkmates and leaves a different position is a different
mate. The pieces already placed must not be moved.


354.--AN AMAZING DILEMMA.

In a game of chess between Mr. Black and Mr. White, Black was in
difficulties, and as usual was obliged to catch a train. So he proposed
that White should complete the game in his absence on condition that no
moves whatever should be made for Black, but only with the White pieces.
Mr. White accepted, but to his dismay found it utterly impossible to win
the game under such conditions. Try as he would, he could not checkmate
his opponent. On which square did Mr. Black leave his king? The other
pieces are in their proper positions in the diagram. White may leave
Black in check as often as he likes, for it makes no difference, as he
can never arrive at a checkmate position.

[Illustration]


355.--CHECKMATE!

[Illustration]

Strolling into one of the rooms of a London club, I noticed a position
left by two players who had gone. This position is shown in the diagram.
It is evident that White has checkmated Black. But how did he do it?
That is the puzzle.


356.--QUEER CHESS.

Can you place two White rooks and a White knight on the board so that
the Black king (who must be on one of the four squares in the middle of
the board) shall be in check with no possible move open to him? "In
other words," the reader will say, "the king is to be shown checkmated."
Well, you can use the term if you wish, though I intentionally do not
employ it myself. The mere fact that there is no White king on the board
would be a sufficient reason for my not doing so.


357.--ANCIENT CHINESE PUZZLE.

[Illustration]

My next puzzle is supposed to be Chinese, many hundreds of years old,
and never fails to interest. White to play and mate, moving each of the
three pieces once, and once only.


358.--THE SIX PAWNS.

In how many different ways may I place six pawns on the chessboard so
that there shall be an even number of unoccupied squares in every row
and every column? We are not here considering the diagonals at all, and
every different six squares occupied makes a different solution, so we
have not to exclude reversals or reflections.


359.--COUNTER SOLITAIRE.

Here is a little game of solitaire that is quite easy, but not so easy
as to be uninteresting. You can either rule out the squares on a sheet
of cardboard or paper, or you can use a portion of your chessboard. I
have shown numbered counters in the illustration so as to make the
solution easy and intelligible to all, but chess pawns or draughts will
serve just as well in practice.

[Illustration]

The puzzle is to remove all the counters except one, and this one that
is left must be No. 1. You remove a counter by jumping over another
counter to the next space beyond, if that square is vacant, but you
cannot make a leap in a diagonal direction. The following moves will
make the play quite clear: 1-9, 2-10, 1-2, and so on. Here 1 jumps over
9, and you remove 9 from the board; then 2 jumps over 10, and you remove
10; then 1 jumps over 2, and you remove 2. Every move is thus a capture,
until the last capture of all is made by No. 1.


360.--CHESSBOARD SOLITAIRE.

[Illustration]

Here is an extension of the last game of solitaire. All you need is a
chessboard and the thirty-two pieces, or the same number of draughts or
counters. In the illustration numbered counters are used. The puzzle is
to remove all the counters except two, and these two must have
originally been on the same side of the board; that is, the two left
must either belong to the group 1 to 16 or to the other group, 17 to 32.
You remove a counter by jumping over it with another counter to the next
square beyond, if that square is vacant, but you cannot make a leap in a
diagonal direction. The following moves will make the play quite clear:
3-11, 4-12, 3-4, 13-3. Here 3 jumps over 11, and you remove 11; 4 jumps
over 12, and you remove 12; and so on. It will be found a fascinating
little game of patience, and the solution requires the exercise of some
ingenuity.


361.--THE MONSTROSITY.

One Christmas Eve I was travelling by rail to a little place in one of
the southern counties. The compartment was very full, and the passengers
were wedged in very tightly. My neighbour in one of the corner seats was
closely studying a position set up on one of those little folding
chessboards that can be carried conveniently in the pocket, and I could
scarcely avoid looking at it myself. Here is the position:--

[Illustration]

My fellow-passenger suddenly turned his head and caught the look of
bewilderment on my face.

"Do you play chess?" he asked.

"Yes, a little. What is that? A problem?"

"Problem? No; a game."

"Impossible!" I exclaimed rather rudely. "The position is a perfect
monstrosity!"

He took from his pocket a postcard and handed it to me. It bore an
address at one side and on the other the words "43. K to Kt 8."

"It is a correspondence game." he exclaimed. "That is my friend's last
move, and I am considering my reply."

"But you really must excuse me; the position seems utterly impossible.
How on earth, for example--"

"Ah!" he broke in smilingly. "I see; you are a beginner; you play to
win."

"Of course you wouldn't play to lose or draw!"

He laughed aloud.

"You have much to learn. My friend and myself do not play for results of
that antiquated kind. We seek in chess the wonderful, the whimsical, the
weird. Did you ever see a position like that?"

I inwardly congratulated myself that I never had.

"That position, sir, materializes the sinuous evolvements and syncretic,
synthetic, and synchronous concatenations of two cerebral
individualities. It is the product of an amphoteric and intercalatory
interchange of--"

"Have you seen the evening paper, sir?" interrupted the man opposite,
holding out a newspaper. I noticed on the margin beside his thumb some
pencilled writing. Thanking him, I took the paper and read--"Insane, but
quite harmless. He is in my charge."

After that I let the poor fellow run on in his wild way until both got
out at the next station.

But that queer position became fixed indelibly in my mind, with Black's
last move 43. K to Kt 8; and a short time afterwards I found it actually
possible to arrive at such a position in forty-three moves. Can the
reader construct such a sequence? How did White get his rooks and king's
bishop into their present positions, considering Black can never have
moved his king's bishop? No odds were given, and every move was
perfectly legitimate.



MEASURING, WEIGHING, AND PACKING PUZZLES.

    "Measure still for measure."
    _Measure for Measure_, v. 1.


Apparently the first printed puzzle involving the measuring of a given
quantity of liquid by pouring from one vessel to others of known
capacity was that propounded by Niccola Fontana, better known as
"Tartaglia" (the stammerer), 1500-1559. It consists in dividing 24 oz.
of valuable balsam into three equal parts, the only measures available
being vessels holding 5, 11, and 13 ounces respectively. There are many
different solutions to this puzzle in six manipulations, or pourings
from one vessel to another. Bachet de Méziriac reprinted this and other
of Tartaglia's puzzles in his _Problèmes plaisans et délectables_
(1612). It is the general opinion that puzzles of this class can only be
solved by trial, but I think formulæ can be constructed for the solution
generally of certain related cases. It is a practically unexplored field
for investigation.

The classic weighing problem is, of course, that proposed by Bachet. It
entails the determination of the least number of weights that would
serve to weigh any integral number of pounds from 1 lb. to 40 lbs.
inclusive, when we are allowed to put a weight in either of the two
pans. The answer is 1, 3, 9, and 27 lbs. Tartaglia had previously
propounded the same puzzle with the condition that the weights may only
be placed in one pan. The answer in that case is 1, 2, 4, 8, 16, 32 lbs.
Major MacMahon has solved the problem quite generally. A full account
will be found in Ball's _Mathematical Recreations_ (5th edition).

Packing puzzles, in which we are required to pack a maximum number of
articles of given dimensions into a box of known dimensions, are, I
believe, of quite recent introduction. At least I cannot recall any
example in the books of the old writers. One would rather expect to find
in the toy shops the idea presented as a mechanical puzzle, but I do not
think I have ever seen such a thing. The nearest approach to it would
appear to be the puzzles of the jig-saw character, where there is only
one depth of the pieces to be adjusted.


362.--THE WASSAIL BOWL.

One Christmas Eve three Weary Willies came into possession of what was
to them a veritable wassail bowl, in the form of a small barrel,
containing exactly six quarts of fine ale. One of the men possessed a
five-pint jug and another a three-pint jug, and the problem for them was
to divide the liquor equally amongst them without waste. Of course, they
are not to use any other vessels or measures. If you can show how it was
to be done at all, then try to find the way that requires the fewest
possible manipulations, every separate pouring from one vessel to
another, or down a man's throat, counting as a manipulation.


363.--THE DOCTOR'S QUERY.

"A curious little point occurred to me in my dispensary this morning,"
said a doctor. "I had a bottle containing ten ounces of spirits of wine,
and another bottle containing ten ounces of water. I poured a quarter of
an ounce of spirits into the water and shook them up together. The
mixture was then clearly forty to one. Then I poured back a
quarter-ounce of the mixture, so that the two bottles should again each
contain the same quantity of fluid. What proportion of spirits to water
did the spirits of wine bottle then contain?"


364.--THE BARREL PUZZLE.

The men in the illustration are disputing over the liquid contents of a
barrel. What the particular liquid is it is impossible to say, for we
are unable to look into the barrel; so we will call it water. One man
says that the barrel is more than half full, while the other insists
that it is not half full. What is their easiest way of settling the
point? It is not necessary to use stick, string, or implement of any
kind for measuring. I give this merely as one of the simplest possible
examples of the value of ordinary sagacity in the solving of puzzles.
What are apparently very difficult problems may frequently be solved in
a similarly easy manner if we only use a little common sense.

[Illustration]


365.--NEW MEASURING PUZZLE.

Here is a new poser in measuring liquids that will be found interesting.
A man has two ten-quart vessels full of wine, and a five-quart and a
four-quart measure. He wants to put exactly three quarts into each of
the two measures. How is he to do it? And how many manipulations
(pourings from one vessel to another) do you require? Of course, waste
of wine, tilting, and other tricks are not allowed.


366.--THE HONEST DAIRYMAN.

An honest dairyman in preparing his milk for public consumption employed
a can marked B, containing milk, and a can marked A, containing water.
From can A he poured enough to double the contents of can B. Then he
poured from can B into can A enough to double its contents. Then he
finally poured from can A into can B until their contents were exactly
equal. After these operations he would send the can A to London, and the
puzzle is to discover what are the relative proportions of milk and
water that he provides for the Londoners' breakfast-tables. Do they get
equal proportions of milk and water--or two parts of milk and one of
water--or what? It is an interesting question, though, curiously enough,
we are not told how much milk or water he puts into the cans at the
start of his operations.


367.--WINE AND WATER.

Mr. Goodfellow has adopted a capital idea of late. When he gives a
little dinner party and the time arrives to smoke, after the departure
of the ladies, he sometimes finds that the conversation is apt to become
too political, too personal, too slow, or too scandalous. Then he always
manages to introduce to the company some new poser that he has secreted
up his sleeve for the occasion. This invariably results in no end of
interesting discussion and debate, and puts everybody in a good humour.

Here is a little puzzle that he propounded the other night, and it is
extraordinary how the company differed in their answers. He filled a
wine-glass half full of wine, and another glass twice the size one-third
full of wine. Then he filled up each glass with water and emptied the
contents of both into a tumbler. "Now," he said, "what part of the
mixture is wine and what part water?" Can you give the correct answer?


368.--THE KEG OF WINE.

Here is a curious little problem. A man had a ten-gallon keg full of
wine and a jug. One day he drew off a jugful of wine and filled up the
keg with water. Later on, when the wine and water had got thoroughly
mixed, he drew off another jugful and again filled up the keg with
water. It was then found that the keg contained equal proportions of
wine and water. Can you find from these facts the capacity of the jug?


369.--MIXING THE TEA.

"Mrs. Spooner called this morning," said the honest grocer to his
assistant. "She wants twenty pounds of tea at 2s. 4½d. per lb. Of
course we have a good 2s. 6d. tea, a slightly inferior at 2s. 3d., and a
cheap Indian at 1s. 9d., but she is very particular always about her
prices."

"What do you propose to do?" asked the innocent assistant.

"Do?" exclaimed the grocer. "Why, just mix up the three teas in
different proportions so that the twenty pounds will work out fairly at
the lady's price. Only don't put in more of the best tea than you can
help, as we make less profit on that, and of course you will use only
our complete pound packets. Don't do any weighing."

How was the poor fellow to mix the three teas? Could you have shown him
how to do it?


370.--A PACKING PUZZLE.

As we all know by experience, considerable ingenuity is often required
in packing articles into a box if space is not to be unduly wasted. A
man once told me that he had a large number of iron balls, all exactly
two inches in diameter, and he wished to pack as many of these as
possible into a rectangular box 24+9/10 inches long, 22+4/5 inches
wide, and 14 inches deep. Now, what is the greatest number of the
balls that he could pack into that box?


371.--GOLD PACKING IN RUSSIA.

The editor of the _Times_ newspaper was invited by a high Russian
official to inspect the gold stored in reserve at St. Petersburg, in
order that he might satisfy himself that it was not another "Humbert
safe." He replied that it would be of no use whatever, for although the
gold might appear to be there, he would be quite unable from a mere
inspection to declare that what he saw was really gold. A correspondent
of the _Daily Mail_ thereupon took up the challenge, but, although he
was greatly impressed by what he saw, he was compelled to confess his
incompetence (without emptying and counting the contents of every box
and sack, and assaying every piece of gold) to give any assurance on the
subject. In presenting the following little puzzle, I wish it to be also
understood that I do not guarantee the real existence of the gold, and
the point is not at all material to our purpose. Moreover, if the reader
says that gold is not usually "put up" in slabs of the dimensions that I
give, I can only claim problematic licence.

Russian officials were engaged in packing 800 gold slabs, each measuring
12½ inches long, 11 inches wide, and 1 inch deep. What are the
interior dimensions of a box of equal length and width, and necessary
depth, that will exactly contain them without any space being left over?
Not more than twelve slabs may be laid on edge, according to the rules
of the government. It is an interesting little problem in packing, and
not at all difficult.


372.--THE BARRELS OF HONEY.

[Illustration]

Once upon a time there was an aged merchant of Bagdad who was much
respected by all who knew him. He had three sons, and it was a rule of
his life to treat them all exactly alike. Whenever one received a
present, the other two were each given one of equal value. One day this
worthy man fell sick and died, bequeathing all his possessions to his
three sons in equal shares.

The only difficulty that arose was over the stock of honey. There were
exactly twenty-one barrels. The old man had left instructions that not
only should every son receive an equal quantity of honey, but should
receive exactly the same number of barrels, and that no honey should be
transferred from barrel to barrel on account of the waste involved. Now,
as seven of these barrels were full of honey, seven were half-full, and
seven were empty, this was found to be quite a puzzle, especially as
each brother objected to taking more than four barrels of, the same
description--full, half-full, or empty. Can you show how they succeeded
in making a correct division of the property?



CROSSING RIVER PROBLEMS

    "My boat is on the shore."
    BYRON.


This is another mediæval class of puzzles. Probably the earliest example
was by Abbot Alcuin, who was born in Yorkshire in 735 and died at Tours
in 804. And everybody knows the story of the man with the wolf, goat,
and basket of cabbages whose boat would only take one of the three at a
time with the man himself. His difficulties arose from his being unable
to leave the wolf alone with the goat, or the goat alone with the
cabbages. These puzzles were considered by Tartaglia and Bachet, and
have been later investigated by Lucas, De Fonteney, Delannoy, Tarry, and
others. In the puzzles I give there will be found one or two new
conditions which add to the complexity somewhat. I also include a pulley
problem that practically involves the same principles.


[Illustration]

373.--CROSSING THE STREAM.

During a country ramble Mr. and Mrs. Softleigh found themselves in a
pretty little dilemma. They had to cross a stream in a small boat which
was capable of carrying only 150 lbs. weight. But Mr. Softleigh and his
wife each weighed exactly 150 lbs., and each of their sons weighed 75
lbs. And then there was the dog, who could not be induced on any terms
to swim. On the principle of "ladies first," they at once sent Mrs.
Softleigh over; but this was a stupid oversight, because she had to come
back again with the boat, so nothing was gained by that operation. How
did they all succeed in getting across? The reader will find it much
easier than the Softleigh family did, for their greatest enemy could not
have truthfully called them a brilliant quartette--while the dog was a
perfect fool.


374--CROSSING THE RIVER AXE.

Many years ago, in the days of the smuggler known as "Rob Roy of the
West," a piratical band buried on the coast of South Devon a quantity of
treasure which was, of course, abandoned by them in the usual
inexplicable way. Some time afterwards its whereabouts was discovered by
three countrymen, who visited the spot one night and divided the spoil
between them, Giles taking treasure to the value of £800, Jasper £500
worth, and Timothy £300 worth. In returning they had to cross the river
Axe at a point where they had left a small boat in readiness. Here,
however, was a difficulty they had not anticipated. The boat would only
carry two men, or one man and a sack, and they had so little confidence
in one another that no person could be left alone on the land or in the
boat with more than his share of the spoil, though two persons (being a
check on each other) might be left with more than their shares. The
puzzle is to show how they got over the river in the fewest possible
crossings, taking their treasure with them. No tricks, such as ropes,
"flying bridges," currents, swimming, or similar dodges, may be
employed.


375.--FIVE JEALOUS HUSBANDS.

During certain local floods five married couples found themselves
surrounded by water, and had to escape from their unpleasant position in
a boat that would only hold three persons at a time. Every husband was
so jealous that he would not allow his wife to be in the boat or on
either bank with another man (or with other men) unless he was himself
present. Show the quickest way of getting these five men and their wives
across into safety.

Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. To
go over and return counts as two crossings. No tricks such as ropes,
swimming, currents, etc., are permitted.


376.--THE FOUR ELOPEMENTS.

Colonel B---- was a widower of a very taciturn disposition. His
treatment of his four daughters was unusually severe, almost cruel, and
they not unnaturally felt disposed to resent it. Being charming girls
with every virtue and many accomplishments, it is not surprising that
each had a fond admirer. But the father forbade the young men to call at
his house, intercepted all letters, and placed his daughters under
stricter supervision than ever. But love, which scorns locks and keys
and garden walls, was equal to the occasion, and the four youths
conspired together and planned a general elopement.

At the foot of the tennis lawn at the bottom of the garden ran the
silver Thames, and one night, after the four girls had been safely
conducted from a dormitory window to _terra firma_, they all crept
softly down to the bank of the river, where a small boat belonging to
the Colonel was moored. With this they proposed to cross to the opposite
side and make their way to a lane where conveyances were waiting to
carry them in their flight. Alas! here at the water's brink their
difficulties already began.

The young men were so extremely jealous that not one of them would allow
his prospective bride to remain at any time in the company of another
man, or men, unless he himself were present also. Now, the boat would
only hold two persons, though it could, of course, be rowed by one, and
it seemed impossible that the four couples would ever get across. But
midway in the stream was a small island, and this seemed to present a
way out of the difficulty, because a person or persons could be left
there while the boat was rowed back or to the opposite shore. If they
had been prepared for their difficulty they could have easily worked out
a solution to the little poser at any other time. But they were now so
hurried and excited in their flight that the confusion they soon got
into was exceedingly amusing--or would have been to any one except
themselves.

As a consequence they took twice as long and crossed the river twice as
often as was really necessary. Meanwhile, the Colonel, who was a very
light sleeper, thought he heard a splash of oars. He quickly raised the
alarm among his household, and the young ladies were found to be
missing. Somebody was sent to the police-station, and a number of
officers soon aided in the pursuit of the fugitives, who, in consequence
of that delay in crossing the river, were quickly overtaken. The four
girls returned sadly to their homes, and afterwards broke off their
engagements in disgust.

For a considerable time it was a mystery how the party of eight managed
to cross the river in that little boat without any girl being ever left
with a man, unless her betrothed was also present. The favourite method
is to take eight counters or pieces of cardboard and mark them A, B, C,
D, a, b, c, d, to represent the four men and their prospective brides,
and carry them from one side of a table to the other in a matchbox (to
represent the boat), a penny being placed in the middle of the table as
the island.

Readers are now asked to find the quickest method of getting the party
across the river. How many passages are necessary from land to land? By
"land" is understood either shore or island. Though the boat would not
necessarily call at the island every time of crossing, the possibility
of its doing so must be provided for. For example, it would not do for a
man to be alone in the boat (though it were understood that he intended
merely to cross from one bank to the opposite one) if there happened to
be a girl alone on the island other than the one to whom he was engaged.


377.--STEALING THE CASTLE TREASURE.

The ingenious manner in which a box of treasure, consisting principally
of jewels and precious stones, was stolen from Gloomhurst Castle has
been handed down as a tradition in the De Gourney family. The thieves
consisted of a man, a youth, and a small boy, whose only mode of escape
with the box of treasure was by means of a high window. Outside the
window was fixed a pulley, over which ran a rope with a basket at each
end. When one basket was on the ground the other was at the window. The
rope was so disposed that the persons in the basket could neither help
themselves by means of it nor receive help from others. In short, the
only way the baskets could be used was by placing a heavier weight in
one than in the other.

Now, the man weighed 195 lbs., the youth 105 lbs., the boy 90 lbs., and
the box of treasure 75 lbs. The weight in the descending basket could
not exceed that in the other by more than 15 lbs. without causing a
descent so rapid as to be most dangerous to a human being, though it
would not injure the stolen property. Only two persons, or one person
and the treasure, could be placed in the same basket at one time. How
did they all manage to escape and take the box of treasure with them?

The puzzle is to find the shortest way of performing the feat, which in
itself is not difficult. Remember, a person cannot help himself by
hanging on to the rope, the only way being to go down "with a bump,"
with the weight in the other basket as a counterpoise.



PROBLEMS CONCERNING GAMES.

    "The little pleasure of the game."
                         MATTHEW PRIOR.

Every game lends itself to the propounding of a variety of puzzles. They
can be made, as we have seen, out of the chessboard and the peculiar
moves of the chess pieces. I will now give just a few examples of
puzzles with playing cards and dominoes, and also go out of doors and
consider one or two little posers in the cricket field, at the football
match, and the horse race and motor-car race.


378.--DOMINOES IN PROGRESSION.

[Illustration]

It will be seen that I have played six dominoes, in the illustration, in
accordance with the ordinary rules of the game, 4 against 4, 1 against
1, and so on, and yet the sum of the spots on the successive dominoes,
4, 5, 6, 7, 8, 9, are in arithmetical progression; that is, the numbers
taken in order have a common difference of 1. In how many different ways
may we play six dominoes, from an ordinary box of twenty-eight, so that
the numbers on them may lie in arithmetical progression? We must always
play from left to right, and numbers in decreasing arithmetical
progression (such as 9, 8, 7, 6, 5, 4) are not admissible.


379.--THE FIVE DOMINOES.

[Illustration]

Here is a new little puzzle that is not difficult, but will probably be
found entertaining by my readers. It will be seen that the five dominoes
are so arranged in proper sequence (that is, with 1 against 1, 2 against
2, and so on), that the total number of pips on the two end dominoes is
five, and the sum of the pips on the three dominoes in the middle is
also five. There are just three other arrangements giving five for the
additions. They are: --

    (1--0) (0--0) (0--2) (2--1) (1--3)
    (4--0) (0--0) (0--2) (2--1) (1--0)
    (2--0) (0--0) (0--1) (1--3) (3--0)

Now, how many similar arrangements are there of five dominoes that shall
give six instead of five in the two additions?


380.--THE DOMINO FRAME PUZZLE.

[Illustration]

It will be seen in the illustration that the full set of twenty-eight
dominoes is arranged in the form of a square frame, with 6 against 6, 2
against 2, blank against blank, and so on, as in the game. It will be
found that the pips in the top row and left-hand column both add up 44.
The pips in the other two sides sum to 59 and 32 respectively. The
puzzle is to rearrange the dominoes in the same form so that all of the
four sides shall sum to 44. Remember that the dominoes must be correctly
placed one against another as in the game.


381.--THE CARD FRAME PUZZLE.

In the illustration we have a frame constructed from the ten playing
cards, ace to ten of diamonds. The children who made it wanted the pips
on all four sides to add up alike, but they failed in their attempt and
gave it up as impossible. It will be seen that the pips in the top row,
the bottom row, and the left-hand side all add up 14, but the right-hand
side sums to 23. Now, what they were trying to do is quite possible. Can
you rearrange the ten cards in the same formation so that all four sides
shall add up alike? Of course they need not add up 14, but any number
you choose to select.

[Illustration]


382.--THE CROSS OF CARDS.

[Illustration]

In this case we use only nine cards--the ace to nine of diamonds. The
puzzle is to arrange them in the form of a cross, exactly in the way
shown in the illustration, so that the pips in the vertical bar and in
the horizontal bar add up alike. In the example given it will be found
that both directions add up 23. What I want to know is, how many
different ways are there of rearranging the cards in order to bring
about this result? It will be seen that, without affecting the solution,
we may exchange the 5 with the 6, the 5 with the 7, the 8 with the 3,
and so on. Also we may make the horizontal and the vertical bars change
places. But such obvious manipulations as these are not to be regarded
as different solutions. They are all mere variations of one fundamental
solution. Now, how many of these fundamentally different solutions are
there? The pips need not, of course, always add up 23.


383.--THE "T" CARD PUZZLE.

[Illustration]

An entertaining little puzzle with cards is to take the nine cards of a
suit, from ace to nine inclusive, and arrange them in the form of the
letter "T," as shown in the illustration, so that the pips in the
horizontal line shall count the same as those in the column. In the
example given they add up twenty-three both ways. Now, it is quite easy
to get a single correct arrangement. The puzzle is to discover in just
how many different ways it may be done. Though the number is high, the
solution is not really difficult if we attack the puzzle in the right
manner. The reverse way obtained by reflecting the illustration in a
mirror we will not count as different, but all other changes in the
relative positions of the cards will here count. How many different ways
are there?


384.--CARD TRIANGLES.

Here you pick out the nine cards, ace to nine of diamonds, and arrange
them in the form of a triangle, exactly as shown in the illustration, so
that the pips add up the same on the three sides. In the example given
it will be seen that they sum to 20 on each side, but the particular
number is of no importance so long as it is the same on all three sides.
The puzzle is to find out in just how many different ways this can be
done.

If you simply turn the cards round so that one of the other two sides is
nearest to you this will not count as different, for the order will be
the same. Also, if you make the 4, 9, 5 change places with the 7, 3, 8,
and at the same time exchange the 1 and the 6, it will not be different.
But if you only change the 1 and the 6 it will be different, because the
order round the triangle is not the same. This explanation will prevent
any doubt arising as to the conditions.

[Illustration]


385.--"STRAND" PATIENCE.

The idea for this came to me when considering the game of Patience that
I gave in the _Strand Magazine_ for December, 1910, which has been
reprinted in Ernest Bergholt's _Second Book of Patience Games_, under
the new name of "King Albert."

Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3 D, 2
S, 1 D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the 9 of
diamonds at the bottom of one pile and the 9 of hearts at the bottom of
the other. The point is to exchange the spades with the clubs, so that
the diamonds and clubs are still in numerical order in one pile and the
hearts and spades in the other. There are four vacant spaces in addition
to the two spaces occupied by the piles, and any card may be laid on a
space, but a card can only be laid on another of the next higher
value--an ace on a two, a two on a three, and so on. Patience is
required to discover the shortest way of doing this. When there are four
vacant spaces you can pile four cards in seven moves, with only three
spaces you can pile them in nine moves, and with two spaces you cannot
pile more than two cards. When you have a grasp of these and similar
facts you will be able to remove a number of cards bodily and write down
7, 9, or whatever the number of moves may be. The gradual shortening of
play is fascinating, and first attempts are surprisingly lengthy.


386.--A TRICK WITH DICE.

[Illustration]

Here is a neat little trick with three dice. I ask you to throw the dice
without my seeing them. Then I tell you to multiply the points of the
first die by 2 and add 5; then multiply the result by 5 and add the
points of the second die; then multiply the result by 10 and add the
points of the third die. You then give me the total, and I can at once
tell you the points thrown with the three dice. How do I do it? As an
example, if you threw 1, 3, and 6, as in the illustration, the result
you would give me would be 386, from which I could at once say what you
had thrown.


387.--THE VILLAGE CRICKET MATCH.

In a cricket match, Dingley Dell v. All Muggleton, the latter had the
first innings. Mr. Dumkins and Mr. Podder were at the wickets, when the
wary Dumkins made a splendid late cut, and Mr. Podder called on him to
run. Four runs were apparently completed, but the vigilant umpires at
each end called, "three short," making six short runs in all. What
number did Mr. Dumkins score? When Dingley Dell took their turn at the
wickets their champions were Mr. Luffey and Mr. Struggles. The latter
made a magnificent off-drive, and invited his colleague to "come along,"
with the result that the observant spectators applauded them for what
was supposed to have been three sharp runs. But the umpires declared
that there had been two short runs at each end--four in all. To what
extent, if any, did this manoeuvre increase Mr. Struggles's total?


388.--SLOW CRICKET.

In the recent county match between Wessex and Nincomshire the former
team were at the wickets all day, the last man being put out a few
minutes before the time for drawing stumps. The play was so slow that
most of the spectators were fast asleep, and, on being awakened by one
of the officials clearing the ground, we learnt that two men had been
put out leg-before-wicket for a combined score of 19 runs; four men were
caught for a combined score or 17 runs; one man was run out for a duck's
egg; and the others were all bowled for 3 runs each. There were no
extras. We were not told which of the men was the captain, but he made
exactly 15 more than the average of his team. What was the captain's
score?


389.--THE FOOTBALL PLAYERS.

"It is a glorious game!" an enthusiast was heard to exclaim. "At the
close of last season, of the footballers of my acquaintance four had
broken their left arm, five had broken their right arm, two had the
right arm sound, and three had sound left arms." Can you discover from
that statement what is the smallest number of players that the speaker
could be acquainted with?

It does not at all follow that there were as many as fourteen men,
because, for example, two of the men who had broken the left arm might
also be the two who had sound right arms.


390.--THE HORSE-RACE PUZZLE.

There are no morals in puzzles. When we are solving the old puzzle of
the captain who, having to throw half his crew overboard in a storm,
arranged to draw lots, but so placed the men that only the Turks were
sacrificed, and all the Christians left on board, we do not stop to
discuss the questionable morality of the proceeding. And when we are
dealing with a measuring problem, in which certain thirsty pilgrims are
to make an equitable division of a barrel of beer, we do not object
that, as total abstainers, it is against our conscience to have anything
to do with intoxicating liquor. Therefore I make no apology for
introducing a puzzle that deals with betting.

Three horses--Acorn, Bluebottle, and Capsule--start in a race. The odds
are 4 to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how much
must I invest on each horse in order to win £13, no matter which horse
comes in first? Supposing, as an example, that I betted £5 on each
horse. Then, if Acorn won, I should receive £20 (four times £5), and
have to pay £5 each for the other two horses; thereby winning £10. But
it will be found that if Bluebottle was first I should only win £5, and
if Capsule won I should gain nothing and lose nothing. This will make
the question perfectly clear to the novice, who, like myself, is not
interested in the calling of the fraternity who profess to be engaged in
the noble task of "improving the breed of horses."



391.--THE MOTOR-CAR RACE.

Sometimes a quite simple statement of fact, if worded in an unfamiliar
manner, will cause considerable perplexity. Here is an example, and it
will doubtless puzzle some of my more youthful readers just a little. I
happened to be at a motor-car race at Brooklands, when one spectator
said to another, while a number of cars were whirling round and round
the circular track:--

"There's Gogglesmith--that man in the white car!"

"Yes, I see," was the reply; "but how many cars are running in this
race?"

Then came this curious rejoinder:--

"One-third of the cars in front of Gogglesmith added to three-quarters
of those behind him will give you the answer."

Now, can you tell how many cars were running in the race?



PUZZLE GAMES.


    "He that is beaten may be said
    To lie in honour's truckle bed."
                              HUDIBRAS.

It may be said generally that a game is a contest of skill for two or
more persons, into which we enter either for amusement or to win a
prize. A puzzle is something to be done or solved by the individual. For
example, if it were possible for us so to master the complexities of the
game of chess that we could be assured of always winning with the first
or second move, as the case might be, or of always drawing, then it
would cease to be a game and would become a puzzle. Of course among the
young and uninformed, when the correct winning play is not understood, a
puzzle may well make a very good game. Thus there is no doubt children
will continue to play "Noughts and Crosses," though I have shown (No.
109, "_Canterbury Puzzles_") that between two players who both
thoroughly understand the play, every game should be drawn. Neither
player could ever win except through the blundering of his opponent. But
I am writing from the point of view of the student of these things.

The examples that I give in this class are apparently games, but, since
I show in every case how one player may win if he only play correctly,
they are in reality puzzles. Their interest, therefore, lies in
attempting to discover the leading method of play.


392.--THE PEBBLE GAME.

Here is an interesting little puzzle game that I used to play with an
acquaintance on the beach at Slocomb-on-Sea. Two players place an odd
number of pebbles, we will say fifteen, between them. Then each takes in
turn one, two, or three pebbles (as he chooses), and the winner is the
one who gets the odd number. Thus, if you get seven and your opponent
eight, you win. If you get six and he gets nine, he wins. Ought the
first or second player to win, and how? When you have settled the
question with fifteen pebbles try again with, say, thirteen.


393.--THE TWO ROOKS.

This is a puzzle game for two players. Each player has a single rook.
The first player places his rook on any square of the board that he may
choose to select, and then the second player does the same. They now
play in turn, the point of each play being to capture the opponent's
rook. But in this game you cannot play through a line of attack without
being captured. That is to say, if in the diagram it is Black's turn to
play, he cannot move his rook to his king's knight's square, or to his
king's rook's square, because he would enter the "line of fire" when
passing his king's bishop's square. For the same reason he cannot move
to his queen's rook's seventh or eighth squares. Now, the game can never
end in a draw. Sooner or later one of the rooks must fall, unless, of
course, both players commit the absurdity of not trying to win. The
trick of winning is ridiculously simple when you know it. Can you solve
the puzzle?

[Illustration]


394.--PUSS IN THE CORNER.

[Illustration]

This variation of the last puzzle is also played by two persons. One
puts a counter on No. 6, and the other puts one on No. 55, and they play
alternately by removing the counter to any other number in a line. If
your opponent moves at any time on to one of the lines you occupy, or
even crosses one of your lines, you immediately capture him and win. We
will take an illustrative game.

A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to
15; A retreats to 26; B retreats to 13; A advances to 21; B retreats to
2; A advances to 7; B goes to 3; A moves to 6; B must now go to 4; A
establishes himself at 11, and B must be captured next move because he
is compelled to cross a line on which A stands. Play this over and you
will understand the game directly. Now, the puzzle part of the game is
this: Which player should win, and how many moves are necessary?


395.--A WAR PUZZLE GAME.

[Illustration]

Here is another puzzle game. One player, representing the British
general, places a counter at B, and the other player, representing the
enemy, places his counter at E. The Britisher makes the first advance
along one of the roads to the next town, then the enemy moves to one of
his nearest towns, and so on in turns, until the British general gets
into the same town as the enemy and captures him. Although each must
always move along a road to the next town only, and the second player
may do his utmost to avoid capture, the British general (as we should
suppose, from the analogy of real life) must infallibly win. But how?
That is the question.


396.--A MATCH MYSTERY.

Here is a little game that is childishly simple in its conditions. But
it is well worth investigation.

Mr. Stubbs pulled a small table between himself and his friend, Mr.
Wilson, and took a box of matches, from which he counted out thirty.

"Here are thirty matches," he said. "I divide them into three unequal
heaps. Let me see. We have 14, 11, and 5, as it happens. Now, the two
players draw alternately any number from any one heap, and he who draws
the last match loses the game. That's all! I will play with you, Wilson.
I have formed the heaps, so you have the first draw."

"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usual
moderation and take all the 14 heap."

"That is the worst you could do, for it loses right away. I take 6 from
the 11, leaving two equal heaps of 5, and to leave two equal heaps is a
certain win (with the single exception of 1, 1), because whatever you do
in one heap I can repeat in the other. If you leave 4 in one heap, I
leave 4 in the other. If you then leave 2 in one heap, I leave 2 in the
other. If you leave only 1 in one heap, then I take all the other heap.
If you take all one heap, I take all but one in the other. No, you must
never leave two heaps, unless they are equal heaps and more than 1, 1.
Let's begin again."

"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and
leave you 8, 11, 5."

Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3;
Mr. Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr.
Stubbs, 2, 3, 1; Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1,
1.

"It is now quite clear that I must win," said Mr. Stubbs, because you
must take 1, and then I take 1, leaving you the last match. You never
had a chance. There are just thirteen different ways in which the
matches may be grouped at the start for a certain win. In fact, the
groups selected, 14, 11, 5, are a certain win, because for whatever your
opponent may play there is another winning group you can secure, and so
on and on down to the last match."


397.--THE MONTENEGRIN DICE GAME.

It is said that the inhabitants of Montenegro have a little dice game
that is both ingenious and well worth investigation. The two players
first select two different pairs of odd numbers (always higher than 3)
and then alternately toss three dice. Whichever first throws the dice so
that they add up to one of his selected numbers wins. If they are both
successful in two successive throws it is a draw and they try again. For
example, one player may select 7 and 15 and the other 5 and 13. Then if
the first player throws so that the three dice add up 7 or 15 he wins,
unless the second man gets either 5 or 13 on his throw.

The puzzle is to discover which two pairs of numbers should be selected
in order to give both players an exactly even chance.


398.--THE CIGAR PUZZLE.

I once propounded the following puzzle in a London club, and for a
considerable period it absorbed the attention of the members. They could
make nothing of it, and considered it quite impossible of solution. And
yet, as I shall show, the answer is remarkably simple.

Two men are seated at a square-topped table. One places an ordinary
cigar (flat at one end, pointed at the other) on the table, then the
other does the same, and so on alternately, a condition being that no
cigar shall touch another. Which player should succeed in placing the
last cigar, assuming that they each will play in the best possible
manner? The size of the table top and the size of the cigar are not
given, but in order to exclude the ridiculous answer that the table
might be so diminutive as only to take one cigar, we will say that the
table must not be less than 2 feet square and the cigar not more than 4½
inches long. With those restrictions you may take any dimensions you
like. Of course we assume that all the cigars are exactly alike in
every respect. Should the first player, or the second player, win?



MAGIC SQUARE PROBLEMS.

    "By magic numbers."
    CONGREVE, _The Mourning Bride._

This is a very ancient branch of mathematical puzzledom, and it has an
immense, though scattered, literature of its own. In their simple form
of consecutive whole numbers arranged in a square so that every column,
every row, and each of the two long diagonals shall add up alike, these
magic squares offer three main lines of investigation: Construction,
Enumeration, and Classification. Of recent years many ingenious methods
have been devised for the construction of magics, and the law of their
formation is so well understood that all the ancient mystery has
evaporated and there is no longer any difficulty in making squares of
any dimensions. Almost the last word has been said on this subject. The
question of the enumeration of all the possible squares of a given order
stands just where it did over two hundred years ago. Everybody knows
that there is only one solution for the third order, three cells by
three; and Frénicle published in 1693 diagrams of all the arrangements
of the fourth order--880 in number--and his results have been verified
over and over again. I may here refer to the general solution for this
order, for numbers not necessarily consecutive, by E. Bergholt in
_Nature_, May 26, 1910, as it is of the greatest importance to students
of this subject. The enumeration of the examples of any higher order is
a completely unsolved problem.

As to classification, it is largely a matter of individual
taste--perhaps an æsthetic question, for there is beauty in the law and
order of numbers. A man once said that he divided the human race into
two great classes: those who take snuff and those who do not. I am not
sure that some of our classifications of magic squares are not almost as
valueless. However, lovers of these things seem somewhat agreed that
Nasik magic squares (so named by Mr. Frost, a student of them, after the
town in India where he lived, and also called Diabolique and
Pandiagonal) and Associated magic squares are of special interest, so I
will just explain what these are for the benefit of the novice.

[Illustration: SIMPLE]

[Illustration: SEMI-NASIK]

[Illustration: ASSOCIATED]

[Illustration: NASIK]


I published in _The Queen_ for January 15, 1910, an article that would
enable the reader to write out, if he so desired, all the 880 magics of
the fourth order, and the following is the complete classification that
I gave. The first example is that of a Simple square that fulfils the
simple conditions and no more. The second example is a Semi-Nasik, which
has the additional property that the opposite short diagonals of two
cells each together sum to 34. Thus, 14 + 4 + 11 + 5 = 34 and 12 + 6 +
13 + 3 = 34. The third example is not only Semi-Nasik but also
Associated, because in it every number, if added to the number that is
equidistant, in a straight line, from the centre gives 17. Thus, 1 + 16,
2 + 15, 3 + 14, etc. The fourth example, considered the most "perfect"
of all, is a Nasik. Here all the broken diagonals sum to 34. Thus, for
example, 15 + 14 + 2 + 3, and 10 + 4 + 7 + 13, and 15 + 5 + 2 + 12. As a
consequence, its properties are such that if you repeat the square in
all directions you may mark off a square, 4 × 4, wherever you please,
and it will be magic.

The following table not only gives a complete enumeration under the four
forms described, but also a classification under the twelve graphic
types indicated in the diagrams. The dots at the end of each line
represent the relative positions of those complementary pairs, 1 + 16, 2
+ 15, etc., which sum to 17. For example, it will be seen that the first
and second magic squares given are of Type VI., that the third square is
of Type III., and that the fourth is of Type I. Edouard Lucas indicated
these types, but he dropped exactly half of them and did not attempt the
classification.

    NASIK (Type I.)  .   .   .   .   .           48
    SEMI-NASIK (Type II., Transpositions
                of Nasik)            .       48
        "      (Type III., Associated)       48
        "      (Type IV.)  .    .    .   96
        "      (Type V.)   .    .    .   96 192
                                        ___
        "      (Type VI.)  .    .    .       96 384
                                            ___
    SIMPLE.    (Type VI.)  .    .    .      208
       "       (Type VII.) .    .    .   56
       "       (Type VIII.).    .    .   56
       "       (Type IX.)  .    .    .   56
       "       (Type X.)   .    .    .   56 224
                                        ___
       "       (Type XI.)  .    .    .    8
       "       (Type XII.) .    .    .    8  16 448
                                        ___ ___ ___
                                                880
                                                ___

It is hardly necessary to say that every one of these squares will
produce seven others by mere reversals and reflections, which we do not
count as different. So that there are 7,040 squares of this order, 880
of which are fundamentally different.

An infinite variety of puzzles may be made introducing new conditions
into the magic square. In _The Canterbury Puzzles_ I have given examples
of such squares with coins, with postage stamps, with cutting-out
conditions, and other tricks. I will now give a few variants involving
further novel conditions.


399.--THE TROUBLESOME EIGHT.

Nearly everybody knows that a "magic square" is an arrangement of
numbers in the form of a square so that every row, every column, and
each of the two long diagonals adds up alike. For example, you would
find little difficulty in merely placing a different number in each of
the nine cells in the illustration so that the rows, columns, and
diagonals shall all add up 15. And at your first attempt you will
probably find that you have an 8 in one of the corners. The puzzle is to
construct the magic square, under the same conditions, with the 8 in the
position shown.

[Illustration]


400.--THE MAGIC STRIPS.

[Illustration]

I happened to have lying on my table a number of strips of cardboard,
with numbers printed on them from 1 upwards in numerical order. The idea
suddenly came to me, as ideas have a way of unexpectedly coming, to make
a little puzzle of this. I wonder whether many readers will arrive at
the same solution that I did.

Take seven strips of cardboard and lay them together as above. Then
write on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so that
the numbers shall form seven rows and seven columns.

Now, the puzzle is to cut these strips into the fewest possible pieces
so that they may be placed together and form a magic square, the seven
rows, seven columns, and two diagonals adding up the same number. No
figures may be turned upside down or placed on their sides--that is, all
the strips must lie in their original direction.

Of course you could cut each strip into seven separate pieces, each
piece containing a number, and the puzzle would then be very easy, but I
need hardly say that forty-nine pieces is a long way from being the
fewest possible.


401.--EIGHT JOLLY GAOL BIRDS.

[Illustration]

The illustration shows the plan of a prison of nine cells all
communicating with one another by doorways. The eight prisoners have
their numbers on their backs, and any one of them is allowed to exercise
himself in whichever cell may happen to be vacant, subject to the rule
that at no time shall two prisoners be in the same cell. The merry
monarch in whose dominions the prison was situated offered them special
comforts one Christmas Eve if, without breaking that rule, they could so
place themselves that their numbers should form a magic square.

Now, prisoner No. 7 happened to know a good deal about magic squares, so
he worked out a scheme and naturally selected the method that was most
expeditious--that is, one involving the fewest possible moves from cell
to cell. But one man was a surly, obstinate fellow (quite unfit for the
society of his jovial companions), and he refused to move out of his
cell or take any part in the proceedings. But No. 7 was quite equal to
the emergency, and found that he could still do what was required in the
fewest possible moves without troubling the brute to leave his cell. The
puzzle is to show how he did it and, incidentally, to discover which
prisoner was so stupidly obstinate. Can you find the fellow?


402.--NINE JOLLY GAOL BIRDS.

[Illustration]

Shortly after the episode recorded in the last puzzle occurred, a ninth
prisoner was placed in the vacant cell, and the merry monarch then
offered them all complete liberty on the following strange conditions.
They were required so to rearrange themselves in the cells that their
numbers formed a magic square without their movements causing any two of
them ever to be in the same cell together, except that at the start one
man was allowed to be placed on the shoulders of another man, and thus
add their numbers together, and move as one man. For example, No. 8
might be placed on the shoulders of No. 2, and then they would move
about together as 10. The reader should seek first to solve the puzzle
in the fewest possible moves, and then see that the man who is burdened
has the least possible amount of work to do.


403.--THE SPANISH DUNGEON.

Not fifty miles from Cadiz stood in the middle ages a castle, all traces
of which have for centuries disappeared. Among other interesting
features, this castle contained a particularly unpleasant dungeon
divided into sixteen cells, all communicating with one another, as shown
in the illustration.

Now, the governor was a merry wight, and very fond of puzzles withal.
One day he went to the dungeon and said to the prisoners, "By my
halidame!" (or its equivalent in Spanish) "you shall all be set free if
you can solve this puzzle. You must so arrange yourselves in the sixteen
cells that the numbers on your backs shall form a magic square in which
every column, every row, and each of the two diagonals shall add up the
same. Only remember this: that in no case may two of you ever be
together in the same cell."

One of the prisoners, after working at the problem for two or three
days, with a piece of chalk, undertook to obtain the liberty of himself
and his fellow-prisoners if they would follow his directions and move
through the doorway from cell to cell in the order in which he should
call out their numbers.

[Illustration]

He succeeded in his attempt, and, what is more remarkable, it would seem
from the account of his method recorded in the ancient manuscript lying
before me, that he did so in the fewest possible moves. The reader is
asked to show what these moves were.


404.--THE SIBERIAN DUNGEONS.

[Illustration]

The above is a trustworthy plan of a certain Russian prison in Siberia.
All the cells are numbered, and the prisoners are numbered the same as
the cells they occupy. The prison diet is so fattening that these
political prisoners are in perpetual fear lest, should their pardon
arrive, they might not be able to squeeze themselves through the narrow
doorways and get out. And of course it would be an unreasonable thing to
ask any government to pull down the walls of a prison just to liberate
the prisoners, however innocent they might be. Therefore these men take
all the healthy exercise they can in order to retard their increasing
obesity, and one of their recreations will serve to furnish us with the
following puzzle.

Show, in the fewest possible moves, how the sixteen men may form
themselves into a magic square, so that the numbers on their backs shall
add up the same in each of the four columns, four rows, and two
diagonals without two prisoners having been at any time in the same cell
together. I had better say, for the information of those who have not
yet been made acquainted with these places, that it is a peculiarity of
prisons that you are not allowed to go outside their walls. Any prisoner
may go any distance that is possible in a single move.


405.--CARD MAGIC SQUARES.

[Illustration]

Take an ordinary pack of cards and throw out the twelve court cards.
Now, with nine of the remainder (different suits are of no consequence)
form the above magic square. It will be seen that the pips add up
fifteen in every row in every column, and in each of the two long
diagonals. The puzzle is with the remaining cards (without disturbing
this arrangement) to form three more such magic squares, so that each of
the four shall add up to a different sum. There will, of course, be four
cards in the reduced pack that will not be used. These four may be any
that you choose. It is not a difficult puzzle, but requires just a
little thought.


406.--THE EIGHTEEN DOMINOES.

The illustration shows eighteen dominoes arranged in the form of a
square so that the pips in every one of the six columns, six rows, and
two long diagonals add up 13. This is the smallest summation possible
with any selection of dominoes from an ordinary box of twenty-eight. The
greatest possible summation is 23, and a solution for this number may be
easily obtained by substituting for every number its complement to 6.
Thus for every blank substitute a 6, for every 1 a 5, for every 2 a 4,
for 3 a 3, for 4 a 2, for 5 a 1, and for 6 a blank. But the puzzle is to
make a selection of eighteen dominoes and arrange them (in exactly the
form shown) so that the summations shall be 18 in all the fourteen
directions mentioned.

[Illustration]



SUBTRACTING, MULTIPLYING, AND DIVIDING MAGICS.

Although the adding magic square is of such great antiquity, curiously
enough the multiplying magic does not appear to have been mentioned
until the end of the eighteenth century, when it was referred to
slightly by one writer and then forgotten until I revived it in
_Tit-Bits_ in 1897. The dividing magic was apparently first discussed by
me in _The Weekly Dispatch_ in June 1898. The subtracting magic is here
introduced for the first time. It will now be convenient to deal with
all four kinds of magic squares together.

[Illustration: ADDING SUBTRACTING MULTIPLYING DIVIDING]

In these four diagrams we have examples in the third order of adding,
subtracting, multiplying, and dividing squares. In the first the
constant, 15, is obtained by the addition of the rows, columns, and two
diagonals. In the second case you get the constant, 5, by subtracting
the first number in a line from the second, and the result from the
third. You can, of course, perform the operation in either direction;
but, in order to avoid negative numbers, it is more convenient simply to
deduct the middle number from the sum of the two extreme numbers. This
is, in effect, the same thing. It will be seen that the constant of the
adding square is n times that of the subtracting square derived from
it, where n is the number of cells in the side of square. And the
manner of derivation here is simply to reverse the two diagonals. Both
squares are "associated"--a term I have explained in the introductory
article to this department.

The third square is a multiplying magic. The constant, 216, is obtained
by multiplying together the three numbers in any line. It is
"associated" by multiplication, instead of by addition. It is here
necessary to remark that in an adding square it is not essential that
the nine numbers should be consecutive. Write down any nine numbers in
this way--

    1    3    5
    4    6    8
    7    9   11

so that the horizontal differences are all alike and the vertical
differences also alike (here 2 and 3), and these numbers will form an
adding magic square. By making the differences 1 and 3 we, of course,
get consecutive numbers--a particular case, and nothing more. Now, in
the case of the multiplying square we must take these numbers in
geometrical instead of arithmetical progression, thus--

    1    3    9
    2    6   18
    4   12   36

Here each successive number in the rows is multiplied by 3, and in the
columns by 2. Had we multiplied by 2 and 8 we should get the regular
geometrical progression, 1, 2, 4, 8, 16, 32, 64, 128, and 256, but I
wish to avoid high numbers. The numbers are arranged in the square in
the same order as in the adding square.

The fourth diagram is a dividing magic square. The constant 6 is here
obtained by dividing the second number in a line by the first (in either
direction) and the third number by the quotient. But, again, the process
is simplified by dividing the product of the two extreme numbers by the
middle number. This square is also "associated" by multiplication. It is
derived from the multiplying square by merely reversing the diagonals,
and the constant of the multiplying square is the cube of that of the
dividing square derived from it.

The next set of diagrams shows the solutions for the fifth order of
square. They are all "associated" in the same way as before. The
subtracting square is derived from the adding square by reversing the
diagonals and exchanging opposite numbers in the centres of the borders,
and the constant of one is again n times that of the other. The
dividing square is derived from the multiplying square in the same way,
and the constant of the latter is the 5th power (that is the nth) of
that of the former.

[Illustration]

These squares are thus quite easy for odd orders. But the reader will
probably find some difficulty over the even orders, concerning which I
will leave him to make his own researches, merely propounding two little
problems.


407.--TWO NEW MAGIC SQUARES.

Construct a subtracting magic square with the first sixteen whole
numbers that shall be "associated" by _subtraction_. The constant is, of
course, obtained by subtracting the first number from the second in
line, the result from the third, and the result again from the fourth.
Also construct a dividing magic square of the same order that shall be
"associated" by _division_. The constant is obtained by dividing the
second number in a line by the first, the third by the quotient, and the
fourth by the next quotient.


408.--MAGIC SQUARES OF TWO DEGREES.

While reading a French mathematical work I happened to come across, the
following statement: "A very remarkable magic square of 8, in two
degrees, has been constructed by M. Pfeffermann. In other words, he has
managed to dispose the sixty-four first numbers on the squares of a
chessboard in such a way that the sum of the numbers in every line,
every column, and in each of the two diagonals, shall be the same; and
more, that if one substitutes for all the numbers their squares, the
square still remains magic." I at once set to work to solve this
problem, and, although it proved a very hard nut, one was rewarded by
the discovery of some curious and beautiful laws that govern it. The
reader may like to try his hand at the puzzle.



MAGIC SQUARES OF PRIMES.

The problem of constructing magic squares with prime numbers only was
first discussed by myself in _The Weekly Dispatch_ for 22nd July and 5th
August 1900; but during the last three or four years it has received
great attention from American mathematicians. First, they have sought to
form these squares with the lowest possible constants. Thus, the first
nine prime numbers, 1 to 23 inclusive, sum to 99, which (being divisible
by 3) is theoretically a suitable series; yet it has been demonstrated
that the lowest possible constant is 111, and the required series as
follows: 1, 7, 13, 31, 37, 43, 61, 67, and 73. Similarly, in the case of
the fourth order, the lowest series of primes that are "theoretically
suitable" will not serve. But in every other order, up to the 12th
inclusive, magic squares have been constructed with the lowest series of
primes theoretically possible. And the 12th is the lowest order in which
a straight series of prime numbers, unbroken, from 1 upwards has been
made to work. In other words, the first 144 odd prime numbers have
actually been arranged in magic form. The following summary is taken
from _The Monist_ (Chicago) for October 1913:--

    Order of    Totals of   Lowest        Squares
    Square.      Series.    Constants.   made by--
                                       (Henry E.
      3rd         333        111       { Dudeney
                                       ( (1900).

                                       (Ernest Bergholt
      4th         408        102       { and C. D.
                                       ( Shuldham.

      5th        1065        213        H. A. Sayles.

                                       (C. D. Shuldham
      6th        2448        408       { and J.
                                       ( N. Muncey.

      7th        4893        699            do.
      8th        8912       1114            do.
      9th       15129       1681            do.
     10th       24160       2416        J. N. Muncey.
     11th       36095       3355            do.
     12th       54168       4514            do.

For further details the reader should consult the article itself, by W.
S. Andrews and H. A. Sayles.

These same investigators have also performed notable feats in
constructing associated and bordered prime magics, and Mr. Shuldham has
sent me a remarkable paper in which he gives examples of Nasik squares
constructed with primes for all orders from the 4th to the 10th, with
the exception of the 3rd (which is clearly impossible) and the 9th,
which, up to the time of writing, has baffled all attempts.


409.--THE BASKETS OF PLUMS.

[Illustration]

This is the form in which I first introduced the question of magic
squares with prime numbers. I will here warn the reader that there is a
little trap.

A fruit merchant had nine baskets. Every basket contained plums (all
sound and ripe), and the number in every basket was different. When
placed as shown in the illustration they formed a magic square, so that
if he took any three baskets in a line in the eight possible directions
there would always be the same number of plums. This part of the puzzle
is easy enough to understand. But what follows seems at first sight a
little queer.

The merchant told one of his men to distribute the contents of any
basket he chose among some children, giving plums to every child so that
each should receive an equal number. But the man found it quite
impossible, no matter which basket he selected and no matter how many
children he included in the treat. Show, by giving contents of the nine
baskets, how this could come about.


410.--THE MANDARIN'S "T" PUZZLE.

[Illustration]

Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in the
Far East, he prided himself on his knowledge of magic squares, a subject
that he had made his special hobby; but he soon discovered that he had
never really touched more than the fringe of the subject, and that the
wily Chinee could beat him easily. I present a little problem that one
learned mandarin propounded to our traveller, as depicted on the last
page.

The Chinaman, after remarking that the construction of the ordinary
magic square of twenty-five cells is "too velly muchee easy," asked our
countryman so to place the numbers 1 to 25 in the square that every
column, every row, and each of the two diagonals should add up 65, with
only prime numbers on the shaded "T." Of course the prime numbers
available are 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23, so you are at
liberty to select any nine of these that will serve your purpose. Can
you construct this curious little magic square?


411.--A MAGIC SQUARE OF COMPOSITES.

As we have just discussed the construction of magic squares with prime
numbers, the following forms an interesting companion problem. Make a
magic square with nine consecutive composite numbers--the smallest
possible.


412.--THE MAGIC KNIGHT'S TOUR.

Here is a problem that has never yet been solved, nor has its
impossibility been demonstrated. Play the knight once to every square of
the chessboard in a complete tour, numbering the squares in the order
visited, so that when completed the square shall be "magic," adding up
to 260 in every column, every row, and each of the two long diagonals. I
shall give the best answer that I have been able to obtain, in which
there is a slight error in the diagonals alone. Can a perfect solution
be found? I am convinced that it cannot, but it is only a "pious
opinion."



MAZES AND HOW TO THREAD THEM.

    "In wandering mazes lost."
                   _Paradise Lost._

The Old English word "maze," signifying a labyrinth, probably comes from
the Scandinavian, but its origin is somewhat uncertain. The late
Professor Skeat thought that the substantive was derived from the verb,
and as in old times to be mazed or amazed was to be "lost in thought,"
the transition to a maze in whose tortuous windings we are lost is
natural and easy.

The word "labyrinth" is derived from a Greek word signifying the
passages of a mine. The ancient mines of Greece and elsewhere inspired
fear and awe on account of their darkness and the danger of getting lost
in their intricate passages. Legend was afterwards built round these
mazes. The most familiar instance is the labyrinth made by Dædalus in
Crete for King Minos. In the centre was placed the Minotaur, and no one
who entered could find his way out again, but became the prey of the
monster. Seven youths and seven maidens were sent regularly by the
Athenians, and were duly devoured, until Theseus slew the monster and
escaped from the maze by aid of the clue of thread provided by Ariadne;
which accounts for our using to-day the expression "threading a maze."

The various forms of construction of mazes include complicated ranges of
caverns, architectural labyrinths, or sepulchral buildings, tortuous
devices indicated by coloured marbles and tiled pavements, winding paths
cut in the turf, and topiary mazes formed by clipped hedges. As a matter
of fact, they may be said to have descended to us in precisely this
order of variety.

Mazes were used as ornaments on the state robes of Christian emperors
before the ninth century, and were soon adopted in the decoration of
cathedrals and other churches. The original idea was doubtless to employ
them as symbols of the complicated folds of sin by which man is
surrounded. They began to abound in the early part of the twelfth
century, and I give an illustration of one of this period in the parish
church at St. Quentin (Fig. 1). It formed a pavement of the nave, and
its diameter is 34½ feet. The path here is the line itself. If you place
your pencil at the point A and ignore the enclosing line, the line leads
you to the centre by a long route over the entire area; but you never
have any option as to direction during your course. As we shall find in
similar cases, these early ecclesiastical mazes were generally not of a
puzzle nature, but simply long, winding paths that took you over
practically all the ground enclosed.

[Illustration: FIG. 1.--Maze at St. Quentin.]

[Illustration: FIG. 2.--Maze in Chartres Cathedral.]

In the abbey church of St. Berlin, at St. Omer, is another of these
curious floors, representing the Temple of Jerusalem, with stations for
pilgrims. These mazes were actually visited and traversed by them as a
compromise for not going to the Holy Land in fulfilment of a vow. They
were also used as a means of penance, the penitent frequently being
directed to go the whole course of the maze on hands and knees.

[Illustration: FIG. 3.--Maze in Lucca Cathedral.]

The maze in Chartres Cathedral, of which I give an illustration (Fig.
2), is 40 feet across, and was used by penitents following the
procession of Calvary. A labyrinth in Amiens Cathedral was octagonal,
similar to that at St. Quentin, measuring 42 feet across. It bore the
date 1288, but was destroyed in 1708. In the chapter-house at Bayeux is
a labyrinth formed of tiles, red, black, and encaustic, with a pattern
of brown and yellow. Dr. Ducarel, in his "_Tour through Part of
Normandy_" (printed in 1767), mentions the floor of the great
guard-chamber in the abbey of St. Stephen, at Caen, "the middle whereof
represents a maze or labyrinth about 10 feet diameter, and so artfully
contrived that, were we to suppose a man following all the intricate
meanders of its volutes, he could not travel less than a mile before he
got from one end to the other."

[Illustration: FIG. 4.--Maze at Saffron Walden, Essex.]

Then these mazes were sometimes reduced in size and represented on a
single tile (Fig. 3). I give an example from Lucca Cathedral. It is on
one of the porch piers, and is 19½ inches in diameter. A writer in
1858 says that, "from the continual attrition it has received from
thousands of tracing fingers, a central group of Theseus and the
Minotaur has now been very nearly effaced." Other examples were, and
perhaps still are, to be found in the Abbey of Toussarts, at
Châlons-sur-Marne, in the very ancient church of St. Michele at Pavia,
at Aix in Provence, in the cathedrals of Poitiers, Rheims, and Arras, in
the church of Santa Maria in Aquiro in Rome, in San Vitale at Ravenna,
in the Roman mosaic pavement found at Salzburg, and elsewhere. These
mazes were sometimes called "Chemins de Jerusalem," as being
emblematical of the difficulties attending a journey to the earthly
Jerusalem and of those encountered by the Christian before he can reach
the heavenly Jerusalem--where the centre was frequently called "Ciel."

Common as these mazes were upon the Continent, it is probable that no
example is to be found in any English church; at least I am not aware of
the existence of any. But almost every county has, or has had, its
specimens of mazes cut in the turf. Though these are frequently known as
"miz-mazes" or "mize-mazes," it is not uncommon to find them locally
called "Troy-towns," "shepherds' races," or "Julian's Bowers"--names
that are misleading, as suggesting a false origin. From the facts alone
that many of these English turf mazes are clearly copied from those in
the Continental churches, and practically all are found close to some
ecclesiastical building or near the site of an ancient one, we may
regard it as certain that they were of church origin and not invented by
the shepherds or other rustics. And curiously enough, these turf mazes
are apparently unknown on the Continent. They are distinctly mentioned
by Shakespeare:--

    "The nine men's morris is filled up with mud,
     And the quaint mazes in the wanton green
     For lack of tread are undistinguishable."

    _A Midsummer Night's Dream_, ii. 1.



    "My old bones ache: here's a maze trod indeed,
     Through forth-rights and meanders!"

    _The Tempest_, iii. 3.


[Illustration: FIG. 5.--Maze at Sneinton, Nottinghamshire.]

There was such a maze at Comberton, in Cambridgeshire, and another,
locally called the "miz-maze," at Leigh, in Dorset. The latter was on
the highest part of a field on the top of a hill, a quarter of a mile
from the village, and was slightly hollow in the middle and enclosed by
a bank about 3 feet high. It was circular, and was thirty paces in
diameter. In 1868 the turf had grown over the little trenches, and it
was then impossible to trace the paths of the maze. The Comberton one
was at the same date believed to be perfect, but whether either or both
have now disappeared I cannot say. Nor have I been able to verify the
existence or non-existence of the other examples of which I am able to
give illustrations. I shall therefore write of them all in the past
tense, retaining the hope that some are still preserved.

[Illustration: FIG. 6.--Maze at Alkborough, Lincolnshire.]

In the next two mazes given--that at Saffron Walden, Essex (110 feet in
diameter, Fig. 4), and the one near St. Anne's Well, at Sneinton,
Nottinghamshire (Fig. 5), which was ploughed up on February 27th, 1797
(51 feet in diameter, with a path 535 yards long)--the paths must in
each case be understood to be on the lines, black or white, as the case
may be.

[Illustration: FIG. 7.--Maze at Boughton Green, Nottinghamshire.]

I give in Fig. 6 a maze that was at Alkborough, Lincolnshire,
overlooking the Humber. This was 44 feet in diameter, and the
resemblance between it and the mazes at Chartres and Lucca (Figs. 2 and
3) will be at once perceived. A maze at Boughton Green, in
Nottinghamshire, a place celebrated at one time for its fair (Fig. 7),
was 37 feet in diameter. I also include the plan (Fig. 8) of one that
used to be on the outskirts of the village of Wing, near Uppingham,
Rutlandshire. This maze was 40 feet in diameter.

[Illustration: FIG. 8.--Maze at Wing, Rutlandshire.]

[Illustration: FIG. 9.--Maze on St. Catherine's Hill, Winchester.]

The maze that was on St. Catherine's Hill, Winchester, in the parish of
Chilcombe, was a poor specimen (Fig. 9), since, as will be seen, there
was one short direct route to the centre, unless, as in Fig. 10 again,
the path is the line itself from end to end. This maze was 86 feet
square, cut in the turf, and was locally known as the "Mize-maze." It
became very indistinct about 1858, and was then recut by the Warden of
Winchester, with the aid of a plan possessed by a lady living in the
neighbourhood.

[Illustration: FIG. 10.--Maze on Ripon Common.]

A maze formerly existed on Ripon Common, in Yorkshire (Fig. 10). It was
ploughed up in 1827, but its plan was fortunately preserved. This
example was 20 yards in diameter, and its path is said to have been 407
yards long.

[Illustration: FIG. 11.--Maze at Theobalds, Hertfordshire.]

In the case of the maze at Theobalds, Hertfordshire, after you have
found the entrance within the four enclosing hedges, the path is forced
(Fig. 11). As further illustrations of this class of maze, I give one
taken from an Italian work on architecture by Serlio, published in 1537
(Fig. 12), and one by London and Wise, the designers of the Hampton
Court maze, from their book, _The Retired Gard'ner_, published in 1706
(Fig. 13). Also, I add a Dutch maze (Fig. 14).

[Illustration: FIG. 12.--Italian Maze of Sixteenth Century.]

[Illustration: FIG. 13.--By the Designers of Hampton Court Maze.]

[Illustration: FIG. 14.--A Dutch Maze.]

So far our mazes have been of historical interest, but they have
presented no difficulty in threading. After the Reformation period we
find mazes converted into mediums for recreation, and they generally
consisted of labyrinthine paths enclosed by thick and carefully trimmed
hedges. These topiary hedges were known to the Romans, with whom the
_topiarius_ was the ornamental gardener. This type of maze has of late
years degenerated into the seaside "Puzzle Gardens. Teas, sixpence,
including admission to the Maze." The Hampton Court Maze, sometimes
called the "Wilderness," at the royal palace, was designed, as I have
said, by London and Wise for William III., who had a liking for such
things (Fig. 15). I have before me some three or four versions of it,
all slightly different from one another; but the plan I select is taken
from an old guide-book to the palace, and therefore ought to be
trustworthy. The meaning of the dotted lines, etc., will be explained
later on.

[Illustration: FIG. 15.--Maze at Hampton Court Palace.]

[Illustration: FIG. 16.--Maze at Hatfield House, Herts.]

[Illustration: FIG. 17.--Maze formerly at South Kensington.]

[Illustration: FIG. 18.--A German Maze.]

The maze at Hatfield House (Fig. 16), the seat of the Marquis of
Salisbury, like so many labyrinths, is not difficult on paper; but both
this and the Hampton Court Maze may prove very puzzling to actually
thread without knowing the plan. One reason is that one is so apt to go
down the same blind alleys over and over again, if one proceeds without
method. The maze planned by the desire of the Prince Consort for the
Royal Horticultural Society's Gardens at South Kensington was allowed to
go to ruin, and was then destroyed--no great loss, for it was a feeble
thing. It will be seen that there were three entrances from the outside
(Fig. 17), but the way to the centre is very easy to discover. I include
a German maze that is curious, but not difficult to thread on paper
(Fig. 18). The example of a labyrinth formerly existing at Pimperne, in
Dorset, is in a class by itself (Fig. 19). It was formed of small ridges
about a foot high, and covered nearly an acre of ground; but it was,
unfortunately, ploughed up in 1730.

[Illustration: FIG. 19.--Maze at Pimperne, Dorset.]

We will now pass to the interesting subject of how to thread any maze.
While being necessarily brief, I will try to make the matter clear to
readers who have no knowledge of mathematics. And first of all we will
assume that we are trying to enter a maze (that is, get to the "centre")
of which we have no plan and about which we know nothing. The first rule
is this: If a maze has no parts of its hedges detached from the rest,
then if we always keep in touch with the hedge with the right hand (or
always touch it with the left), going down to the stop in every blind
alley and coming back on the other side, we shall pass through every
part of the maze and make our exit where we went in. Therefore we must
at one time or another enter the centre, and every alley will be
traversed twice.

[Illustration: FIG. 20.--M. Tremaux's Method of Solution.]

[Illustration: FIG. 21.--How to thread the Hatfield Maze.]

Now look at the Hampton Court plan. Follow, say to the right, the path
indicated by the dotted line, and what I have said is clearly correct if
we obliterate the two detached parts, or "islands," situated on each
side of the star. But as these islands are there, you cannot by this
method traverse every part of the maze; and if it had been so planned
that the "centre" was, like the star, between the two islands, you would
never pass through the "centre" at all. A glance at the Hatfield maze
will show that there are three of these detached hedges or islands at
the centre, so this method will never take you to the "centre" of that
one. But the rule will at least always bring you safely out again unless
you blunder in the following way. Suppose, when you were going in the
direction of the arrow in the Hampton Court Maze, that you could not
distinctly see the turning at the bottom, that you imagined you were in
a blind alley and, to save time, crossed at once to the opposite hedge,
then you would go round and round that U-shaped island with your right
hand still always on the hedge--for ever after!

[Illustration: FIG. 22. The Philadelphia Maze, and its Solution.]

This blunder happened to me a few years ago in a little maze on the isle
of Caldy, South Wales. I knew the maze was a small one, but after a very
long walk I was amazed to find that I did not either reach the "centre"
or get out again. So I threw a piece of paper on the ground, and soon
came round to it; from which I knew that I had blundered over a supposed
blind alley and was going round and round an island. Crossing to the
opposite hedge and using more care, I was quickly at the centre and out
again. Now, if I had made a similar mistake at Hampton Court, and
discovered the error when at the star, I should merely have passed from
one island to another! And if I had again discovered that I was on a
detached part, I might with ill luck have recrossed to the first island
again! We thus see that this "touching the hedge" method should always
bring us safely out of a maze that we have entered; it may happen to
take us through the "centre," and if we miss the centre we shall know
there must be islands. But it has to be done with a little care, and in
no case can we be sure that we have traversed every alley or that there
are no detached parts.

[Illustration: FIG. 23.--Simplified Diagram of Fig. 22.]

If the maze has many islands, the traversing of the whole of it may be a
matter of considerable difficulty. Here is a method for solving any
maze, due to M. Trémaux, but it necessitates carefully marking in some
way your entrances and exits where the galleries fork. I give a diagram
of an imaginary maze of a very simple character that will serve our
purpose just as well as something more complex (Fig. 20). The circles at
the regions where we have a choice of turnings we may call nodes. A
"new" path or node is one that has not been entered before on the route;
an "old" path or node is one that has already been entered, 1. No path
may be traversed more than twice. 2. When you come to a new node, take
any path you like. 3. When by a new path you come to an old node or to
the stop of a blind alley, return by the path you came. 4. When by an
old path you come to an old node, take a new path if there is one; if
not, an old path. The route indicated by the dotted line in the diagram
is taken in accordance with these simple rules, and it will be seen
that it leads us to the centre, although the maze consists of four
islands.

[Illustration: FIG. 24.--Can you find the Shortest Way to Centre?]

Neither of the methods I have given will disclose to us the shortest way
to the centre, nor the number of the different routes. But we can easily
settle these points with a plan. Let us take the Hatfield maze (Fig.
21). It will be seen that I have suppressed all the blind alleys by the
shading. I begin at the stop and work backwards until the path forks.
These shaded parts, therefore, can never be entered without our having
to retrace our steps. Then it is very clearly seen that if we enter at A
we must come out at B; if we enter at C we must come out at D. Then we
have merely to determine whether A, B, E, or C, D, E, is the shorter
route. As a matter of fact, it will be found by rough measurement or
calculation that the shortest route to the centre is by way of C, D, E,
F.

[Illustration: FIG. 25.--Rosamund's Bower.]

I will now give three mazes that are simply puzzles on paper, for, so
far as I know, they have never been constructed in any other way. The
first I will call the Philadelphia maze (Fig. 22). Fourteen years ago a
travelling salesman, living in Philadelphia, U.S.A., developed a
curiously unrestrained passion for puzzles. He neglected his business,
and soon his position was taken from him. His days and nights were now
passed with the subject that fascinated him, and this little maze seems
to have driven him into insanity. He had been puzzling over it for some
time, and finally it sent him mad and caused him to fire a bullet
through his brain. Goodness knows what his difficulties could have been!
But there can be little doubt that he had a disordered mind, and that if
this little puzzle had not caused him to lose his mental balance some
other more or less trivial thing would in time have done so. There is no
moral in the story, unless it be that of the Irish maxim, which applies
to every occupation of life as much as to the solving of puzzles: "Take
things aisy; if you can't take them aisy, take them as aisy as you can."
And it is a bad and empirical way of solving any puzzle--by blowing your
brains out.

Now, how many different routes are there from A to B in this maze if we
must never in any route go along the same passage twice? The four open
spaces where four passages end are not reckoned as "passages." In the
diagram (Fig. 22) it will be seen that I have again suppressed the blind
alleys. It will be found that, in any case, we must go from A to C, and
also from F to B. But when we have arrived at C there are three ways,
marked 1, 2, 3, of getting to D. Similarly, when we get to E there are
three ways, marked 4, 5, 6, of getting to F. We have also the dotted
route from C to E, the other dotted route from D to F, and the passage
from D to E, indicated by stars. We can, therefore, express the position
of affairs by the little diagram annexed (Fig. 23). Here every
condition of route exactly corresponds to that in the circular maze,
only it is much less confusing to the eye. Now, the number of routes,
under the conditions, from A to B on this simplified diagram is 640, and
that is the required answer to the maze puzzle.

Finally, I will leave two easy maze puzzles (Figs. 24, 25) for my
readers to solve for themselves. The puzzle in each case is to find the
shortest possible route to the centre. Everybody knows the story of Fair
Rosamund and the Woodstock maze. What the maze was like or whether it
ever existed except in imagination is not known, many writers believing
that it was simply a badly-constructed house with a large number of
confusing rooms and passages. At any rate, my sketch lacks the authority
of the other mazes in this article. My "Rosamund's Bower" is simply
designed to show that where you have the plan before you it often
happens that the easiest way to find a route into a maze is by working
backwards and first finding a way out.



THE PARADOX PARTY.

    "Is not life itself a paradox?"
        C.L. DODGSON, _Pillow Problems_.


"It is a wonderful age!" said Mr. Allgood, and everybody at the table
turned towards him and assumed an attitude of expectancy.

This was an ordinary Christmas dinner of the Allgood family, with a
sprinkling of local friends. Nobody would have supposed that the above
remark would lead, as it did, to a succession of curious puzzles and
paradoxes, to which every member of the party contributed something of
interest. The little symposium was quite unpremeditated, so we must not
be too critical respecting a few of the posers that were forthcoming.
The varied character of the contributions is just what we would expect
on such an occasion, for it was a gathering not of expert mathematicians
and logicians, but of quite ordinary folk.

"It is a wonderful age!" repeated Mr. Allgood. "A man has just designed
a square house in such a cunning manner that all the windows on the four
sides have a south aspect."

"That would appeal to me," said Mrs. Allgood, "for I cannot endure a
room with a north aspect."

"I cannot conceive how it is done," Uncle John confessed. "I suppose he
puts bay windows on the east and west sides; but how on earth can be
contrive to look south from the north side? Does he use mirrors, or
something of that kind?"

"No," replied Mr. Allgood, "nothing of the sort. All the windows are
flush with the walls, and yet you get a southerly prospect from every
one of them. You see, there is no real difficulty in designing the house
if you select the proper spot for its erection. Now, this house is
designed for a gentleman who proposes to build it exactly at the North
Pole. If you think a moment you will realize that when you stand at the
North Pole it is impossible, no matter which way you may turn, to look
elsewhere than due south! There are no such directions as north, east,
or west when you are exactly at the North Pole. Everything is due
south!"

"I am afraid, mother," said her son George, after the laughter had
subsided, "that, however much you might like the aspect, the situation
would be a little too bracing for you."

"Ah, well!" she replied. "Your Uncle John fell also into the trap. I am
no good at catches and puzzles. I suppose I haven't the right sort of
brain. Perhaps some one will explain this to me. Only last week I
remarked to my hairdresser that it had been said that there are more
persons in the world than any one of them has hairs on his head. He
replied, 'Then it follows, madam, that two persons, at least, must have
exactly the same number of hairs on their heads.' If this is a fact, I
confess I cannot see it."

"How do the bald-headed affect the question?" asked Uncle John.

"If there are such persons in existence," replied Mrs. Allgood, "who
haven't a solitary hair on their heads discoverable under a
magnifying-glass, we will leave them out of the question. Still, I
don't see how you are to prove that at least two persons have exactly
the same number to a hair."

"I think I can make it clear," said Mr. Filkins, who had dropped in for
the evening. "Assume the population of the world to be only one million.
Any number will do as well as another. Then your statement was to the
effect that no person has more than nine hundred and ninety-nine
thousand nine hundred and ninety-nine hairs on his head. Is that so?"

"Let me think," said Mrs. Allgood. "Yes--yes--that is correct."

"Very well, then. As there are only nine hundred and ninety-nine
thousand nine hundred and ninety-nine _different_ ways of bearing hair,
it is clear that the millionth person must repeat one of those ways. Do
you see?"

"Yes; I see that--at least I think I see it."

"Therefore two persons at least must have the same number of hairs on
their heads; and as the number of people on the earth so greatly exceeds
the number of hairs on any one person's head, there must, of course, be
an immense number of these repetitions."

"But, Mr. Filkins," said little Willie Allgood, "why could not the
millionth man have, say, ten thousand hairs and a half?"

"That is mere hair-splitting, Willie, and does not come into the
question."

"Here is a curious paradox," said George. "If a thousand soldiers are
drawn up in battle array on a plane"--they understood him to mean
"plain"--"only one man will stand upright."

Nobody could see why. But George explained that, according to Euclid, a
plane can touch a sphere only at one point, and that person only who
stands at that point, with respect to the centre of the earth, will
stand upright.

"In the same way," he remarked, "if a billiard-table were quite
level--that is, a perfect plane--the balls ought to roll to the centre."

Though he tried to explain this by placing a visiting-card on an orange
and expounding the law of gravitation, Mrs. Allgood declined to accept
the statement. She could not see that the top of a true billiard-table
must, theoretically, be spherical, just like a portion of the
orange-peel that George cut out. Of course, the table is so small in
proportion to the surface of the earth that the curvature is not
appreciable, but it is nevertheless true in theory. A surface that we
call level is not the same as our idea of a true geometrical plane.

"Uncle John," broke in Willie Allgood, "there is a certain island
situated between England and France, and yet that island is farther from
France than England is. What is the island?"

"That seems absurd, my boy; because if I place this tumbler, to
represent the island, between these two plates, it seems impossible that
the tumbler can be farther from either of the plates than they are from
each other."

"But isn't Guernsey between England and France?" asked Willie.

"Yes, certainly."

"Well, then, I think you will find, uncle, that Guernsey is about
twenty-six miles from France, and England is only twenty-one miles from
France, between Calais and Dover."

"My mathematical master," said George, "has been trying to induce me to
accept the axiom that 'if equals be multiplied by equals the products
are equal.'"

"It is self-evident," pointed out Mr. Filkins. "For example, if 3 feet
equal 1 yard, then twice 3 feet will equal 2 yards. Do you see?"

"But, Mr. Filkins," asked George, "is this tumbler half full of water
equal to a similar glass half empty?"

"Certainly, George."

"Then it follows from the axiom that a glass full must equal a glass
empty. Is that correct?"

"No, clearly not. I never thought of it in that light."

"Perhaps," suggested Mr. Allgood, "the rule does not apply to liquids."

"Just what I was thinking, Allgood. It would seem that we must make an
exception in the case of liquids."

"But it would be awkward," said George, with a smile, "if we also had to
except the case of solids. For instance, let us take the solid earth.
One mile square equals one square mile. Therefore two miles square must
equal two square miles. Is this so?"

"Well, let me see! No, of course not," Mr. Filkins replied, "because two
miles square is four square miles."

"Then," said George, "if the axiom is not true in these cases, when is
it true?"

Mr. Filkins promised to look into the matter, and perhaps the reader
will also like to give it consideration at leisure.

"Look here, George," said his cousin Reginald Woolley: "by what
fractional part does four-fourths exceed three-fourths?"

"By one-fourth!" shouted everybody at once.

"Try another one," George suggested.

"With pleasure, when you have answered that one correctly," was
Reginald's reply.

"Do you mean to say that it isn't one-fourth?"

"Certainly I do."

Several members of the company failed to see that the correct answer is
"one-third," although Reginald tried to explain that three of anything,
if increased by one-third, becomes four.

"Uncle John, how do you pronounce 't-o-o'?" asked Willie.

"'Too," my boy."

"And how do you pronounce 't-w-o'?"

"That is also 'too.'"

"Then how do you pronounce the second day of the week?"

"Well, that I should pronounce 'Tuesday,' not 'Toosday.'"

"Would you really? I should pronounce it 'Monday.'"

"If you go on like this, Willie," said Uncle John, with mock severity,
"you will soon be without a friend in the world."

"Can any of you write down quickly in figures 'twelve thousand twelve
hundred and twelve pounds'?" asked Mr. Allgood.

His eldest daughter, Miss Mildred, was the only person who happened to
have a pencil at hand.

"It can't be done," she declared, after making an attempt on the white
table-cloth; but Mr. Allgood showed her that it should be written,
"£13,212."

"Now it is my turn," said Mildred. "I have been waiting to ask you all a
question. In the Massacre of the Innocents under Herod, a number of poor
little children were buried in the sand with only their feet sticking
out. How might you distinguish the boys from the girls?"

"I suppose," said Mrs. Allgood, "it is a conundrum--something to do with
their poor little 'souls.'"

But after everybody had given it up, Mildred reminded the company that
only boys were put to death.

"Once upon a time," began George, "Achilles had a race with a
tortoise--"

"Stop, George!" interposed Mr. Allgood. "We won't have that one. I knew
two men in my youth who were once the best of friends, but they
quarrelled over that infernal thing of Zeno's, and they never spoke to
one another again for the rest of their lives. I draw the line at that,
and the other stupid thing by Zeno about the flying arrow. I don't
believe anybody understands them, because I could never do so myself."

"Oh, very well, then, father. Here is another. The Post-Office people
were about to erect a line of telegraph-posts over a high hill from
Turmitville to Wurzleton; but as it was found that a railway company was
making a deep level cutting in the same direction, they arranged to put
up the posts beside the line. Now, the posts were to be a hundred yards
apart, the length of the road over the hill being five miles, and the
length of the level cutting only four and a half miles. How many posts
did they save by erecting them on the level?"

"That is a very simple matter of calculation," said Mr. Filkins. "Find
how many times one hundred yards will go in five miles, and how many
times in four and a half miles. Then deduct one from the other, and you
have the number of posts saved by the shorter route."

"Quite right," confirmed Mr. Allgood. "Nothing could be easier."

"That is just what the Post-Office people said," replied George, "but it
is quite wrong. If you look at this sketch that I have just made, you
will see that there is no difference whatever. If the posts are a
hundred yards apart, just the same number will be required on the level
as over the surface of the hill."

[Illustration]

"Surely you must be wrong, George," said Mrs. Allgood, "for if the posts
are a hundred yards apart and it is half a mile farther over the hill,
you have to put up posts on that extra half-mile."

"Look at the diagram, mother. You will see that the distance from post
to post is not the distance from base to base measured along the ground.
I am just the same distance from you if I stand on this spot on the
carpet or stand immediately above it on the chair."

But Mrs. Allgood was not convinced.

Mr. Smoothly, the curate, at the end of the table, said at this point
that he had a little question to ask.

"Suppose the earth were a perfect sphere with a smooth surface, and a
girdle of steel were placed round the Equator so that it touched at
every point."

"'I'll put a girdle round about the earth in forty minutes,'" muttered
George, quoting the words of Puck in _A Midsummer Night's Dream_.

"Now, if six yards were added to the length of the girdle, what would
then be the distance between the girdle and the earth, supposing that
distance to be equal all round?"

"In such a great length," said Mr. Allgood, "I do not suppose the
distance would be worth mentioning."

"What do you say, George?" asked Mr. Smoothly.

"Well, without calculating I should imagine it would be a very minute
fraction of an inch."

Reginald and Mr. Filkins were of the same opinion.

"I think it will surprise you all," said the curate, "to learn that
those extra six yards would make the distance from the earth all round
the girdle very nearly a yard!"

"Very nearly a yard!" everybody exclaimed, with astonishment; but Mr.
Smoothly was quite correct. The increase is independent of the original
length of the girdle, which may be round the earth or round an orange;
in any case the additional six yards will give a distance of nearly a
yard all round. This is apt to surprise the non-mathematical mind.

"Did you hear the story of the extraordinary precocity of Mrs. Perkins's
baby that died last week?" asked Mrs. Allgood. "It was only three months
old, and lying at the point of death, when the grief-stricken mother
asked the doctor if nothing could save it. 'Absolutely nothing!' said
the doctor. Then the infant looked up pitifully into its mother's face
and said--absolutely nothing!"

"Impossible!" insisted Mildred. "And only three months old!"

"There have been extraordinary cases of infantile precocity," said Mr.
Filkins, "the truth of which has often been carefully attested. But are
you sure this really happened, Mrs. Allgood?"

"Positive," replied the lady. "But do you really think it astonishing
that a child of three months should say absolutely nothing? What would
you expect it to say?"

"Speaking of death," said Mr. Smoothly, solemnly, "I knew two men,
father and son, who died in the same battle during the South African
War. They were both named Andrew Johnson and buried side by side, but
there was some difficulty in distinguishing them on the headstones. What
would you have done?"

"Quite simple," said Mr. Allgood. "They should have described one as
'Andrew Johnson, Senior,' and the other as 'Andrew Johnson, Junior.'"

"But I forgot to tell you that the father died first."

"What difference can that make?"

"Well, you see, they wanted to be absolutely exact, and that was the
difficulty."

"But I don't see any difficulty," said Mr. Allgood, nor could anybody
else.

"Well," explained Mr. Smoothly, "it is like this. If the father died
first, the son was then no longer 'Junior.' Is that so?"

"To be strictly exact, yes."

"That is just what they wanted--to be strictly exact. Now, if he was no
longer 'Junior,' then he did not die 'Junior." Consequently it must be
incorrect so to describe him on the headstone. Do you see the point?"

"Here is a rather curious thing," said Mr. Filkins, "that I have just
remembered. A man wrote to me the other day that he had recently
discovered two old coins while digging in his garden. One was dated '51
B.C.,' and the other one marked 'George I.' How do I know that he was
not writing the truth?"

"Perhaps you know the man to be addicted to lying," said Reginald.

"But that would be no proof that he was not telling the truth in this
instance."

"Perhaps," suggested Mildred, "you know that there were no coins made at
those dates.

"On the contrary, they were made at both periods."

"Were they silver or copper coins?" asked Willie.

"My friend did not state, and I really cannot see, Willie, that it makes
any difference."

"I see it!" shouted Reginald. "The letters 'B.C.' would never be used on
a coin made before the birth of Christ. They never anticipated the event
in that way. The letters were only adopted later to denote dates
previous to those which we call 'A.D.' That is very good; but I cannot
see why the other statement could not be correct."

"Reginald is quite right," said Mr. Filkins, "about the first coin. The
second one could not exist, because the first George would never be
described in his lifetime as 'George I.'"

"Why not?" asked Mrs. Allgood. "He _was_ George I."

"Yes; but they would not know it until there was a George II."

"Then there was no George II. until George III. came to the throne?"

"That does not follow. The second George becomes 'George II.' on account
of there having been a 'George I.'"

"Then the first George was 'George I.' on account of there having been
no king of that name before him."

"Don't you see, mother," said George Allgood, "we did not call Queen
Victoria 'Victoria I.;' but if there is ever a 'Victoria II.,' then she
will be known that way."

"But there _have_ been several Georges, and therefore he was 'George I.'
There _haven't_ been several Victorias, so the two cases are not
similar."

They gave up the attempt to convince Mrs. Allgood, but the reader will,
of course, see the point clearly.

"Here is a question," said Mildred Allgood, "that I should like some of
you to settle for me. I am accustomed to buy from our greengrocer
bundles of asparagus, each 12 inches in circumference. I always put a
tape measure round them to make sure I am getting the full quantity. The
other day the man had no large bundles in stock, but handed me instead
two small ones, each 6 inches in circumference. 'That is the same
thing,' I said, 'and, of course, the price will be the same;' but he
insisted that the two bundles together contained more than the large
one, and charged me a few pence extra. Now, what I want to know is,
which of us was correct? Would the two small bundles contain the same
quantity as the large one? Or would they contain more?"

"That is the ancient puzzle," said Reginald, laughing, "of the sack of
corn that Sempronius borrowed from Caius, which your greengrocer,
perhaps, had been reading about somewhere. He caught you beautifully."

"Then they were equal?"

"On the contrary, you were both wrong, and you were badly cheated. You
only got half the quantity that would have been contained in a large
bundle, and therefore ought to have been charged half the original
price, instead of more."

Yes, it was a bad swindle, undoubtedly. A circle with a circumference
half that of another must have its area a quarter that of the other.
Therefore the two small bundles contained together only half as much
asparagus as a large one.

"Mr. Filkins, can you answer this?" asked Willie. "There is a man in the
next village who eats two eggs for breakfast every morning."

"Nothing very extraordinary in that," George broke in. "If you told us
that the two eggs ate the man it would be interesting."

"Don't interrupt the boy, George," said his mother.

"Well," Willie continued, "this man neither buys, borrows, barters,
begs, steals, nor finds the eggs. He doesn't keep hens, and the eggs are
not given to him. How does he get the eggs?"

"Does he take them in exchange for something else?" asked Mildred.

"That would be bartering them," Willie replied.

"Perhaps some friend sends them to him," suggested Mrs. Allgood.

"I said that they were not given to him."

"I know," said George, with confidence. "A strange hen comes into his
place and lays them."

"But that would be finding them, wouldn't it?"

"Does he hire them?" asked Reginald.

"If so, he could not return them after they were eaten, so that would be
stealing them."

"Perhaps it is a pun on the word 'lay,'" Mr. Filkins said. "Does he lay
them on the table?"

"He would have to get them first, wouldn't he? The question was, How
does he get them?"

"Give it up!" said everybody. Then little Willie crept round to the
protection of his mother, for George was apt to be rough on such
occasions.

"The man keeps ducks!" he cried, "and his servant collects the eggs
every morning."

"But you said he doesn't keep birds!" George protested.

"I didn't, did I, Mr. Filkins? I said he doesn't keep hens."

"But he finds them," said Reginald.

"No; I said his servant finds them."

"Well, then," Mildred interposed, "his servant gives them to him."

"You cannot give a man his own property, can you?"

All agreed that Willie's answer was quite satisfactory. Then Uncle John
produced a little fallacy that "brought the proceedings to a close," as
the newspapers say.


413.--A CHESSBOARD FALLACY.

[Illustration]

"Here is a diagram of a chessboard," he said. "You see there are
sixty-four squares--eight by eight. Now I draw a straight line from the
top left-hand corner, where the first and second squares meet, to the
bottom right-hand corner. I cut along this line with the scissors, slide
up the piece that I have marked B, and then clip off the little corner C
by a cut along the first upright line. This little piece will exactly
fit into its place at the top, and we now have an oblong with seven
squares on one side and nine squares on the other. There are, therefore,
now only sixty-three squares, because seven multiplied by nine makes
sixty-three. Where on earth does that lost square go to? I have tried
over and over again to catch the little beggar, but he always eludes me.
For the life of me I cannot discover where he hides himself."

"It seems to be like the other old chessboard fallacy, and perhaps the
explanation is the same," said Reginald--"that the pieces do not exactly
fit."

"But they _do_ fit," said Uncle John. "Try it, and you will see."

Later in the evening Reginald and George, were seen in a corner with
their heads together, trying to catch that elusive little square, and it
is only fair to record that before they retired for the night they
succeeded in securing their prey, though some others of the company
failed to see it when captured. Can the reader solve the little mystery?



UNCLASSIFIED PROBLEMS.


    "A snapper up of unconsidered trifles."
              _Winter's Tale_, iv. 2.


414.--WHO WAS FIRST?

Anderson, Biggs, and Carpenter were staying together at a place by the
seaside. One day they went out in a boat and were a mile at sea when a
rifle was fired on shore in their direction. Why or by whom the shot was
fired fortunately does not concern us, as no information on these points
is obtainable, but from the facts I picked up we can get material for a
curious little puzzle for the novice.

It seems that Anderson only heard the report of the gun, Biggs only saw
the smoke, and Carpenter merely saw the bullet strike the water near
them. Now, the question arises: Which of them first knew of the
discharge of the rifle?


415.--A WONDERFUL VILLAGE.

There is a certain village in Japan, situated in a very low valley, and
yet the sun is nearer to the inhabitants every noon, by 3,000 miles and
upwards, than when he either rises or sets to these people. In what part
of the country is the village situated?


416.--A CALENDAR PUZZLE.

If the end of the world should come on the first day of a new century,
can you say what are the chances that it will happen on a Sunday?


417.--THE TIRING IRONS.

[Illustration]

The illustration represents one of the most ancient of all mechanical
puzzles. Its origin is unknown. Cardan, the mathematician, wrote about
it in 1550, and Wallis in 1693; while it is said still to be found in
obscure English villages (sometimes deposited in strange places, such as
a church belfry), made of iron, and appropriately called "tiring-irons,"
and to be used by the Norwegians to-day as a lock for boxes and bags. In
the toyshops it is sometimes called the "Chinese rings," though there
seems to be no authority for the description, and it more frequently
goes by the unsatisfactory name of "the puzzling rings." The French call
it "Baguenaudier."

The puzzle will be seen to consist of a simple _loop_ of wire fixed in a
handle to be held in the left hand, and a certain number of _rings_
secured by _wires_ which pass through holes in the _bar_ and are kept
there by their blunted ends. The wires work freely in the bar, but
cannot come apart from it, nor can the wires be removed from the rings.
The general puzzle is to detach the loop completely from all the rings,
and then to put them all on again.

Now, it will be seen at a glance that the first ring (to the right) can
be taken off at any time by sliding it over the end and dropping it
through the loop; or it may be put on by reversing the operation. With
this exception, the only ring that can ever be removed is the one that
happens to be a contiguous second on the loop at the right-hand end.
Thus, with all the rings on, the second can be dropped at once; with the
first ring down, you cannot drop the second, but may remove the third;
with the first three rings down, you cannot drop the fourth, but may
remove the fifth; and so on. It will be found that the first and second
rings can be dropped together or put on together; but to prevent
confusion we will throughout disallow this exceptional double move, and
say that only one ring may be put on or removed at a time.

We can thus take off one ring in 1 move; two rings in 2 moves; three
rings in 5 moves; four rings in 10 moves; five rings in 21 moves; and if
we keep on doubling (and adding one where the number of rings is odd) we
may easily ascertain the number of moves for completely removing any
number of rings. To get off all the seven rings requires 85 moves. Let
us look at the five moves made in removing the first three rings, the
circles above the line standing for rings on the loop and those under
for rings off the loop.

Drop the first ring; drop the third; put up the first; drop the second;
and drop the first--5 moves, as shown clearly in the diagrams. The dark
circles show at each stage, from the starting position to the finish,
which rings it is possible to drop. After move 2 it will be noticed that
no ring can be dropped until one has been put on, because the first and
second rings from the right now on the loop are not together. After the
fifth move, if we wish to remove all seven rings we must now drop the
fifth. But before we can then remove the fourth it is necessary to put
on the first three and remove the first two. We shall then have 7, 6, 4,
3 on the loop, and may therefore drop the fourth. When we have put on 2
and 1 and removed 3, 2, 1, we may drop the seventh ring. The next
operation then will be to get 6, 5, 4, 3, 2, 1 on the loop and remove 4,
3, 2, 1, when 6 will come off; then get 5, 4, 3, 2, 1 on the loop, and
remove 3, 2, 1, when 5 will come off; then get 4, 3, 2, 1 on the loop
and remove 2, 1, when 4 will come off; then get 3, 2, 1 on the loop and
remove 1, when 3 will come off; then get 2, 1 on the loop, when 2 will
come off; and 1 will fall through on the 85th move, leaving the loop
quite free. The reader should now be able to understand the puzzle,
whether or not he has it in his hand in a practical form.

[Illustration]

[Illustration:


      o o o o o * *
     {-------------

      o o o o * o
    1{------------- o

      o o o o   o
    2{-------------
              o   o

      o o o o   * *
    3{-------------
              o

      o o o o     *
    4{-------------
              o o

      o o * o
    5{-------------
              o o o

]

The particular problem I propose is simply this. Suppose there are
altogether fourteen rings on the tiring-irons, and we proceed to take
them all off in the correct way so as not to waste any moves. What will
be the position of the rings after the 9,999th move has been made?



418.--SUCH A GETTING UPSTAIRS.

In a suburban villa there is a small staircase with eight steps, not
counting the landing. The little puzzle with which Tommy Smart perplexed
his family is this. You are required to start from the bottom and land
twice on the floor above (stopping there at the finish), having returned
once to the ground floor. But you must be careful to use every tread the
same number of times. In how few steps can you make the ascent? It seems
a very simple matter, but it is more than likely that at your first
attempt you will make a great many more steps than are necessary. Of
course you must not go more than one riser at a time.

Tommy knows the trick, and has shown it to his father, who professes to
have a contempt for such things; but when the children are in bed the
pater will often take friends out into the hall and enjoy a good laugh
at their bewilderment. And yet it is all so very simple when you know
how it is done.


419.--THE FIVE PENNIES.

Here is a really hard puzzle, and yet its conditions are so absurdly
simple. Every reader knows how to place four pennies so that they are
equidistant from each other. All you have to do is to arrange three of
them flat on the table so that they touch one another in the form of a
triangle, and lay the fourth penny on top in the centre. Then, as every
penny touches every other penny, they are all at equal distances from
one another. Now try to do the same thing with five pennies--place them
so that every penny shall touch every other penny--and you will find it
a different matter altogether.


420.--THE INDUSTRIOUS BOOKWORM.

[Illustration]

Our friend Professor Rackbrane is seen in the illustration to be
propounding another of his little posers. He is explaining that since he
last had occasion to take down those three volumes of a learned book
from their place on his shelves a bookworm has actually bored a hole
straight through from the first page to the last. He says that the
leaves are together three inches thick in each volume, and that every
cover is exactly one-eighth of an inch thick, and he asks how long a
tunnel had the industrious worm to bore in preparing his new tube
railway. Can you tell him?


421.--A CHAIN PUZZLE.

[Illustration]

This is a puzzle based on a pretty little idea first dealt with by the
late Mr. Sam Loyd. A man had nine pieces of chain, as shown in the
illustration. He wanted to join these fifty links into one endless
chain. It will cost a penny to open any link and twopence to weld a link
together again, but he could buy a new endless chain of the same
character and quality for 2s. 2d. What was the cheapest course for him
to adopt? Unless the reader is cunning he may find himself a good way
out in his answer.


422.--THE SABBATH PUZZLE.

I have come across the following little poser in an old book. I wonder
how many readers will see the author's intended solution to the riddle.

    Christians the week's _first_ day for Sabbath hold;
    The Jews the _seventh_, as they did of old;
    The Turks the _sixth_, as we have oft been told.
    How can these three, in the same place and day,
    Have each his own true Sabbath? tell, I pray.


423.--THE RUBY BROOCH.

The annals of Scotland Yard contain some remarkable cases of jewel
robberies, but one of the most perplexing was the theft of Lady
Littlewood's rubies. There have, of course, been many greater robberies
in point of value, but few so artfully conceived. Lady Littlewood, of
Romley Manor, had a beautiful but rather eccentric heirloom in the form
of a ruby brooch. While staying at her town house early in the eighties
she took the jewel to a shop in Brompton for some slight repairs.

"A fine collection of rubies, madam," said the shopkeeper, to whom her
ladyship was a stranger.

"Yes," she replied; "but curiously enough I have never actually counted
them. My mother once pointed out to me that if you start from the centre
and count up one line, along the outside and down the next line, there
are always eight rubies. So I should always know if a stone were
missing."

[Illustration]

Six months later a brother of Lady Littlewood's, who had returned from
his regiment in India, noticed that his sister was wearing the ruby
brooch one night at a county ball, and on their return home asked to
look at it more closely. He immediately detected the fact that four of
the stones were gone.

"How can that possibly be?" said Lady Littlewood. "If you count up one
line from the centre, along the edge, and down the next line, in any
direction, there are always eight stones. This was always so and is so
now. How, therefore, would it be possible to remove a stone without my
detecting it?"

"Nothing could be simpler," replied the brother. "I know the brooch
well. It originally contained forty-five stones, and there are now only
forty-one. Somebody has stolen four rubies, and then reset as small a
number of the others as possible in such a way that there shall always
be eight in any of the directions you have mentioned."

There was not the slightest doubt that the Brompton jeweller was the
thief, and the matter was placed in the hands of the police. But the man
was wanted for other robberies, and had left the neighbourhood some time
before. To this day he has never been found.

The interesting little point that at first baffled the police, and which
forms the subject of our puzzle, is this: How were the forty-five rubies
originally arranged on the brooch? The illustration shows exactly how
the forty-one were arranged after it came back from the jeweller; but
although they count eight correctly in any of the directions mentioned,
there are four stones missing.


424.--THE DOVETAILED BLOCK.

[Illustration]

Here is a curious mechanical puzzle that was given to me some years ago,
but I cannot say who first invented it. It consists of two solid blocks
of wood securely dovetailed together. On the other two vertical sides
that are not visible the appearance is precisely the same as on those
shown. How were the pieces put together? When I published this little
puzzle in a London newspaper I received (though they were unsolicited)
quite a stack of models, in oak, in teak, in mahogany, rosewood,
satinwood, elm, and deal; some half a foot in length, and others varying
in size right down to a delicate little model about half an inch square.
It seemed to create considerable interest.


425.--JACK AND THE BEANSTALK.

[Illustration]

The illustration, by a British artist, is a sketch of Jack climbing the
beanstalk. Now, the artist has made a serious blunder in this drawing.
Can you find out what it is?


426.--THE HYMN-BOARD POSER.

The worthy vicar of Chumpley St. Winifred is in great distress. A little
church difficulty has arisen that all the combined intelligence of the
parish seems unable to surmount. What this difficulty is I will state
hereafter, but it may add to the interest of the problem if I first give
a short account of the curious position that has been brought about. It
all has to do with the church hymn-boards, the plates of which have
become so damaged that they have ceased to fulfil the purpose for which
they were devised. A generous parishioner has promised to pay for a new
set of plates at a certain rate of cost; but strange as it may seem, no
agreement can be come to as to what that cost should be. The proposed
maker of the plates has named a price which the donor declares to be
absurd. The good vicar thinks they are both wrong, so he asks the
schoolmaster to work out the little sum. But this individual declares
that he can find no rule bearing on the subject in any of his arithmetic
books. An application having been made to the local medical
practitioner, as a man of more than average intellect at Chumpley, he
has assured the vicar that his practice is so heavy that he has not had
time even to look at it, though his assistant whispers that the doctor
has been sitting up unusually late for several nights past. Widow Wilson
has a smart son, who is reputed to have once won a prize for
puzzle-solving. He asserts that as he cannot find any solution to the
problem it must have something to do with the squaring of the circle,
the duplication of the cube, or the trisection of an angle; at any rate,
he has never before seen a puzzle on the principle, and he gives it up.

[Illustration]

This was the state of affairs when the assistant curate (who, I should
say, had frankly confessed from the first that a profound study of
theology had knocked out of his head all the knowledge of mathematics he
ever possessed) kindly sent me the puzzle.

A church has three hymn-boards, each to indicate the numbers of five
different hymns to be sung at a service. All the boards are in use at
the same service. The hymn-book contains 700 hymns. A new set of numbers
is required, and a kind parishioner offers to present a set painted on
metal plates, but stipulates that only the smallest number of plates
necessary shall be purchased. The cost of each plate is to be 6d., and
for the painting of each plate the charges are to be: For one plate,
1s.; for two plates alike, 11¾d. each; for three plates alike,
11½d. each, and so on, the charge being one farthing less per plate
for each similarly painted plate. Now, what should be the lowest cost?

Readers will note that they are required to use every legitimate and
practical method of economy. The illustration will make clear the nature
of the three hymn-boards and plates. The five hymns are here indicated
by means of twelve plates. These plates slide in separately at the back,
and in the illustration there is room, of course, for three more plates.


427.--PHEASANT-SHOOTING.

A Cockney friend, who is very apt to draw the long bow, and is evidently
less of a sportsman than he pretends to be, relates to me the following
not very credible yarn:--

"I've just been pheasant-shooting with my friend the duke. We had
splendid sport, and I made some wonderful shots. What do you think of
this, for instance? Perhaps you can twist it into a puzzle. The duke and
I were crossing a field when suddenly twenty-four pheasants rose on the
wing right in front of us. I fired, and two-thirds of them dropped dead
at my feet. Then the duke had a shot at what were left, and brought down
three-twenty-fourths of them, wounded in the wing. Now, out of those
twenty-four birds, how many still remained?"

It seems a simple enough question, but can the reader give a correct
answer?


428.--THE GARDENER AND THE COOK.

A correspondent, signing himself "Simple Simon," suggested that I should
give a special catch puzzle in the issue of _The Weekly Dispatch_ for
All Fools' Day, 1900. So I gave the following, and it caused
considerable amusement; for out of a very large body of competitors,
many quite expert, not a single person solved it, though it ran for
nearly a month.

[Illustration]

"The illustration is a fancy sketch of my correspondent, 'Simple Simon,'
in the act of trying to solve the following innocent little arithmetical
puzzle. A race between a man and a woman that I happened to witness one
All Fools' Day has fixed itself indelibly on my memory. It happened at a
country-house, where the gardener and the cook decided to run a race to
a point 100 feet straight away and return. I found that the gardener ran
3 feet at every bound and the cook only 2 feet, but then she made three
bounds to his two. Now, what was the result of the race?"

A fortnight after publication I added the following note: "It has been
suggested that perhaps there is a catch in the 'return,' but there is
not. The race is to a point 100 feet away and home again--that is, a
distance of 200 feet. One correspondent asks whether they take exactly
the same time in turning, to which I reply that they do. Another seems
to suspect that it is really a conundrum, and that the answer is that
'the result of the race was a (matrimonial) tie.' But I had no such
intention. The puzzle is an arithmetical one, as it purports to be."


429.--PLACING HALFPENNIES.

[Illustration]

Here is an interesting little puzzle suggested to me by Mr. W. T. Whyte.
Mark off on a sheet of paper a rectangular space 5 inches by 3 inches,
and then find the greatest number of halfpennies that can be placed
within the enclosure under the following conditions. A halfpenny is
exactly an inch in diameter. Place your first halfpenny where you like,
then place your second coin at exactly the distance of an inch from the
first, the third an inch distance from the second, and so on. No
halfpenny may touch another halfpenny or cross the boundary. Our
illustration will make the matter perfectly clear. No. 2 coin is an inch
from No. 1; No. 3 an inch from No. 2; No. 4 an inch from No. 3; but
after No. 10 is placed we can go no further in this attempt. Yet several
more halfpennies might have been got in. How many can the reader place?


430.--FIND THE MAN'S WIFE.

[Illustration]

One summer day in 1903 I was loitering on the Brighton front, watching
the people strolling about on the beach, when the friend who was with me
suddenly drew my attention to an individual who was standing alone, and
said, "Can you point out that man's wife? They are stopping at the same
hotel as I am, and the lady is one of those in view." After a few
minutes' observation, I was successful in indicating the lady correctly.
My friend was curious to know by what method of reasoning I had arrived
at the result. This was my answer:--

"We may at once exclude that Sister of Mercy and the girl in the short
frock; also the woman selling oranges. It cannot be the lady in widows'
weeds. It is not the lady in the bath chair, because she is not staying
at your hotel, for I happened to see her come out of a private house
this morning assisted by her maid. The two ladies in red breakfasted at
my hotel this morning, and as they were not wearing outdoor dress I
conclude they are staying there. It therefore rests between the lady in
blue and the one with the green parasol. But the left hand that holds
the parasol is, you see, ungloved and bears no wedding-ring.
Consequently I am driven to the conclusion that the lady in blue is the
man's wife--and you say this is correct."

Now, as my friend was an artist, and as I thought an amusing puzzle
might be devised on the lines of his question, I asked him to make me a
drawing according to some directions that I gave him, and I have
pleasure in presenting his production to my readers. It will be seen
that the picture shows six men and six ladies: Nos. 1, 3, 5, 7, 9, and
11 are ladies, and Nos. 2, 4, 6, 8, 10, and 12 are men. These twelve
individuals represent six married couples, all strangers to one another,
who, in walking aimlessly about, have got mixed up. But we are only
concerned with the man that is wearing a straw hat--Number 10. The
puzzle is to find this man's wife. Examine the six ladies carefully, and
see if you can determine which one of them it is.

I showed the picture at the time to a few friends, and they expressed
very different opinions on the matter. One said, "I don't believe he
would marry a girl like Number 7." Another said, "I am sure a nice girl
like Number 3 would not marry such a fellow!" Another said, "It must be
Number 1, because she has got as far away as possible from the brute!"
It was suggested, again, that it must be Number 11, because "he seems to
be looking towards her;" but a cynic retorted, "For that very reason, if
he is really looking at her, I should say that she is not his wife!"

I now leave the question in the hands of my readers. Which is really
Number 10's wife?

The illustration is of necessity considerably reduced from the large
scale on which it originally appeared in _The Weekly Dispatch_ (24th May
1903), but it is hoped that the details will be sufficiently clear to
allow the reader to derive entertainment from its examination. In any
case the solution given will enable him to follow the points with
interest.



SOLUTIONS.


1.--A POST-OFFICE PERPLEXITY.

The young lady supplied 5 twopenny stamps, 30 penny stamps, and 8
twopence-halfpenny stamps, which delivery exactly fulfils the conditions
and represents a cost of five shillings.


2.--YOUTHFUL PRECOCITY.

The price of the banana must have been one penny farthing. Thus, 960
bananas would cost £5, and 480 sixpences would buy 2,304 bananas.


3.--AT A CATTLE MARKET.

Jakes must have taken 7 animals to market, Hodge must have taken 11, and
Durrant must have taken 21. There were thus 39 animals altogether.


4.--THE BEANFEAST PUZZLE.

The cobblers spent 35s., the tailors spent also 35s., the hatters spent
42s., and the glovers spent 21s. Thus, they spent altogether £6,13s.,
while it will be found that the five cobblers spent as much as four
tailors, twelve tailors as much as nine hatters, and six hatters as much
as eight glovers.


5.--A QUEER COINCIDENCE.

Puzzles of this class are generally solved in the old books by the
tedious process of "working backwards." But a simple general solution is
as follows: If there are n players, the amount held by every player at
the end will be m(2^n), the last winner must have held m(n + 1)
at the start, the next m(2n + 1), the next m(4n + 1), the next
m(8n + 1), and so on to the first player, who must have held
m(2^{n - 1}n + 1).

Thus, in this case, n = 7, and the amount held by every player at the
end was 2^7 farthings. Therefore m = 1, and G started with 8 farthings,
F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449
farthings.


6.--A CHARITABLE BEQUEST.

There are seven different ways in which the money may be distributed: 5
women and 19 men, 10 women and 16 men, 15 women and 13 men, 20 women and
10 men, 25 women and 7 men, 30 women and 4 men, and 35 women and 1 man.
But the last case must not be counted, because the condition was that
there should be "men," and a single man is not men. Therefore the answer
is six years.


7.--THE WIDOW'S LEGACY.

The widow's share of the legacy must be £205, 2s. 6d. and 10/13 of a
penny.


8.--INDISCRIMINATE CHARITY

The gentleman must have had 3s. 6d. in his pocket when he set out for
home.


9.--THE TWO AEROPLANES.

The man must have paid £500 and £750 for the two machines, making
together £1,250; but as he sold them for only £1,200, he lost £50 by the
transaction.


10.--BUYING PRESENTS.

Jorkins had originally £19, 18s. in his pocket, and spent £9, 19s.


11.--THE CYCLISTS' FEAST.

There were ten cyclists at the feast. They should have paid 8s. each;
but, owing to the departure of two persons, the remaining eight would
pay 10s. each.


12.--A QUEER THING IN MONEY.

The answer is as follows: £44,444, 4s. 4d. = 28, and, reduced to pence,
10,666,612=28.

It is a curious little coincidence that in the answer 10,666,612 the
four central figures indicate the only other answer, £66, 6s. 6d.



13.--A NEW MONEY PUZZLE.

The smallest sum of money, in pounds, shillings, pence, and farthings,
containing all the nine digits once, and once only, is £2,567, 18s.
9¾d.


14.--SQUARE MONEY.

The answer is 1½d. and 3d. Added together they make 4½d., and
1½d. multiplied by 3 is also 4½d.


15.--POCKET MONEY.

The largest possible sum is 15s. 9d., composed of a crown and a
half-crown (or three half-crowns), four florins, and a threepenny piece.


16.--THE MILLIONAIRE'S PERPLEXITY.

The answer to this quite easy puzzle may, of course, be readily obtained
by trial, deducting the largest power of 7 that is contained in one
million dollars, then the next largest power from the remainder, and so
on. But the little problem is intended to illustrate a simple direct
method. The answer is given at once by converting 1,000,000 to the
septenary scale, and it is on this subject of scales of notation that I
propose to write a few words for the benefit of those who have never
sufficiently considered the matter.

Our manner of figuring is a sort of perfected arithmetical shorthand, a
system devised to enable us to manipulate numbers as rapidly and
correctly as possible by means of symbols. If we write the number 2,341
to represent two thousand three hundred and forty-one dollars, we wish
to imply 1 dollar, added to four times 10 dollars, added to three times
100 dollars, added to two times 1,000 dollars. From the number in the
units place on the right, every figure to the left is understood to
represent a multiple of the particular power of 10 that its position
indicates, while a cipher (0) must be inserted where necessary in order
to prevent confusion, for if instead of 207 we wrote 27 it would be
obviously misleading. We thus only require ten figures, because directly
a number exceeds 9 we put a second figure to the left, directly it
exceeds 99 we put a third figure to the left, and so on. It will be seen
that this is a purely arbitrary method. It is working in the denary (or
ten) scale of notation, a system undoubtedly derived from the fact that
our forefathers who devised it had ten fingers upon which they were
accustomed to count, like our children of to-day. It is unnecessary for
us ordinarily to state that we are using the denary scale, because this
is always understood in the common affairs of life.

But if a man said that he had 6,553 dollars in the septenary (or seven)
scale of notation, you will find that this is precisely the same amount
as 2,341 in our ordinary denary scale. Instead of using powers of ten,
he uses powers of 7, so that he never needs any figure higher than 6,
and 6,553 really stands for 3, added to five times 7, added to five
times 49, added to six times 343 (in the ordinary notation), or 2,341.
To reverse the operation, and convert 2,341 from the denary to the
septenary scale, we divide it by 7, and get 334 and remainder 3; divide
334 by 7, and get 47 and remainder 5; and so keep on dividing by 7 as
long as there is anything to divide. The remainders, read backwards, 6,
5, 5, 3, give us the answer, 6,553.

Now, as I have said, our puzzle may be solved at once by merely
converting 1,000,000 dollars to the septenary scale. Keep on dividing
this number by 7 until there is nothing more left to divide, and the
remainders will be found to be 11333311 which is 1,000,000 expressed in
the septenary scale. Therefore, 1 gift of 1 dollar, 1 gift of 7 dollars,
3 gifts of 49 dollars, 3 gifts of 343 dollars, 3 gifts of 2,401 dollars,
3 gifts of 16,807 dollars, 1 gift of 117,649 dollars, and one
substantial gift of 823,543 dollars, satisfactorily solves our problem.
And it is the only possible solution. It is thus seen that no "trials"
are necessary; by converting to the septenary scale of notation we go
direct to the answer.


17.--THE PUZZLING MONEY BOXES.

The correct answer to this puzzle is as follows: John put into his
money-box two double florins (8s.), William a half-sovereign and a
florin (12s.), Charles a crown (5s.), and Thomas a sovereign (20s.).
There are six coins in all, of a total value of 45s. If John had 2s.
more, William 2s. less, Charles twice as much, and Thomas half as much
as they really possessed, they would each have had exactly 10s.


18.--THE MARKET WOMEN.

The price received was in every case 105 farthings. Therefore the
greatest number of women is eight, as the goods could only be sold at
the following rates: 105 lbs. at 1 farthing, 35 at 3, 21 at 5, 15 at 7,
7 at 15, 5 at 21, 3 at 35, and 1 lb. at 105 farthings.


19.--THE NEW YEAR'S EVE SUPPERS.

The company present on the occasion must have consisted of seven pairs,
ten single men, and one single lady. Thus, there were twenty-five
persons in all, and at the prices stated they would pay exactly £5
together.


20.--BEEF AND SAUSAGES.

The lady bought 48 lbs. of beef at 2s., and the same quantity of
sausages at 1s. 6d., thus spending £8, 8s. Had she bought 42 lbs. of
beef and 56 lbs. of sausages she would have spent £4, 4s. on each, and
have obtained 98 lbs. instead of 96 lbs.--a gain in weight of 2 lbs.


21.--A DEAL IN APPLES.

I was first offered sixteen apples for my shilling, which would be at
the rate of ninepence a dozen. The two extra apples gave me eighteen for
a shilling, which is at the rate of eightpence a dozen, or one penny a
dozen less than the first price asked.


22.--A DEAL IN EGGS.

The man must have bought ten eggs at fivepence, ten eggs at one penny,
and eighty eggs at a halfpenny. He would then have one hundred eggs at a
cost of eight shillings and fourpence, and the same number of eggs of
two of the qualities.


23.--THE CHRISTMAS-BOXES.

The distribution took place "some years ago," when the fourpenny-piece
was in circulation. Nineteen persons must each have received nineteen
pence. There are five different ways in which this sum may have been
paid in silver coins. We need only use two of these ways. Thus if
fourteen men each received four four-penny-pieces and one
threepenny-piece, and five men each received five threepenny-pieces and
one fourpenny-piece, each man would receive nineteen pence, and there
would be exactly one hundred coins of a total value of £1, 10s. 1d.


24.--A SHOPPING PERPLEXITY.

The first purchase amounted to 1s. 5¾d., the second to 1s. 11½d.,
and together they make 3s. 5¼d. Not one of these three amounts can be
paid in fewer than six current coins of the realm.


25.--CHINESE MONEY.

As a ching-chang is worth twopence and four-fifteenths of a ching-chang,
the remaining eleven-fifteenths of a ching-chang must be worth twopence.
Therefore eleven ching-changs are worth exactly thirty pence, or half a
crown. Now, the exchange must be made with seven round-holed coins and
one square-holed coin. Thus it will be seen that 7 round-holed coins are
worth seven-elevenths of 15 ching-changs, and 1 square-holed coin is
worth one-eleventh of 16 ching-changs--that is, 77 rounds equal 105
ching-changs and 11 squares equal 16 ching-changs. Therefore 77 rounds
added to 11 squares equal 121 ching-changs; or 7 rounds and 1 square
equal 11 ching-changs, or its equivalent, half a crown. This is more
simple in practice than it looks here.


26.--THE JUNIOR CLERKS' PUZZLE.

Although Snoggs's _reason_ for wishing to take his rise at £2, 10s.
half-yearly did not concern our puzzle, the _fact_ that he was duping
his employer into paying him more than was intended did concern it. Many
readers will be surprised to find that, although Moggs only received
£350 in five years, the artful Snoggs actually obtained £362, 10s. in
the same time. The rest is simplicity itself. It is evident that if
Moggs saved £87, 10s. and Snoggs £181, 5s., the latter would be saving
twice as great a proportion of his salary as the former (namely,
one-half as against one-quarter), and the two sums added together make
£268, 15s.


27.--GIVING CHANGE.

The way to help the American tradesman out of his dilemma is this.
Describing the coins by the number of cents that they represent, the
tradesman puts on the counter 50 and 25; the buyer puts down 100, 3, and
2; the stranger adds his 10, 10, 5, 2, and 1. Now, considering that the
cost of the purchase amounted to 34 cents, it is clear that out of this
pooled money the tradesman has to receive 109, the buyer 71, and the
stranger his 28 cents. Therefore it is obvious at a glance that the
100-piece must go to the tradesman, and it then follows that the
50-piece must go to the buyer, and then the 25-piece can only go to the
stranger. Another glance will now make it clear that the two 10-cent
pieces must go to the buyer, because the tradesman now only wants 9 and
the stranger 3. Then it becomes obvious that the buyer must take the 1
cent, that the stranger must take the 3 cents, and the tradesman the 5,
2, and 2. To sum up, the tradesman takes 100, 5, 2, and 2; the buyer,
50, 10, 10, and 1; the stranger, 25 and 3. It will be seen that not one
of the three persons retains any one of his own coins.


28.--DEFECTIVE OBSERVATION.

Of course the date on a penny is on the same side as Britannia--the
"tail" side. Six pennies may be laid around another penny, all flat on
the table, so that every one of them touches the central one. The number
of threepenny-pieces that may be laid on the surface of a half-crown, so
that no piece lies on another or overlaps the edge of the half-crown, is
one. A second threepenny-piece will overlap the edge of the larger coin.
Few people guess fewer than three, and many persons give an absurdly
high number.


29.--THE BROKEN COINS.

If the three broken coins when perfect were worth 253 pence, and are now
in their broken condition worth 240 pence, it should be obvious that
13/253 of the original value has been lost. And as the same fraction of
each coin has been broken away, each coin has lost 13/253 of its
original bulk.


30.--TWO QUESTIONS IN PROBABILITIES.

In tossing with the five pennies all at the same time, it is obvious
that there are 32 different ways in which the coins may fall, because
the first coin may fall in either of two ways, then the second coin may
also fall in either of two ways, and so on. Therefore five 2's
multiplied together make 32. Now, how are these 32 ways made up? Here
they are:--

    (a)   5 heads                 1 way
    (b)   5 tails                 1 way
    (c)   4 heads and 1 tail      5 ways
    (d)   4 tails and 1 head      5 ways
    (e)   3 heads and 2 tails    10 ways
    (f)   3 tails and 2 heads    10 ways

Now, it will be seen that the only favourable cases are a, b, c,
and d--12 cases. The remaining 20 cases are unfavourable, because they
do not give at least four heads or four tails. Therefore the chances are
only 12 to 20 in your favour, or (which is the same thing) 3 to 5. Put
another way, you have only 3 chances out of 8.

The amount that should be paid for a draw from the bag that contains
three sovereigns and one shilling is 15s. 3d. Many persons will say
that, as one's chances of drawing a sovereign were 3 out of 4, one
should pay three-fourths of a pound, or 15s., overlooking the fact that
one must draw at least a shilling--there being no blanks.



31.--DOMESTIC ECONOMY.

Without the hint that I gave, my readers would probably have been
unanimous in deciding that Mr. Perkins's income must have been £1,710.
But this is quite wrong. Mrs. Perkins says, "We have spent a third of
his yearly income in rent," etc., etc.--that is, in two years they have
spent an amount in rent, etc., equal to one-third of his yearly income.
Note that she does _not_ say that they have spent _each year_ this sum,
whatever it is, but that _during the two years_ that amount has been
spent. The only possible answer, according to the exact reading of her
words, is, therefore, that his income was £180 per annum. Thus the
amount spent in two years, during which his income has amounted to £360,
will be £60 in rent, etc., £90 in domestic expenses, £20 in other ways,
leaving the balance of £190 in the bank as stated.


32.--THE EXCURSION TICKET PUZZLE.

Nineteen shillings and ninepence may be paid in 458,908,622 different
ways.

I do not propose to give my method of solution. Any such explanation
would occupy an amount of space out of proportion to its interest or
value. If I could give within reasonable limits a general solution for
all money payments, I would strain a point to find room; but such a
solution would be extremely complex and cumbersome, and I do not
consider it worth the labour of working out.

Just to give an idea of what such a solution would involve, I will
merely say that I find that, dealing only with those sums of money that
are multiples of threepence, if we only use bronze coins any sum can be
paid in (n + 1)² ways where n always represents the number of
pence. If threepenny-pieces are admitted, there are

              2n³ + 15n² + 33n
            --------------------- + 1 ways.
                    18

If sixpences are also used there are

          n^{4} + 22n³ + 159n² + 414n + 216
          ---------------------------------
                        216

ways, when the sum is a multiple of sixpence, and the constant, 216,
changes to 324 when the money is not such a multiple. And so the
formulas increase in complexity in an accelerating ratio as we go on to
the other coins.

I will, however, add an interesting little table of the possible ways of
changing our current coins which I believe has never been given in a
book before. Change may be given for a

    Farthing in           0 way.
    Halfpenny in          1 way.
    Penny in              3 ways.
    Threepenny-piece in   16 ways.
    Sixpence in           66 ways.
    Shilling in           402 ways.
    Florin in             3,818 ways.
    Half-crown in         8,709 ways.
    Double florin in      60,239 ways.
    Crown in              166,651 ways.
    Half-sovereign in     6,261,622 ways.
    Sovereign in          500,291,833 ways.

It is a little surprising to find that a sovereign may be changed in
over five hundred million different ways. But I have no doubt as to the
correctness of my figures.


33.--A PUZZLE IN REVERSALS.

(i) £13. (2) £23, 19s. 11d. The words "the number of pounds exceeds that
of the pence" exclude such sums of money as £2, 16s. 2d. and all sums
under £1.


34.--THE GROCER AND DRAPER.

The grocer was delayed half a minute and the draper eight minutes and a
half (seventeen times as long as the grocer), making together nine
minutes. Now, the grocer took twenty-four minutes to weigh out the
sugar, and, with the half-minute delay, spent 24 min. 30 sec. over the
task; but the draper had only to make _forty-seven_ cuts to divide the
roll of cloth, containing forty-eight yards, into yard pieces! This took
him 15 min. 40 sec., and when we add the eight minutes and a half delay
we get 24 min. 10 sec., from which it is clear that the draper won the
race by twenty seconds. The majority of solvers make forty-eight cuts to
divide the roll into forty-eight pieces!


35.--JUDKINS'S CATTLE.

As there were five droves with an equal number of animals in each drove,
the number must be divisible by 5; and as every one of the eight dealers
bought the same number of animals, the number must be divisible by 8.
Therefore the number must be a multiple of 40. The highest possible
multiple of 40 that will work will be found to be 120, and this number
could be made up in one of two ways--1 ox, 23 pigs, and 96 sheep, or 3
oxen, 8 pigs, and 109 sheep. But the first is excluded by the statement
that the animals consisted of "oxen, pigs, and sheep," because a single
ox is not oxen. Therefore the second grouping is the correct answer.


36.--BUYING APPLES.

As there were the same number of boys as girls, it is clear that the
number of children must be even, and, apart from a careful and exact
reading of the question, there would be three different answers. There
might be two, six, or fourteen children. In the first of these cases
there are ten different ways in which the apples could be bought. But we
were told there was an equal number of "boys and girls," and one boy and
one girl are not boys and girls, so this case has to be excluded. In the
case of fourteen children, the only possible distribution is that each
child receives one halfpenny apple. But we were told that each child was
to receive an equal distribution of "apples," and one apple is not
apples, so this case has also to be excluded. We are therefore driven
back on our third case, which exactly fits in with all the conditions.
Three boys and three girls each receive 1 halfpenny apple and 2
third-penny apples. The value of these 3 apples is one penny and
one-sixth, which multiplied by six makes sevenpence. Consequently, the
correct answer is that there were six children--three girls and three
boys.


37.--BUYING CHESTNUTS.

In solving this little puzzle we are concerned with the exact
interpretation of the words used by the buyer and seller. I will give
the question again, this time adding a few words to make the matter more
clear. The added words are printed in italics.

"A man went into a shop to buy chestnuts. He said he wanted a
pennyworth, and was given five chestnuts. 'It is not enough; I ought to
have a sixth _of a chestnut more_,' he remarked. 'But if I give you one
chestnut more,' the shopman replied, 'you will have _five-sixths_ too
many.' Now, strange to say, they were both right. How many chestnuts
should the buyer receive for half a crown?"

The answer is that the price was 155 chestnuts for half a crown. Divide
this number by 30, and we find that the buyer was entitled to 5+1/6
chestnuts in exchange for his penny. He was, therefore, right when he
said, after receiving five only, that he still wanted a sixth. And the
salesman was also correct in saying that if he gave one chestnut more
(that is, six chestnuts in all) he would be giving five-sixths of a
chestnut in excess.


38.--THE BICYCLE THIEF.

People give all sorts of absurd answers to this question, and yet it is
perfectly simple if one just considers that the salesman cannot possibly
have lost more than the cyclist actually stole. The latter rode away
with a bicycle which cost the salesman eleven pounds, and the ten pounds
"change;" he thus made off with twenty-one pounds, in exchange for a
worthless bit of paper. This is the exact amount of the salesman's loss,
and the other operations of changing the cheque and borrowing from a
friend do not affect the question in the slightest. The loss of
prospective profit on the sale of the bicycle is, of course, not direct
loss of money out of pocket.


39.--THE COSTERMONGER'S PUZZLE.

Bill must have paid 8s. per hundred for his oranges--that is, 125 for
10s. At 8s. 4d. per hundred, he would only have received 120 oranges for
10s. This exactly agrees with Bill's statement.


40.--MAMMA'S AGE.

The age of Mamma must have been 29 years 2 months; that of Papa, 35
years; and that of the child, Tommy, 5 years 10 months. Added together,
these make seventy years. The father is six times the age of the son,
and, after 23 years 4 months have elapsed, their united ages will amount
to 140 years, and Tommy will be just half the age of his father.


41.--THEIR AGES.

The gentleman's age must have been 54 years and that of his wife 45
years.


42.--THE FAMILY AGES.

The ages were as follows: Billie, 3½ years; Gertrude, 1¾ year;
Henrietta, 5¼ years; Charlie, 10½; years; and Janet, 21 years.


43.--MRS. TIMPKINS'S AGE.

The age of the younger at marriage is always the same as the number of
years that expire before the elder becomes twice her age, if he was
three times as old at marriage. In our case it was eighteen years
afterwards; therefore Mrs. Timpkins was eighteen years of age on the
wedding-day, and her husband fifty-four.


44.--A CENSUS PUZZLE.

Miss Ada Jorkins must have been twenty-four and her little brother
Johnnie three years of age, with thirteen brothers and sisters between.
There was a trap for the solver in the words "seven times older than
little Johnnie." Of course, "seven times older" is equal to eight times
as old. It is surprising how many people hastily assume that it is the
same as "seven times as old." Some of the best writers have committed
this blunder. Probably many of my readers thought that the ages 24½
and 3½ were correct.


45.--MOTHER AND DAUGHTER.

In four and a half years, when the daughter will be sixteen years and a
half and the mother forty-nine and a half years of age.


46.--MARY AND MARMADUKE.

Marmaduke's age must have been twenty-nine years and two-fifths, and
Mary's nineteen years and three-fifths. When Marmaduke was aged nineteen
and three-fifths, Mary was only nine and four-fifths; so Marmaduke was
at that time twice her age.


47.--ROVER'S AGE.

Rover's present age is ten years and Mildred's thirty years. Five years
ago their respective ages were five and twenty-five. Remember that we
said "four times older than the dog," which is the same as "five times
as old." (See answer to No. 44.)


48.--CONCERNING TOMMY'S AGE.

Tommy Smart's age must have been nine years and three-fifths. Ann's age
was sixteen and four-fifths, the mother's thirty-eight and two-fifths,
and the father's fifty and two-fifths.


49.--NEXT-DOOR NEIGHBOURS.

Mr. Jupp 39, Mrs. Jupp 34, Julia 14, and Joe 13; Mr. Simkin 42; Mrs.
Simkin 40; Sophy 10; and Sammy 8.


50.--THE BAG OF NUTS.

It will be found that when Herbert takes twelve, Robert and Christopher
will take nine and fourteen respectively, and that they will have
together taken thirty-five nuts. As 35 is contained in 770 twenty-two
times, we have merely to multiply 12, 9, and 14 by 22 to discover that
Herbert's share was 264, Robert's 198, and Christopher's 308. Then, as
the total of their ages is 17½ years or half the sum of 12, 9, and 14,
their respective ages must be 6, 4½, and 7 years.


51.--HOW OLD WAS MARY?

The age of Mary to that of Ann must be as 5 to 3. And as the sum of
their ages was 44, Mary was 27½ and Ann 16½. One is exactly 11 years
older than the other. I will now insert in brackets in the original
statement the various ages specified: "Mary is (27½) twice as old as Ann
was (13¾) when Mary was half as old (24¾) as Ann will be (49½) when Ann
is three times as old (49½) as Mary was (16½) when Mary was (16½) three
times as old as Ann (5½)." Now, check this backwards. When Mary was
three times as old as Ann, Mary was 16½ and Ann 5½ (11 years younger).
Then we get 49½ for the age Ann will be when she is three times as old
as Mary was then. When Mary was half this she was 24¾. And at that time
Ann must have been 13¾ (11 years younger). Therefore Mary is now twice
as old--27½, and Ann 11 years younger--16½.


52.--QUEER RELATIONSHIPS.

If a man marries a woman, who dies, and he then marries his deceased
wife's sister and himself dies, it may be correctly said that he had
(previously) married the sister of his widow.

The youth was not the nephew of Jane Brown, because he happened to be
her son. Her surname was the same as that of her brother, because she
had married a man of the same name as herself.


53.--HEARD ON THE TUBE RAILWAY.

The gentleman was the second lady's uncle.


54.--A FAMILY PARTY.

The party consisted of two little girls and a boy, their father and
mother, and their father's father and mother.


55.--A MIXED PEDIGREE.

[Illustration:

                                     Thos. Bloggs  m . . . . .
                                                   |
                    +------------------------+------------+
                    |                        |            |
                    |                        |            |
                    |    W. Snoggs m Kate Bloggs.         |
                    |              |                      |
                    |              |                      |
       . . m Henry Bloggs.         |                Joseph Bloggs m
           |                       |                              |
           |              +--------+-------------+                |
           |              |                      |                |
           |              |                      |                |
         Jane            John                    Alf.            Mary
        Bloggs    m     Snoggs                  Snoggs    m     Bloggs

]

The letter m stands for "married." It will be seen that John Snoggs
can say to Joseph Bloggs, "You are my _father's brother-in-law_, because
my father married your sister Kate; you are my _brother's
father-in-law_, because my brother Alfred married your daughter Mary;
and you are my _father-in-law's brother_, because my wife Jane was your
brother Henry's daughter."


56.--WILSON'S POSER.

If there are two men, each of whom marries the mother of the other, and
there is a son of each marriage, then each of such sons will be at the
same time uncle and nephew of the other. There are other ways in which
the relationship may be brought about, but this is the simplest.


57.--WHAT WAS THE TIME?

The time must have been 9.36 p.m. A quarter of the time since noon is 2
hr. 24 min., and a half of the time till noon next day is 7 hr. 12 min.
These added together make 9 hr. 36 min.


58.--A TIME PUZZLE.

Twenty-six minutes.


59.--A PUZZLING WATCH.

If the 65 minutes be counted on the face of the same watch, then the
problem would be impossible: for the hands must coincide every 65+5/11
minutes as shown by its face, and it matters not whether it runs fast or
slow; but if it is measured by true time, it gains 5/11 of a minute in
65 minutes, or 60/143 of a minute per hour.


60.--THE WAPSHAW'S WHARF MYSTERY.

There are eleven different times in twelve hours when the hour and
minute hands of a clock are exactly one above the other. If we divide 12
hours by 11 we get 1 hr. 5 min. 27+3/11 sec., and this is the time after
twelve o'clock when they are first together, and also the time that
elapses between one occasion of the hands being together and the next.
They are together for the second time at 2 hr. 10 min. 54+6/11 sec.
(twice the above time); next at 3 hr. 16 min. 21+9/11 sec.; next at 4
hr. 21 min. 49+1/11 sec. This last is the only occasion on which the two
hands are together with the second hand "just past the forty-ninth
second." This, then, is the time at which the watch must have stopped.
Guy Boothby, in the opening sentence of his _Across the World for a
Wife_, says, "It was a cold, dreary winter's afternoon, and by the time
the hands of the clock on my mantelpiece joined forces and stood at
twenty minutes past four, my chambers were well-nigh as dark as
midnight." It is evident that the author here made a slip, for, as we
have seen above, he is 1 min. 49+1/11 sec. out in his reckoning.


61.--CHANGING PLACES.

There are thirty-six pairs of times when the hands exactly change places
between three p.m. and midnight. The number of pairs of times from any
hour (n) to midnight is the sum of 12 - (n + 1) natural numbers. In
the case of the puzzle n = 3; therefore 12 - (3 + 1) = 8 and 1 + 2 + 3
+ 4 + 5 + 6 + 7 + 8 = 36, the required answer.

The first pair of times is 3 hr. 21+57/143 min. and 4 hr. 16+112/143
min., and the last pair is 10 hr. 59+83/143 min. and 11 hr. 54+138/143
min. I will not give all the remainder of the thirty-six pairs of times,
but supply a formula by which any of the sixty-six pairs that occur from
midday to midnight may be at once found:--

          720b + 60a                720a + 60b min.
    a hr  ---------- min. and b hr. ---------------
             143                         143

For the letter a may be substituted any hour from 0, 1, 2, 3 up to 10
(where nought stands for 12 o'clock midday); and b may represent any
hour, later than a, up to 11.

By the aid of this formula there is no difficulty in discovering the
answer to the second question: a = 8 and b = 11 will give the pair 8 hr.
58+106/143 min. and 11 hr. 44+128/143 min., the latter being the time
when the minute hand is nearest of all to the point IX--in fact, it is
only 15/143 of a minute distant.

Readers may find it instructive to make a table of all the sixty-six
pairs of times when the hands of a clock change places. An easy way is
as follows: Make a column for the first times and a second column for
the second times of the pairs. By making a = 0 and b = 1 in the above
expressions we find the first case, and enter hr. 5+5/143 min. at the
head of the first column, and 1 hr. 0+60/143 min. at the head of the
second column. Now, by successively adding 5+5/143 min. in the first,
and 1 hr. 0+60/143 min. in the second column, we get all the _eleven_
pairs in which the first time is a certain number of minutes after
nought, or mid-day. Then there is a "jump" in the times, but you can
find the next pair by making a = 1 and b = 2, and then by successively
adding these two times as before you will get all the _ten_ pairs after
1 o'clock. Then there is another "jump," and you will be able to get by
addition all the _nine_ pairs after 2 o'clock. And so on to the end. I
will leave readers to investigate for themselves the nature and cause of
the "jumps." In this way we get under the successive hours, 11 + 10 + 9
+ 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 66 pairs of times, which result agrees
with the formula in the first paragraph of this article.

Some time ago the principal of a Civil Service Training College, who
conducts a "Civil Service Column" in one of the periodicals, had the
query addressed to him, "How soon after XII o'clock will a clock with
both hands of the same length be ambiguous?" His first answer was, "Some
time past one o'clock," but he varied the answer from issue to issue. At
length some of his readers convinced him that the answer is, "At 5+5/143
min. past XII;" and this he finally gave as correct, together with the
reason for it that at that time _the time indicated is the same
whichever hand you may assume as hour hand!_


62.--THE CLUB CLOCK.

The positions of the hands shown in the illustration could only indicate
that the clock stopped at 44 min. 51+1143/1427 sec. after eleven
o'clock. The second hand would next be "exactly midway between the other
two hands" at 45 min. 52+496/1427 sec. after eleven o'clock. If we had
been dealing with the points on the circle to which the three hands are
directed, the answer would be 45 min. 22+106/1427 sec. after eleven; but
the question applied to the hands, and the second hand would not be
between the others at that time, but outside them.


63.--THE STOP-WATCH.

The time indicated on the watch was 5+5/11 min. past 9, when the second
hand would be at 27+3/11 sec. The next time the hands would be similar
distances apart would be 54+6/11 min. past 2, when the second hand would
be at 32+8/11 sec. But you need only hold the watch (or our previous
illustration of it) in front of a mirror, when you will see the second
time reflected in it! Of course, when reflected, you will read XI as I,
X as II, and so on.


64.--THE THREE CLOCKS.

As a mere arithmetical problem this question presents no difficulty. In
order that the hands shall all point to twelve o'clock at the same time,
it is necessary that B shall gain at least twelve hours and that C shall
lose twelve hours. As B gains a minute in a day of twenty-four hours,
and C loses a minute in precisely the same time, it is evident that one
will have gained 720 minutes (just twelve hours) in 720 days, and the
other will have lost 720 minutes in 720 days. Clock A keeping perfect
time, all three clocks must indicate twelve o'clock simultaneously at
noon on the 720th day from April 1, 1898. What day of the month will
that be?

I published this little puzzle in 1898 to see how many people were aware
of the fact that 1900 would not be a leap year. It was surprising how
many were then ignorant on the point. Every year that can be divided by
four without a remainder is bissextile or leap year, with the exception
that one leap year is cut off in the century. 1800 was not a leap year,
nor was 1900. On the other hand, however, to make the calendar more
nearly agree with the sun's course, every fourth hundred year is still
considered bissextile. Consequently, 2000, 2400, 2800, 3200, etc., will
all be leap years. May my readers live to see them. We therefore find
that 720 days from noon of April 1, 1898, brings us to noon of March 22,
1900.


65.--THE RAILWAY STATION CLOCK.

The time must have been 43+7/11 min. past two o'clock.


66.--THE VILLAGE SIMPLETON.

The day of the week on which the conversation took place was Sunday. For
when the day after to-morrow (Tuesday) is "yesterday," "to-day" will be
Wednesday; and when the day before yesterday (Friday) was "to-morrow,"
"to-day" was Thursday. There are two days between Thursday and Sunday,
and between Sunday and Wednesday.


67.--AVERAGE SPEED.

The average speed is twelve miles an hour, not twelve and a half, as
most people will hastily declare. Take any distance you like, say sixty
miles. This would have taken six hours going and four hours returning.
The double journey of 120 miles would thus take ten hours, and the
average speed is clearly twelve miles an hour.


68.--THE TWO TRAINS.

One train was running just twice as fast as the other.


69.--THE THREE VILLAGES.

Calling the three villages by their initial letters, it is clear that
the three roads form a triangle, A, B, C, with a perpendicular,
measuring twelve miles, dropped from C to the base A, B. This divides
our triangle into two right-angled triangles with a twelve-mile side in
common. It is then found that the distance from A to C is 15 miles, from
C to B 20 miles, and from A to B 25 (that is 9 and 16) miles. These
figures are easily proved, for the square of 12 added to the square of 9
equals the square of 15, and the square of 12 added to the square of 16
equals the square of 20.


70.--DRAWING HER PENSION.

The distance must be 6¾ miles.


71.--SIR EDWYN DE TUDOR.

The distance must have been sixty miles. If Sir Edwyn left at noon and
rode 15 miles an hour, he would arrive at four o'clock--an hour too
soon. If he rode 10 miles an hour, he would arrive at six o'clock--an
hour too late. But if he went at 12 miles an hour, he would reach the
castle of the wicked baron exactly at five o'clock--the time appointed.


72.--THE HYDROPLANE QUESTION.

The machine must have gone at the rate of seven-twenty-fourths of a mile
per minute and the wind travelled five-twenty-fourths of a mile per
minute. Thus, going, the wind would help, and the machine would do
twelve-twenty-fourths, or half a mile a minute, and returning only
two-twenty-fourths, or one-twelfth of a mile per minute, the wind being
against it. The machine without any wind could therefore do the ten
miles in thirty-four and two-sevenths minutes, since it could do seven
miles in twenty-four minutes.


73.--DONKEY RIDING.

The complete mile was run in nine minutes. From the facts stated we
cannot determine the time taken over the first and second quarter-miles
separately, but together they, of course, took four and a half minutes.
The last two quarters were run in two and a quarter minutes each.


74.--THE BASKET OF POTATOES.

Multiply together the number of potatoes, the number less one, and twice
the number less one, then divide by 3. Thus 50, 49, and 99 multiplied
together make 242,550, which, divided by 3, gives us 80,850 yards as the
correct answer. The boy would thus have to travel 45 miles and
fifteen-sixteenths--a nice little recreation after a day's work.


75.--THE PASSENGER'S FARE.

Mr. Tompkins should have paid fifteen shillings as his correct share of
the motor-car fare. He only shared half the distance travelled for £3,
and therefore should pay half of thirty shillings, or fifteen shillings.


76.--THE BARREL OF BEER.

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which
together sum to 29, whose digital root is 2. As the contents of the
barrels sold must be a number divisible by 3, if one buyer purchased
twice as much as the other, we must find a barrel with root 2, 5, or 8
to set on one side. There is only one barrel, that containing 20
gallons, that fulfils these conditions. So the man must have kept these
20 gallons of beer for his own use and sold one man 33 gallons (the
18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the
16, 19, and 31 gallon barrels).


77.--DIGITS AND SQUARES.

The top row must be one of the four following numbers: 192, 219, 273,
327. The first was the example given.



78.--ODD AND EVEN DIGITS.

As we have to exclude complex and improper fractions and recurring
decimals, the simplest solution is this: 79 + 5+1/3 and 84 + 2/6, both
equal 84+1/3. Without any use of fractions it is obviously impossible.


79.--THE LOCKERS PUZZLE.

The smallest possible total is 356 = 107 + 249, and the largest sum
possible is 981 = 235 + 746, or 657+324. The middle sum may be either
720 = 134 + 586, or 702 = 134 + 568, or 407 = 138 + 269. The total in
this case must be made up of three of the figures 0, 2, 4, 7, but no
sum other than the three given can possibly be obtained. We have
therefore no choice in the case of the first locker, an alternative in
the case of the third, and any one of three arrangements in the case
of the middle locker. Here is one solution:--

    107     134     235
    249     586     746
    ---     ---     ---
    356     720     981


Of course, in each case figures in the first two lines may be exchanged
vertically without altering the total, and as a result there are just
3,072 different ways in which the figures might be actually placed on
the locker doors. I must content myself with showing one little
principle involved in this puzzle. The sum of the digits in the total is
always governed by the digit omitted. 9/9 - 7/10 - 5/11 -3/12 - 1/13 -
8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in the
upper line we omit, the sum of the digits in the total will be found
beneath it. Thus in the case of locker A we omitted 8, and the figures
in the total sum up to 14. If, therefore, we wanted to get 356, we may
know at once to a certainty that it can only be obtained (if at all) by
dropping the 8.


80.--THE THREE GROUPS.

There are nine solutions to this puzzle, as follows, and no more:--

    12 × 483 = 5,796     27 × 198 = 5,346
    42 × 138 = 5,796     39 × 186 = 7,254
    18 × 297 = 5,346     48 × 159 = 7,632
           28 ×   157 = 4,396
            4 × 1,738 = 6,952
            4 × 1,963 = 7,852

The seventh answer is the one that is most likely to be overlooked by
solvers of the puzzle.


81.--THE NINE COUNTERS.

In this case a certain amount of mere "trial" is unavoidable. But there
are two kinds of "trials"--those that are purely haphazard, and those
that are methodical. The true puzzle lover is never satisfied with mere
haphazard trials. The reader will find that by just reversing the
figures in 23 and 46 (making the multipliers 32 and 64) both products
will be 5,056. This is an improvement, but it is not the correct answer.
We can get as large a product as 5,568 if we multiply 174 by 32 and 96
by 58, but this solution is not to be found without the exercise of some
judgment and patience.


82.--THE TEN COUNTERS.

As I pointed out, it is quite easy so to arrange the counters that they
shall form a pair of simple multiplication sums, each of which will give
the same product--in fact, this can be done by anybody in five minutes
with a little patience. But it is quite another matter to find that pair
which gives the largest product and that which gives the smallest
product.

Now, in order to get the smallest product, it is necessary to select as
multipliers the two smallest possible numbers. If, therefore, we place 1
and 2 as multipliers, all we have to do is to arrange the remaining
eight counters in such a way that they shall form two numbers, one of
which is just double the other; and in doing this we must, of course,
try to make the smaller number as low as possible. Of course the lowest
number we could get would be 3,045; but this will not work, neither will
3,405, 3,45O, etc., and it may be ascertained that 3,485 is the lowest
possible. One of the required answers is 3,485 × 2 = 6,970, and 6,970 ×
1 = 6,970.

The other part of the puzzle (finding the pair with the highest product)
is, however, the real knotty point, for it is not at all easy to
discover whether we should let the multiplier consist of one or of two
figures, though it is clear that we must keep, so far as we can, the
largest figures to the left in both multiplier and multiplicand. It will
be seen that by the following arrangement so high a number as 58,560 may
be obtained. Thus, 915 × 64 = 58,560, and 732 × 80 = 58,560.


83.--DIGITAL MULTIPLICATION.

The solution that gives the smallest possible sum of digits in the
common product is 23 × 174 = 58 × 69 = 4,002, and the solution that
gives the largest possible sum of digits, 9×654 =18×327=5,886. In the
first case the digits sum to 6 and in the second case to 27. There is no
way of obtaining the solution but by actual trial.


84.--THE PIERROT'S PUZZLE.

There are just six different solutions to this puzzle, as follows:--

     8 multiplied by 473 equals 3784
     9      "        351   "    3159
    15      "         93   "    1395
    21      "         87   "    1287
    27      "         81   "    2187
    35      "         41   "    1435

It will be seen that in every case the two multipliers contain exactly
the same figures as the product.


85.--THE CAB NUMBERS.

The highest product is, I think, obtained by multiplying 8,745,231 by
96--namely, 839,542,176.

Dealing here with the problem generally, I have shown in the last puzzle
that with three digits there are only two possible solutions, and with
four digits only six different solutions.

These cases have all been given. With five digits there are just
twenty-two solutions, as follows:--

       3 × 4128 = 12384
       3 × 4281 = 12843
       3 × 7125 = 21375
       3 × 7251 = 21753
    2541 × 6    = 15246
     651 × 24   = 15624
     678 × 42   = 28476
     246 × 51   = 12546
      57 × 834  = 47538
      75 × 231  = 17325
     624 × 78   = 48672
     435 × 87   = 37845
           ------
       9 × 7461 = 67149
      72 × 936  = 67392
           ------
       2 × 8714 = 17428
       2 × 8741 = 17482
      65 × 281  = 18265
      65 × 983  = 63985
           ------
    4973 × 8    = 39784
    6521 × 8    = 52168
      14 × 926  = 12964
      86 × 251  = 21586

Now, if we took every possible combination and tested it by
multiplication, we should need to make no fewer than 30,240 trials, or,
if we at once rejected the number 1 as a multiplier, 28,560 trials--a
task that I think most people would be inclined to shirk. But let us
consider whether there be no shorter way of getting at the results
required. I have already explained that if you add together the digits
of any number and then, as often as necessary, add the digits of the
result, you must ultimately get a number composed of one figure. This
last number I call the "digital root." It is necessary in every solution
of our problem that the root of the sum of the digital roots of our
multipliers shall be the same as the root of their product. There are
only four ways in which this can happen: when the digital roots of the
multipliers are 3 and 6, or 9 and 9, or 2 and 2, or 5 and 8. I have
divided the twenty-two answers above into these four classes. It is thus
evident that the digital root of any product in the first two classes
must be 9, and in the second two classes 4.

Owing to the fact that no number of five figures can have a digital sum
less than 15 or more than 35, we find that the figures of our product
must sum to either 18 or 27 to produce the root 9, and to either 22 or
31 to produce the root 4. There are 3 ways of selecting five different
figures that add up to 18, there are 11 ways of selecting five figures
that add up to 27, there are 9 ways of selecting five figures that add
up to 22, and 5 ways of selecting five figures that add up to 31. There
are, therefore, 28 different groups, and no more, from any one of which
a product may be formed.

We next write out in a column these 28 sets of five figures, and proceed
to tabulate the possible factors, or multipliers, into which they may be
split. Roughly speaking, there would now appear to be about 2,000
possible cases to be tried, instead of the 30,240 mentioned above; but
the process of elimination now begins, and if the reader has a quick eye
and a clear head he can rapidly dispose of the large bulk of these
cases, and there will be comparatively few test multiplications
necessary. It would take far too much space to explain my own method in
detail, but I will take the first set of figures in my table and show
how easily it is done by the aid of little tricks and dodges that should
occur to everybody as he goes along.

My first product group of five figures is 84,321. Here, as we have seen,
the root of each factor must be 3 or a multiple of 3. As there is no 6
or 9, the only single multiplier is 3. Now, the remaining four figures
can be arranged in 24 different ways, but there is no need to make 24
multiplications. We see at a glance that, in order to get a five-figure
product, either the 8 or the 4 must be the first figure to the left. But
unless the 2 is preceded on the right by the 8, it will produce when
multiplied either a 6 or a 7, which must not occur. We are, therefore,
reduced at once to the two cases, 3 × 4,128 and 3 x 4,281, both of which
give correct solutions. Suppose next that we are trying the two-figure
factor, 21. Here we see that if the number to be multiplied is under 500
the product will either have only four figures or begin with 10.
Therefore we have only to examine the cases 21 × 843 and 21 × 834. But
we know that the first figure will be repeated, and that the second
figure will be twice the first figure added to the second. Consequently,
as twice 3 added to 4 produces a nought in our product, the first case
is at once rejected. It only remains to try the remaining case by
multiplication, when we find it does not give a correct answer. If we
are next trying the factor 12, we see at the start that neither the 8
nor the 3 can be in the units place, because they would produce a 6, and
so on. A sharp eye and an alert judgment will enable us thus to run
through our table in a much shorter time than would be expected. The
process took me a little more than three hours.

I have not attempted to enumerate the solutions in the cases of six,
seven, eight, and nine digits, but I have recorded nearly fifty examples
with nine digits alone.


86.--QUEER MULTIPLICATION.

If we multiply 32547891 by 6, we get the product, 195287346. In both
cases all the nine digits are used once and once only.


87.--THE NUMBER CHECKS PUZZLE.

Divide the ten checks into the following three groups: 7 1 5--4 6--3 2 8
9 0, and the first multiplied by the second produces the third.


88.--DIGITAL DIVISION.

It is convenient to consider the digits as arranged to form fractions of
the respective values, one-half, one-third, one-fourth, one-fifth,
one-sixth, one-seventh, one-eighth, and one-ninth. I will first give the
eight answers, as follows:--

    6729/13458 = 1/2

    5823/17469 = 1/3

    3942/15768 = 1/4

    2697/13485 = 1/5

    2943/17658 = 1/6

    2394/16758 = 1/7

    3187/25496 = 1/8

    6381/57429 = 1/9

The sum of the numerator digits and the denominator digits will, of
course, always be 45, and the "digital root" is 9. Now, if we separate
the nine digits into any two groups, the sum of the two digital roots
will always be 9. In fact, the two digital roots must be either 9--9,
8--1, 7--2, 6--3, or 5--4. In the first case the actual sum is 18, but
then the digital root of this number is itself 9. The solutions in the
cases of one-third, one-fourth, one-sixth, one-seventh, and one-ninth
must be of the form 9--9; that is to say, the digital roots of both
numerator and denominator will be 9. In the cases of one-half and
one-fifth, however, the digital roots are 6--3, but of course the higher
root may occur either in the numerator or in the denominator; thus
2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first two
arrangements, the roots of the numerator and denominator are
respectively 6--3, and in the last two 3--6. The most curious case of
all is, perhaps, one-eighth, for here the digital roots may be of any
one of the five forms given above.

The denominators of the fractions being regarded as the numerators
multiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must pay
attention to the "carryings over." In order to get five figures in the
product there will, of course, always be a carry-over after multiplying
the last figure to the left, and in every case higher than 4 we must
carry over at least three times. Consequently in cases from one-fifth to
one-ninth we cannot produce different solutions by a mere change of
position of pairs of figures, as, for example, we may with 5832/17496
and 5823/17469, where the 2/6 and 3/9 change places. It is true that the
same figures may often be differently arranged, as shown in the two
pairs of values for one-fifth that I have given in the last paragraph,
but here it will be found there is a general readjustment of figures and
not a simple changing of the positions of pairs. There are other little
points that would occur to every solver--such as that the figure 5
cannot ever appear to the extreme right of the numerator, as this would
result in our getting either a nought or a second 5 in the denominator.
Similarly 1 cannot ever appear in the same position, nor 6 in the
fraction one-sixth, nor an even figure in the fraction one-fifth, and so
on. The preliminary consideration of such points as I have touched upon
will not only prevent our wasting a lot of time in trying to produce
impossible forms, but will lead us more or less directly to the desired
solutions.


89.--ADDING THE DIGITS.

The smallest possible sum of money is £1, 8s. 9¾d., the digits of which
add to 25.


90.--THE CENTURY PUZZLE.

The problem of expressing the number 100 as a mixed number or fraction,
using all the nine digits once, and once only, has, like all these
digital puzzles, a fascinating side to it. The merest tyro can by
patient trial obtain correct results, and there is a singular pleasure
in discovering and recording each new arrangement akin to the delight of
the botanist in finding some long-sought plant. It is simply a matter of
arranging those nine figures correctly, and yet with the thousands of
possible combinations that confront us the task is not so easy as might
at first appear, if we are to get a considerable number of results. Here
are eleven answers, including the one I gave as a specimen:--

       2148     1752     1428     1578
    96 ----, 96 ----, 96 ----, 94 ----,
        537      438      357      263


       7524     5823     5742    3546
    91 ----, 91 ----, 91 ----, 82 ----,
        836      647      638     197


        7524    5643    69258
     81 ----, 81 ----, 3 -----.
         396      297     714

Now, as all the fractions necessarily represent whole numbers, it will
be convenient to deal with them in the following form: 96 + 4, 94 + 6,
91 + 9, 82 + 18, 81 + 19, and 3 + 97.

With any whole number the digital roots of the fraction that brings it
up to 100 will always be of one particular form. Thus, in the case of 96
+ 4, one can say at once that if any answers are obtainable, then the
roots of both the numerator and the denominator of the fraction will be
6. Examine the first three arrangements given above, and you will find
that this is so. In the case of 94 + 6 the roots of the numerator and
denominator will be respectively 3--2, in the case of 91 + 9 and of 82 +
18 they will be 9--8, in the case of 81 + 19 they will be 9--9, and in
the case of 3 + 97 they will be 3--3. Every fraction that can be
employed has, therefore, its particular digital root form, and you are
only wasting your time in unconsciously attempting to break through this
law.

Every reader will have perceived that certain whole numbers are
evidently impossible. Thus, if there is a 5 in the whole number, there
will also be a nought or a second 5 in the fraction, which are barred by
the conditions. Then multiples of 10, such as 90 and 80, cannot of
course occur, nor can the whole number conclude with a 9, like 89 and
79, because the fraction, equal to 11 or 21, will have 1 in the last
place, and will therefore repeat a figure. Whole numbers that repeat a
figure, such as 88 and 77, are also clearly useless. These cases, as I
have said, are all obvious to every reader. But when I declare that such
combinations as 98 + 2, 92 + 8, 86 + 14, 83 + 17, 74 + 26, etc., etc.,
are to be at once dismissed as impossible, the reason is not so evident,
and I unfortunately cannot spare space to explain it.

But when all those combinations have been struck out that are known to
be impossible, it does not follow that all the remaining "possible
forms" will actually work. The elemental form may be right enough, but
there are other and deeper considerations that creep in to defeat our
attempts. For example, 98 + 2 is an impossible combination, because we
are able to say at once that there is no possible form for the digital
roots of the fraction equal to 2. But in the case of 97 + 3 there is a
possible form for the digital roots of the fraction, namely, 6--5, and
it is only on further investigation that we are able to determine that
this form cannot in practice be obtained, owing to curious
considerations. The working is greatly simplified by a process of
elimination, based on such considerations as that certain
multiplications produce a repetition of figures, and that the whole
number cannot be from 12 to 23 inclusive, since in every such case
sufficiently small denominators are not available for forming the
fractional part.


91.--MORE MIXED FRACTIONS.

The point of the present puzzle lies in the fact that the numbers 15 and
18 are not capable of solution. There is no way of determining this
without trial. Here are answers for the ten possible numbers:--

    9+5472/1368 = 13;
    9+6435/1287 = 14;
    12+3576/894 = 16;
    6+13258/947 = 20;
    15+9432/786 = 27;
    24+9756/813 = 36;
    27+5148/396 = 40;
    65+1892/473 = 69;
    59+3614/278 = 72;
    75+3648/192 = 94.

I have only found the one arrangement for each of the numbers 16, 20,
and 27; but the other numbers are all capable of being solved in more
than one way. As for 15 and 18, though these may be easily solved as a
simple fraction, yet a "mixed fraction" assumes the presence of a whole
number; and though my own idea for dodging the conditions is the
following, where the fraction is both complex and mixed, it will be
fairer to keep exactly to the form indicated:--

      3952
      ----
       746  = 15;
    3 ----
        1

      5742
      ----
       638  = 18.
    9 ----
        1

I have proved the possibility of solution for all numbers up to 100,
except 1, 2, 3, 4, 15, and 18. The first three are easily shown to be
impossible. I have also noticed that numbers whose digital root is
8--such as 26, 35, 44, 53, etc.--seem to lend themselves to the greatest
number of answers. For the number 26 alone I have recorded no fewer than
twenty-five different arrangements, and I have no doubt that there are
many more.


92.--DIGITAL SQUARE NUMBERS.

So far as I know, there are no published tables of square numbers that
go sufficiently high to be available for the purposes of this puzzle.
The lowest square number containing all the nine digits once, and once
only, is 139,854,276, the square of 11,826. The highest square number
under the same conditions is, 923,187,456, the square of 30,384.


93.--THE MYSTIC ELEVEN.

Most people know that if the sum of the digits in the odd places of any
number is the same as the sum of the digits in the even places, then the
number is divisible by 11 without remainder. Thus in 896743012 the odd
digits, 20468, add up 20, and the even digits, 1379, also add up 20.
Therefore the number may be divided by 11. But few seem to know that if
the difference between the sum of the odd and the even digits is 11, or
a multiple of 11, the rule equally applies. This law enables us to find,
with a very little trial, that the smallest number containing nine of
the ten digits (calling nought a digit) that is divisible by 11 is
102,347,586, and the highest number possible, 987,652,413.


94.--THE DIGITAL CENTURY.

There is a very large number of different ways in which arithmetical
signs may be placed between the nine digits, arranged in numerical
order, so as to give an expression equal to 100. In fact, unless the
reader investigated the matter very closely, he might not suspect that
so many ways are possible. It was for this reason that I added the
condition that not only must the fewest possible signs be used, but also
the fewest possible strokes. In this way we limit the problem to a
single solution, and arrive at the simplest and therefore (in this case)
the best result.

Just as in the case of magic squares there are methods by which we may
write down with the greatest ease a large number of solutions, but not
all the solutions, so there are several ways in which we may quickly
arrive at dozens of arrangements of the "Digital Century," without
finding all the possible arrangements. There is, in fact, very little
principle in the thing, and there is no certain way of demonstrating
that we have got the best possible solution. All I can say is that the
arrangement I shall give as the best is the best I have up to the
present succeeded in discovering. I will give the reader a few
interesting specimens, the first being the solution usually published,
and the last the best solution that I know.

                                              Signs. Strokes.
    1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 × 9) = 100 ( 9       18)

    - (1 × 2) - 3 - 4 - 5 + (6 × 7) + (8 × 9)
        = 100                                 (12       20)

    1 + (2 × 3) + (4 × 5) - 6 + 7 + (8 × 9)
        = 100                                 (11       21)

    (1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100  ( 9       12)

    1 + (2 × 3) + 4 + 5 + 67 + 8 + 9 =100     (8        16)

    (1 × 2) + 34 + 56 + 7 - 8 + 9 = 100       (7        13)

    12 + 3 - 4 + 5 + 67 + 8 + 9 = 100         (6        11)

    123 - 4 - 5 - 6 - 7 + 8 - 9 = 100         (6         7)

    123 + 4 - 5 + 67 - 8 - 9 = 100            (4         6)

    123 + 45 - 67 + 8 - 9 = 100               (4         6)

    123 - 45 - 67 + 89 = 100                  (3         4)

It will be noticed that in the above I have counted the bracket as one
sign and two strokes. The last solution is singularly simple, and I do
not think it will ever be beaten.


95.--THE FOUR SEVENS.

The way to write four sevens with simple arithmetical signs so that they
represent 100 is as follows:--

     7    7
    -- × -- = 100.
    .7   .7

Of course the fraction, 7 over decimal 7, equals 7 divided by 7/10,
which is the same as 70 divided by 7, or 10. Then 10 multiplied by 10 is
100, and there you are! It will be seen that this solution applies
equally to any number whatever that you may substitute for 7.


96.--THE DICE NUMBERS.

The sum of all the numbers that can be formed with any given set of four
different figures is always 6,666 multiplied by the sum of the four
figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is 66,660. Now,
there are thirty-five different ways of selecting four figures from the
seven on the dice--remembering the 6 and 9 trick. The figures of all
these thirty-five groups add up to 600. Therefore 6,666 multiplied by
600 gives us 3,999,600 as the correct answer.

Let us discard the dice and deal with the problem generally, using the
nine digits, but excluding nought. Now, if you were given simply the sum
of the digits--that is, if the condition were that you could use any
four figures so long as they summed to a given amount--then we have to
remember that several combinations of four digits will, in many cases,
make the same sum.

    10 11 12 13 14 15 16 17 18 19 20
     1  1  2  3  5  6  8  9 11 11 12

    21 22 23 24 25 26 27 28 29 30
    11 11  9  8  6  5  3  2  1  1

Here the top row of numbers gives all the possible sums of four
different figures, and the bottom row the number of different ways in
which each sum may be made. For example 13 may be made in three ways:
1237, 1246, and 1345. It will be found that the numbers in the bottom
row add up to 126, which is the number of combinations of nine figures
taken four at a time. From this table we may at once calculate the
answer to such a question as this: What is the sum of all the numbers
composed of our different digits (nought excluded) that add up to 14?
Multiply 14 by the number beneath t in the table, 5, and multiply the
result by 6,666, and you will have the answer. It follows that, to know
the sum of all the numbers composed of four different digits, if you
multiply all the pairs in the two rows and then add the results
together, you will get 2,520, which, multiplied by 6,666, gives the
answer 16,798,320.

The following general solution for any number of digits will doubtless
interest readers. Let n represent number of digits, then 5 (10^n - 1) 8!
divided by (9 - n)! equals the required sum. Note that 0! equals 1. This
may be reduced to the following practical rule: Multiply together 4 × 7
× 6 × 5 ... to (n - 1) factors; now add (n + 1) ciphers to the right,
and from this result subtract the same set of figures with a single
cipher to the right. Thus for n = 4 (as in the case last mentioned), 4 ×
7 × 6 = 168. Therefore 16,800,000 less 1,680 gives us 16,798,320 in
another way.


97.--THE SPOT ON THE TABLE.

The ordinary schoolboy would correctly treat this as a quadratic
equation. Here is the actual arithmetic. Double the product of the two
distances from the walls. This gives us 144, which is the square of 12.
The sum of the two distances is 17. If we add these two numbers, 12 and
17, together, and also subtract one from the other, we get the two
answers that 29 or 5 was the radius, or half-diameter, of the table.
Consequently, the full diameter was 58 in. or 10 in. But a table of the
latter dimensions would be absurd, and not at all in accordance with the
illustration. Therefore the table must have been 58 in. in diameter. In
this case the spot was on the edge nearest to the corner of the room--to
which the boy was pointing. If the other answer were admissible, the
spot would be on the edge farthest from the corner of the room.


98.--ACADEMIC COURTESIES.

There must have been ten boys and twenty girls. The number of bows girl
to girl was therefore 380, of boy to boy 90, of girl with boy 400, and
of boys and girls to teacher 30, making together 900, as stated. It will
be remembered that it was not said that the teacher himself returned the
bows of any child.


99.--THE THIRTY-THREE PEARLS.

The value of the large central pearl must have been £3,000. The pearl at
one end (from which they increased in value by £100) was £1,400; the
pearl at the other end, £600.


100.--THE LABOURER'S PUZZLE.

The man said, "I am going twice as deep," not "as deep again." That is
to say, he was still going twice as deep as he had gone already, so that
when finished the hole would be three times its present depth. Then the
answer is that at present the hole is 3 ft. 6 in. deep and the man 2 ft.
4 in. above ground. When completed the hole will be 10 ft. 6 in. deep,
and therefore the man will then be 4 ft. 8 in. below the surface, or
twice the distance that he is now above ground.


101.--THE TRUSSES OF HAY.

Add together the ten weights and divide by 4, and we get 289 lbs. as the
weight of the five trusses together. If we call the five trusses in the
order of weight A, B, C, D, and E, the lightest being A and the heaviest
E, then the lightest, no lbs., must be the weight of A and B; and the
next lightest, 112 lbs., must be the weight of A and C. Then the two
heaviest, D and E, must weigh 121 lbs., and C and E must weigh 120 lbs.
We thus know that A, B, D, and E weigh together 231 lbs., which,
deducted from 289 lbs. (the weight of the five trusses), gives us the
weight of C as 58 lbs. Now, by mere subtraction, we find the weight of
each of the five trusses--54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62
lbs. respectively.


102.--MR. GUBBINS IN A FOG.

The candles must have burnt for three hours and three-quarters. One
candle had one-sixteenth of its total length left and the other
four-sixteenths.


103.--PAINTING THE LAMP-POSTS.

Pat must have painted six more posts than Tim, no matter how many
lamp-posts there were. For example, suppose twelve on each side; then
Pat painted fifteen and Tim nine. If a hundred on each side, Pat painted
one hundred and three, and Tim only ninety-seven


104.--CATCHING THE THIEF.

The constable took thirty steps. In the same time the thief would take
forty-eight, which, added to his start of twenty-seven, carried him
seventy-five steps. This distance would be exactly equal to thirty steps
of the constable.


105.--THE PARISH COUNCIL ELECTION,

The voter can vote for one candidate in 23 ways, for two in 253 ways,
for three in 1,771, for four in 8,855, for five in 33,649, for six in
100,947, for seven in 245,157, for eight in 490,314, and for nine
candidates in 817,190 different ways. Add these together, and we get the
total of 1,698,159 ways of voting.


106.--THE MUDDLETOWN ELECTION.

The numbers of votes polled respectively by the Liberal, the
Conservative, the Independent, and the Socialist were 1,553, 1,535,
1,407, and 978 All that was necessary was to add the sum of the three
majorities (739) to the total poll of 5,473 (making 6,212) and divide by
4, which gives us 1,553 as the poll of the Liberal. Then the polls of
the other three candidates can, of course, be found by deducting the
successive majorities from the last-mentioned number.


107.--THE SUFFRAGISTS' MEETING.

Eighteen were present at the meeting and eleven left. If twelve had
gone, two-thirds would have retired. If only nine had gone, the meeting
would have lost half its members.


108.--THE LEAP-YEAR LADIES.

The correct and only answer is that 11,616 ladies made proposals of
marriage. Here are all the details, which the reader can check for
himself with the original statements. Of 10,164 spinsters, 8,085 married
bachelors, 627 married widowers, 1,221 were declined by bachelors, and
231 declined by widowers. Of the 1,452 widows, 1,155 married bachelors,
and 297 married widowers. No widows were declined. The problem is not
difficult, by algebra, when once we have succeeded in correctly stating
it.


109.--THE GREAT SCRAMBLE.

The smallest number of sugar plums that will fulfil the conditions is
26,880. The five boys obtained respectively: Andrew, 2,863; Bob, 6,335;
Charlie, 2,438; David, 10,294; Edgar, 4,950. There is a little trap
concealed in the words near the end, "one-fifth of the same," that seems
at first sight to upset the whole account of the affair. But a little
thought will show that the words could only mean "one-fifth of
five-eighths", the fraction last mentioned--that is, one-eighth of the
three-quarters that Bob and Andrew had last acquired.


110.--THE ABBOT'S PUZZLE.

The only answer is that there were 5 men, 25 women, and 70 children.
There were thus 100 persons in all, 5 times as many women as men, and as
the men would together receive 15 bushels, the women 50 bushels, and the
children 35 bushels, exactly 100 bushels would be distributed.


111.--REAPING THE CORN.

The whole field must have contained 46.626 square rods. The side of the
central square, left by the farmer, is 4.8284 rods, so it contains
23.313 square rods. The area of the field was thus something more than a
quarter of an acre and less than one-third; to be more precise, .2914 of
an acre.


112.--A PUZZLING LEGACY.

As the share of Charles falls in through his death, we have merely to
divide the whole hundred acres between Alfred and Benjamin in the
proportion of one-third to one-fourth--that is in the proportion of
four-twelfths to three-twelfths, which is the same as four to three.
Therefore Alfred takes four-sevenths of the hundred acres and Benjamin
three-sevenths.


113.--THE TORN NUMBER.

The other number that answers all the requirements of the puzzle is
9,801. If we divide this in the middle into two numbers and add them
together we get 99, which, multiplied by itself, produces 9,801. It is
true that 2,025 may be treated in the same way, only this number is
excluded by the condition which requires that no two figures should be
alike.

The general solution is curious. Call the number of figures in each half
of the torn label n. Then, if we add 1 to each of the exponents of the
prime factors (other than 3) of 10^n - 1 (1 being regarded as a factor
with the constant exponent, 1), their product will be the number of
solutions. Thus, for a label of six figures, n = 3. The factors of 10^n
- 1 are 1¹ × 37¹ (not considering the 3³), and the product of 2 × 2 =
4, the number of solutions. This always includes the special cases 98 -
01, 00 - 01, 998 - 01, 000 - 001, etc. The solutions are obtained as
follows:--Factorize 10³ - 1 in all possible ways, always keeping the
powers of 3 together, thus, 37 × 27, 999 × 1. Then solve the equation
37x = 27y + 1. Here x = 19 and y = 26. Therefore, 19 × 37 = 703, the
square of which gives one label, 494,209. A complementary solution
(through 27x = 37x + 1) can at once be found by 10^n - 703 = 297, the
square of which gives 088,209 for second label. (These non-significant
noughts to the left must be included, though they lead to peculiar cases
like 00238 - 04641 = 4879², where 0238 - 4641 would not work.) The
special case 999 × 1 we can write at once 998,001, according to the law
shown above, by adding nines on one half and noughts on the other, and
its complementary will be 1 preceded by five noughts, or 000001. Thus we
get the squares of 999 and 1. These are the four solutions.


114.--CURIOUS NUMBERS.

The three smallest numbers, in addition to 48, are 1,680, 57,120, and
1,940,448. It will be found that 1,681 and 841, 57,121 and 28,561,
1,940,449 and 970,225, are respectively the squares of 41 and 29, 239
and 169, 1,393 and 985.


115.--A PRINTER'S ERROR.

The answer is that 2^5 .9^2 is the same as 2592, and this is the only
possible solution to the puzzle.


116.--THE CONVERTED MISER.

As we are not told in what year Mr. Jasper Bullyon made the generous
distribution of his accumulated wealth, but are required to find the
lowest possible amount of money, it is clear that we must look for a
year of the most favourable form.

There are four cases to be considered--an ordinary year with fifty-two
Sundays and with fifty-three Sundays, and a leap-year with fifty-two and
fifty-three Sundays respectively. Here are the lowest possible amounts
in each case:--

    313  weekdays, 52 Sundays               £112,055
    312  weekdays, 53 Sundays                 19,345
    314  weekdays, 52 Sundays  No solution possible.
    313  weekdays, 53 Sundays                £69,174

The lowest possible amount, and therefore the correct answer, is
£19,345, distributed in an ordinary year that began on a Sunday. The
last year of this kind was 1911. He would have paid £53 on every day of
the year, or £62 on every weekday, with £1 left over, as required, in
the latter event.


117.--A FENCE PROBLEM.

Though this puzzle presents no great difficulty to any one possessing a
knowledge of algebra, it has perhaps rather interesting features.

Seeing, as one does in the illustration, just one corner of the proposed
square, one is scarcely prepared for the fact that the field, in order
to comply with the conditions, must contain exactly 501,760 acres, the
fence requiring the same number of rails. Yet this is the correct
answer, and the only answer, and if that gentleman in Iowa carries out
his intention, his field will be twenty-eight miles long on each side,
and a little larger than the county of Westmorland. I am not aware that
any limit has ever been fixed to the size of a "field," though they do
not run so large as this in Great Britain. Still, out in Iowa, where my
correspondent resides, they do these things on a very big scale. I have,
however, reason to believe that when he finds the sort of task he has
set himself, he will decide to abandon it; for if that cow decides to
roam to fresh woods and pastures new, the milkmaid may have to start out
a week in advance in order to obtain the morning's milk.

Here is a little rule that will always apply where the length of the
rail is half a pole. Multiply the number of rails in a hurdle by four,
and the result is the exact number of miles in the side of a square
field containing the same number of acres as there are rails in the
complete fence. Thus, with a one-rail fence the field is four miles
square; a two-rail fence gives eight miles square; a three-rail fence,
twelve miles square; and so on, until we find that a seven-rail fence
multiplied by four gives a field of twenty-eight miles square. In the
case of our present problem, if the field be made smaller, then the
number of rails will exceed the number of acres; while if the field be
made larger, the number of rails will be less than the acres of the
field.


118.--CIRCLING THE SQUARES.

Though this problem might strike the novice as being rather difficult,
it is, as a matter of fact, quite easy, and is made still easier by
inserting four out of the ten numbers.

First, it will be found that squares that are diametrically opposite
have a common difference. For example, the difference between the square
of 14 and the square of 2, in the diagram, is 192; and the difference
between the square of 16 and the square of 8 is also 192. This must be
so in every case. Then it should be remembered that the difference
between squares of two consecutive numbers is always twice the smaller
number plus 1, and that the difference between the squares of any two
numbers can always be expressed as the difference of the numbers
multiplied by their sum. Thus the square of 5 (25) less the square of 4
(16) equals (2 × 4) + 1, or 9; also, the square of 7 (49) less the
square of 3 (9) equals (7 + 3) × (7 - 3), or 40.

Now, the number 192, referred to above, may be divided into five
different pairs of even factors: 2 × 96, 4 × 48, 6 × 32, 8 × 24, and 12
× 16, and these divided by 2 give us, 1 × 48, 2 × 24, 3 × 16, 4 × 12,
and 6 × 8. The difference and sum respectively of each of these pairs in
turn produce 47, 49; 22, 26; 13, 19; 8, 16; and 2, 14. These are the
required numbers, four of which are already placed. The six numbers that
have to be added may be placed in just six different ways, one of which
is as follows, reading round the circle clockwise: 16, 2, 49, 22, 19, 8,
14, 47, 26, 13.

I will just draw the reader's attention to one other little point. In
all circles of this kind, the difference between diametrically opposite
numbers increases by a certain ratio, the first numbers (with the
exception of a circle of 6) being 4 and 6, and the others formed by
doubling the next preceding but one. Thus, in the above case, the first
difference is 2, and then the numbers increase by 4, 6, 8, and 12. Of
course, an infinite number of solutions may be found if we admit
fractions. The number of squares in a circle of this kind must, however,
be of the form 4n + 6; that is, it must be a number composed of 6 plus a
multiple of 4.


119.--RACKBRANE'S LITTLE LOSS.

The professor must have started the game with thirteen shillings, Mr.
Potts with four shillings, and Mrs. Potts with seven shillings.


120.--THE FARMER AND HIS SHEEP.

The farmer had one sheep only! If he divided this sheep (which is best
done by weight) into two parts, making one part two-thirds and the other
part one-third, then the difference between these two numbers is the
same as the difference between their squares--that is, one-third. Any
two fractions will do if the denominator equals the sum of the two
numerators.


121.--HEADS OR TAILS.

Crooks must have lost, and the longer he went on the more he would lose.
In two tosses he would be left with three-quarters of his money, in four
tosses with nine-sixteenths of his money, in six tosses with
twenty-seven sixty-fourths of his money, and so on. The order of the
wins and losses makes no difference, so long as their number is in the
end equal.


122.--THE SEE-SAW PUZZLE.

The boy's weight must have been about 39.79 lbs. A brick weighed 3 lbs.
Therefore 16 bricks weighed 48 lbs. and 11 bricks 33 lbs. Multiply 48 by
33 and take the square root.


123.--A LEGAL DIFFICULTY.

It was clearly the intention of the deceased to give the son twice as
much as the mother, or the daughter half as much as the mother.
Therefore the most equitable division would be that the mother should
take two-sevenths, the son four-sevenths, and the daughter one-seventh.


124.--A QUESTION OF DEFINITION.

There is, of course, no difference in _area_ between a mile square and a
square mile. But there may be considerable difference in _shape_. A mile
square can be no other shape than square; the expression describes a
surface of a certain specific size and shape. A square mile may be of
any shape; the expression names a unit of area, but does not prescribe
any particular shape.


125.--THE MINERS' HOLIDAY.

Bill Harris must have spent thirteen shillings and sixpence, which would
be three shillings more than the average for the seven men--half a
guinea.


126.--SIMPLE MULTIPLICATION.

The number required is 3,529,411,764,705,882, which may be multiplied by
3 and divided by 2, by the simple expedient of removing the 3 from one
end of the row to the other. If you want a longer number, you can
increase this one to any extent by repeating the sixteen figures in the
same order.


127.--SIMPLE DIVISION.

Subtract every number in turn from every other number, and we get 358
(twice), 716, 1,611, 1,253, and 895. Now, we see at a glance that, as
358 equals 2 × 179, the only number that can divide in every case
without a remainder will be 179. On trial we find that this is such a
divisor. Therefore, 179 is the divisor we want, which always leaves a
remainder 164 in the case of the original numbers given.


128.--A PROBLEM IN SQUARES.

The sides of the three boards measure 31 in., 41 in., and 49 in. The
common difference of area is exactly five square feet. Three numbers
whose squares are in A.P., with a common difference of 7, are 113/120,
337/120, 463/120; and with a common difference of 13 are 80929/19380,
106921/19380, and 127729/19380. In the case of whole square numbers the
common difference will always be divisible by 24, so it is obvious that
our squares must be fractional. Readers should now try to solve the case
where the common difference is 23. It is rather a hard nut.


129.--THE BATTLE OF HASTINGS.

Any number (not itself a square number) may be multiplied by a square
that will give a product 1 less than another square. The given number
must not itself be a square, because a square multiplied by a square
produces a square, and no square plus 1 can be a square. My remarks
throughout must be understood to apply to whole numbers, because
fractional soldiers are not of much use in war.

Now, of all the numbers from 2 to 99 inclusive, 61 happens to be the
most awkward one to work, and the lowest possible answer to our puzzle
is that Harold's army consisted of 3,119,882,982,860,264,400 men. That
is, there would be 51,145,622,669,840,400 men (the square of
226,153,980) in each of the sixty-one squares. Add one man (Harold), and
they could then form one large square with 1,766,319,049 men on every
side. The general problem, of which this is a particular case, is known
as the "Pellian Equation"--apparently because Pell neither first
propounded the question nor first solved it! It was issued as a
challenge by Fermat to the English mathematicians of his day. It is
readily solved by the use of continued fractions.

Next to 61, the most difficult number under 100 is 97, where 97 ×
6,377,352² + 1 = a square.

The reason why I assumed that there must be something wrong with the
figures in the chronicle is that we can confidently say that Harold's
army did not contain over three trillion men! If this army (not to
mention the Normans) had had the whole surface of the earth (sea
included) on which to encamp, each man would have had slightly more than
a quarter of a square inch of space in which to move about! Put another
way: Allowing one square foot of standing-room per man, each small
square would have required all the space allowed by a globe three times
the diameter of the earth.


130.--THE SCULPTOR'S PROBLEM.

A little thought will make it clear that the answer must be fractional,
and that in one case the numerator will be greater and in the other case
less than the denominator. As a matter of fact, the height of the larger
cube must be 8/7 ft., and of the smaller 3/7 ft., if we are to have the
answer in the smallest possible figures. Here the lineal measurement is
11/7 ft.--that is, 1+4/7 ft. What are the cubic contents of the two
cubes? First 8/7 × 3/7 × 8/7 = 512/343, and secondly 3/7 × 3/7 × 3/7 =
27/343. Add these together and the result is 539/343, which reduces to
11/7 or 1+4/7 ft. We thus see that the answers in cubic feet and lineal
feet are precisely the same.

The germ of the idea is to be found in the works of Diophantus of
Alexandria, who wrote about the beginning of the fourth century. These
fractional numbers appear in triads, and are obtained from three
generators, a, b, c, where a is the largest and c the smallest.

Then ab + c² = denominator, and a² - c², b² - c², and a² - b² will be
the three numerators. Thus, using the generators 3, 2, 1, we get 8/7,
3/7, 5/7 and we can pair the first and second, as in the above
solution, or the first and third for a second solution. The
denominator must always be a prime number of the form 6n + 1, or
composed of such primes. Thus you can have 13, 19, etc., as
denominators, but not 25, 55, 187, etc.

When the principle is understood there is no difficulty in writing down
the dimensions of as many sets of cubes as the most exacting collector
may require. If the reader would like one, for example, with plenty of
nines, perhaps the following would satisfy him: 99999999/99990001 and
19999/99990001.


131.--THE SPANISH MISER.

There must have been 386 doubloons in one box, 8,450 in another, and
16,514 in the third, because 386 is the smallest number that can occur.
If I had asked for the smallest aggregate number of coins, the answer
would have been 482, 3,362, and 6,242. It will be found in either case
that if the contents of any two of the three boxes be combined, they
form a square number of coins. It is a curious coincidence (nothing
more, for it will not always happen) that in the first solution the
digits of the three numbers add to 17 in every case, and in the second
solution to 14. It should be noted that the middle one of the three
numbers will always be half a square.


132.--THE NINE TREASURE BOXES.

Here is the answer that fulfils the conditions:--

    A = 4         B = 3,364      C = 6,724
    D = 2,116     E = 5,476      F = 8,836
    G = 9,409     H = 12,769     I = 16,129


Each of these is a square number, the roots, taken in alphabetical
order, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the required
difference between A and B, B and C, D and E. etc., is in every case
3,360.


133.--THE FIVE BRIGANDS.

The sum of 200 doubloons might have been held by the five brigands in
any one of 6,627 different ways. Alfonso may have held any number from 1
to 11. If he held 1 doubloon, there are 1,005 different ways of
distributing the remainder; if he held 2, there are 985 ways; if 3,
there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways;
if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388
ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso
held 11 doubloons, the remainder could be distributed in 3 different
ways. More than 11 doubloons he could not possibly have had. It will
scarcely be expected that I shall give all these 6,627 ways at length.
What I propose to do is to enable the reader, if he should feel so
disposed, to write out all the answers where Alfonso has one and the
same amount. Let us take the cases where Alfonso has 6 doubloons, and
see how we may obtain all the 704 different ways indicated above. Here
are two tables that will serve as keys to all these answers:--

         Table I.              Table II.
    A = 6.                A = 6.
    B = n.                B = n.
    C = (63 - 5n) + m.    C = 1 + m.
    D = (128 + 4n) - 4m.  D = (376 - 16n) - 4m.
    E = 3 + 3m.           E = (15n - 183) + 3m.

In the first table we may substitute for n any whole number from 1 to 12
inclusive, and m may be nought or any whole number from 1 to (31 + n)
inclusive. In the second table n may have the value of any whole number
from 13 to 23 inclusive, and m may be nought or any whole number from 1
to (93 - 4n) inclusive. The first table thus gives (32 + n) answers for
every value of n; and the second table gives (94 - 4n) answers for every
value of n. The former, therefore, produces 462 and the latter 242
answers, which together make 704, as already stated.

Let us take Table I., and say n = 5 and m = 2; also in Table II. take n
= 13 and m = 0. Then we at once get these two answers:--

    A =   6           A =   6
    B =   5           B =  13
    C =  40           C =   1
    D = 140           D = 168
    E =   9           E =  12
        ---               ---
        200 doubloons     200 doubloons.

These will be found to work correctly. All the rest of the 704 answers,
where Alfonso always holds six doubloons, may be obtained in this way
from the two tables by substituting the different numbers for the
letters m and n.

Put in another way, for every holding of Alfonso the number of answers
is the sum of two arithmetical progressions, the common difference in
one case being 1 and in the other -4. Thus in the case where Alfonso
holds 6 doubloons one progression is 33 + 34 + 35 + 36 + ... + 43 + 44,
and the other 42 + 38 + 34 + 30 + ... + 6 + 2. The sum of the first
series is 462, and of the second 242--results which again agree with the
figures already given. The problem may be said to consist in finding the
first and last terms of these progressions. I should remark that where
Alfonso holds 9, 10, or 11 there is only one progression, of the second
form.


134.--THE BANKER'S PUZZLE.

In order that a number of sixpences may not be divisible into a number
of equal piles, it is necessary that the number should be a prime. If
the banker can bring about a prime number, he will win; and I will show
how he can always do this, whatever the customer may put in the box, and
that therefore the banker will win to a certainty. The banker must first
deposit forty sixpences, and then, no matter how many the customer may
add, he will desire the latter to transfer from the counter the square
of the number next below what the customer put in. Thus, banker puts 40,
customer, we will say, adds 6, then transfers from the counter 25 (the
square of 5), which leaves 71 in all, a prime number. Try again. Banker
puts 40, customer adds 12, then transfers 121 (the square of 11), as
desired, which leaves 173, a prime number. The key to the puzzle is the
curious fact that any number up to 39, if added to its square and the
sum increased by 41, makes a prime number. This was first discovered by
Euler, the great mathematician. It has been suggested that the banker
might desire the customer to transfer sufficient to raise the contents
of the box to a given number; but this would not only make the thing an
absurdity, but breaks the rule that neither knows what the other puts
in.


135.--THE STONEMASON'S PROBLEM.

The puzzle amounts to this. Find the smallest square number that may be
expressed as the sum of more than three consecutive cubes, the cube 1
being barred. As more than three heaps were to be supplied, this
condition shuts out the otherwise smallest answer, 23³ + 24³ + 25³ =
204². But it admits the answer, 25³ + 26³ + 27³ + 28³ + 29³ = 315². The
correct answer, however, requires more heaps, but a smaller aggregate
number of blocks. Here it is: 14³ + 15³ + ... up to 25³ inclusive, or
twelve heaps in all, which, added together, make 97,344 blocks of stone
that may be laid out to form a square 312 × 312. I will just remark that
one key to the solution lies in what are called triangular numbers. (See
pp. 13, 25, and 166.)


136.--THE SULTAN'S ARMY.

The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, and
the smallest of the form 4n - 1 are 3, 7, 11, 19, and 23. Now, primes of
the first form can always be expressed as the sum of two squares, and in
only one way. Thus, 5 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37
= 36 + 1. But primes of the second form can never be expressed as the
sum of two squares in any way whatever.

In order that a number may be expressed as the sum of two squares in
several different ways, it is necessary that it shall be a composite
number containing a certain number of primes of our first form. Thus, 5
or 13 alone can only be so expressed in one way; but 65, (5 × 13), can
be expressed in two ways, 1,105, (5 × 13 × 17), in four ways, 32,045, (5
× 13 × 17 × 29), in eight ways. We thus get double as many ways for
every new factor of this form that we introduce. Note, however, that I
say _new_ factor, for the _repetition_ of factors is subject to another
law. We cannot express 25, (5 × 5), in two ways, but only in one; yet
125, (5 × 5 × 5), can be given in two ways, and so can 625, (5 × 5 × 5 ×
5); while if we take in yet another 5 we can express the number as the
sum of two squares in three different ways.

If a prime of the second form gets into your composite number, then that
number cannot be the sum of two squares. Thus 15, (3 × 5), will not
work, nor will 135, (3 × 3 × 3 × 5); but if we take in an even number of
3's it will work, because these 3's will themselves form a square
number, but you will only get one solution. Thus, 45, (3 × 3 × 5, or 9 ×
5) = 36 + 9. Similarly, the factor 2 may always occur, or any power of
2, such as 4, 8, 16, 32; but its introduction or omission will never
affect the number of your solutions, except in such a case as 50, where
it doubles a square and therefore gives you the two answers, 49 + 1 and
25 + 25.

Now, directly a number is decomposed into its prime factors, it is
possible to tell at a glance whether or not it can be split into two
squares; and if it can be, the process of discovery in how many ways is
so simple that it can be done in the head without any effort. The number
I gave was 130. I at once saw that this was 2 × 5 × 13, and consequently
that, as 65 can be expressed in two ways (64 + 1 and 49 + 16), 130 can
also be expressed in two ways, the factor 2 not affecting the question.

The smallest number that can be expressed as the sum of two squares in
twelve different ways is 160,225, and this is therefore the smallest
army that would answer the Sultan's purpose. The number is composed of
the factors 5 × 5 × 13 × 17 × 29, each of which is of the required form.
If they were all different factors, there would be sixteen ways; but as
one of the factors is repeated, there are just twelve ways. Here are the
sides of the twelve pairs of squares: (400 and 15), (399 and 32), (393
and 76), (392 and 81), (384 and 113), (375 and 140), (360 and 175), (356
and 183), (337 and 216), (329 and 228), (311 and 252), (265 and 300).
Square the two numbers in each pair, add them together, and their sum
will in every case be 160,225.


137.--A STUDY IN THRIFT.

Mrs. Sandy McAllister will have to save a tremendous sum out of her
housekeeping allowance if she is to win that sixth present that her
canny husband promised her. And the allowance must be a very liberal one
if it is to admit of such savings. The problem required that we should
find five numbers higher than 36 the units of which may be displayed so
as to form a square, a triangle, two triangles, and three triangles,
using the complete number in every one of the four cases.

Every triangular number is such that if we multiply it by 8 and add 1
the result is an odd square number. For example, multiply 1, 3, 6, 10,
15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which are
the squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every case
where 8x² + 1 = a square number, x² is also a triangular. This point
is dealt with in our puzzle, "The Battle of Hastings." I will now merely
show again how, when the first solution is found, the others may be
discovered without any difficulty. First of all, here are the figures:--

    8 ×     1² + 1 =      3²
    8 ×     6² + 1 =     17²
    8 ×    35² + 1 =     99²
    8 ×   204² + 1 =    577²
    8 ×  1189² + 1 =   3363²
    8 ×  6930² + 1 =  19601²
    8 × 40391² + 1 = 114243²

The successive pairs of numbers are found in this way:--

    (1 × 3) +  (3 × 1)  =   6    (8 × 1) +  (3 × 3) =  17
    (1 × 17) + (3 × 6)  =  35    (8 × 6) + (3 × 17) =  99
    (1 × 99) + (3 × 35) = 204   (8 × 35) + (3 × 99) = 577

and so on. Look for the numbers in the table above, and the method will
explain itself.

Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and
1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and
40391; and they will also form single triangles with sides of 8, 49,
288, 1681, 9800, and 57121. These numbers may be obtained from the last
column in the first table above in this way: simply divide the numbers
by 2 and reject the remainder. Thus the integral halves of 17, 99, and
577 are 8, 49, and 288.

All the numbers we have found will form either two or three triangles at
will. The following little diagram will show you graphically at a glance
that every square number must necessarily be the sum of two triangulars,
and that the side of one triangle will be the same as the side of the
corresponding square, while the other will be just 1 less.

[Illustration

                 +-----------+
    +---------+  |. . . . ./.|
    |. . . ./.|  |. . . ./. .|
    |. . ./. .|  |. . ./. . .|
    |. ./. . .|  |. ./. . . .|
    |./. . . .|  |./. . . . .|
    /. . . . .|  /. . . . . .|
    +---------+  +-----------+

]

Thus a square may always be divided easily into two triangles, and the
sum of two consecutive triangulars will always make a square. In numbers
it is equally clear, for if we examine the first triangulars--1, 3, 6,
10, 15, 21, 28--we find that by adding all the consecutive pairs in turn
we get the series of square numbers--9, 16, 25, 36, 49, etc.

The method of forming three triangles from our numbers is equally
direct, and not at all a matter of trial. But I must content myself with
giving actual figures, and just stating that every triangular higher
than 6 will form three triangulars. I give the sides of the triangles,
and readers will know from my remarks when stating the puzzle how to
find from these sides the number of counters or coins in each, and so
check the results if they so wish.

 +----------------------+-----------+---------------+-----------------------+
 | Number     | Side of |  Side of  | Sides of Two  |   Sides of Three      |
 |            | Square. | Triangle. |  Triangles.   |     Triangles.        |
 +------------+---------+-----------+---------------+-----------------------+
 |         36 |      6  |       8   |     6 + 5     |      5 + 5 + 3        |
 |       1225 |     35  |      49   |    36 + 34    |     33 + 32 + 16      |
 |      41616 |    204  |     288   |   204 + 203   |    192 + 192 + 95     |
 |    1413721 |   1189  |    1681   |  1189 + 1188  |  1121 + 1120 + 560    |
 |   48024900 |   6930  |    9800   |  6930 + 6929  |  6533 + 6533 + 3267   |
 | 1631432881 |  40391  |   57121   | 40391 + 40390 | 38081 + 38080 + 19040 |
 +------------+---------+-----------+---------------+-----------------------+

I should perhaps explain that the arrangements given in the last two
columns are not the only ways of forming two and three triangles. There
are others, but one set of figures will fully serve our purpose. We thus
see that before Mrs. McAllister can claim her sixth £5 present she must
save the respectable sum of £1,631,432,881.


138.--THE ARTILLERYMEN'S DILEMMA.

We were required to find the smallest number of cannon balls that we
could lay on the ground to form a perfect square, and could pile into a
square pyramid. I will try to make the matter clear to the merest
novice.

        1   2   3   4   5   6   7
        1   3   6  10  15  21  28
        1   4  10  20  35  56  84
        1   5  14  30  55  91 140

Here in the first row we place in regular order the natural numbers.
Each number in the second row represents the sum of the numbers in the
row above, from the beginning to the number just over it. Thus 1, 2, 3,
4, added together, make 10. The third row is formed in exactly the same
way as the second. In the fourth row every number is formed by adding
together the number just above it and the preceding number. Thus 4 and
10 make 14, 20 and 35 make 55. Now, all the numbers in the second row
are triangular numbers, which means that these numbers of cannon balls
may be laid out on the ground so as to form equilateral triangles. The
numbers in the third row will all form our triangular pyramids, while
the numbers in the fourth row will all form square pyramids.

Thus the very process of forming the above numbers shows us that every
square pyramid is the sum of two triangular pyramids, one of which has
the same number of balls in the side at the base, and the other one ball
fewer. If we continue the above table to twenty-four places, we shall
reach the number 4,900 in the fourth row. As this number is the square
of 70, we can lay out the balls in a square, and can form a square
pyramid with them. This manner of writing out the series until we come
to a square number does not appeal to the mathematical mind, but it
serves to show how the answer to the particular puzzle may be easily
arrived at by anybody. As a matter of fact, I confess my failure to
discover any number other than 4,900 that fulfils the conditions, nor
have I found any rigid proof that this is the only answer. The problem
is a difficult one, and the second answer, if it exists (which I do not
believe), certainly runs into big figures.

For the benefit of more advanced mathematicians I will add that the
general expression for square pyramid numbers is (2n³ + 3n² + n)/6.
For this expression to be also a square number (the special case of 1
excepted) it is necessary that n = p² - 1 = 6t², where 2p² - 1 = q²
(the "Pellian Equation"). In the case of our solution above, n = 24, p =
5, t = 2, q = 7.


139.--THE DUTCHMEN'S WIVES.

The money paid in every case was a square number of shillings, because
they bought 1 at 1s., 2 at 2s., 3 at 3s., and so on. But every husband
pays altogether 63s. more than his wife, so we have to find in how many
ways 63 may be the difference between two square numbers. These are the
three only possible ways: the square of 8 less the square of 1, the
square of 12 less the square of 9, and the square of 32 less the square
of 31. Here 1, 9, and 31 represent the number of pigs bought and the
number of shillings per pig paid by each woman, and 8, 12, and 32 the
same in the case of their respective husbands. From the further
information given as to their purchases, we can now pair them off as
follows: Cornelius and Gurtrün bought 8 and 1; Elas and Katrün bought 12
and 9; Hendrick and Anna bought 32 and 31. And these pairs represent
correctly the three married couples.

The reader may here desire to know how we may determine the maximum
number of ways in which a number may be expressed as the difference
between two squares, and how we are to find the actual squares. Any
integer except 1, 4, and twice any odd number, may be expressed as the
difference of two integral squares in as many ways as it can be split up
into pairs of factors, counting 1 as a factor. Suppose the number to be
5,940. The factors are 2².3³.5.11. Here the exponents are 2, 3, 1, 1.
Always deduct 1 from the exponents of 2 and add 1 to all the other
exponents; then we get 1, 4, 2, 2, and half the product of these four
numbers will be the required number of ways in which 5,940 may be the
difference of two squares--that is, 8. To find these eight squares, as
it is an _even_ number, we first divide by 4 and get 1485, the eight
pairs of factors of which are 1 × 1485, 3 × 495, 5 × 297, 9 × 165, 11 ×
135, 15 × 99, 27 × 55, and 33 × 45. The sum and difference of any one of
these pairs will give the required numbers. Thus, the square of 1,486
less the square of 1,484 is 5,940, the square of 498 less the square of
492 is the same, and so on. In the case of 63 above, the number is
_odd_; so we factorize at once, 1 × 63, 3 × 21, 7 × 9. Then we find that
_half_ the sum and difference will give us the numbers 32 and 31, 12 and
9, and 8 and 1, as shown in the solution to the puzzle.

The reverse problem, to find the factors of a number when you have
expressed it as the difference of two squares, is obvious. For example,
the sum and difference of any pair of numbers in the last sentence will
give us the factors of 63. Every prime number (except 1 and 2) may be
expressed as the difference of two squares in one way, and in one way
only. If a number can be expressed as the difference of two squares in
more than one way, it is composite; and having so expressed it, we may
at once obtain the factors, as we have seen. Fermat showed in a letter
to Mersenne or Frénicle, in 1643, how we may discover whether a number
may be expressed as the difference of two squares in more than one way,
or proved to be a prime. But the method, when dealing with large
numbers, is necessarily tedious, though in practice it may be
considerably shortened. In many cases it is the shortest method known
for factorizing large numbers, and I have always held the opinion that
Fermat used it in performing a certain feat in factorizing that is
historical and wrapped in mystery.


140.--FIND ADA'S SURNAME.

The girls' names were Ada Smith, Annie Brown, Emily Jones, Mary
Robinson, and Bessie Evans.


141.--SATURDAY MARKETING.

As every person's purchase was of the value of an exact number of
shillings, and as the party possessed when they started out forty
shilling coins altogether, there was no necessity for any lady to have
any smaller change, or any evidence that they actually had such change.
This being so, the only answer possible is that the women were named
respectively Anne Jones, Mary Robinson, Jane Smith, and Kate Brown. It
will now be found that there would be exactly eight shillings left,
which may be divided equally among the eight persons in coin without any
change being required.


142.--THE SILK PATCHWORK.

[Illustration]

Our illustration will show how to cut the stitches of the patchwork so
as to get the square F entire, and four equal pieces, G, H, I, K, that
will form a perfect Greek cross. The reader will know how to assemble
these four pieces from Fig. 13 in the article.

[Illustration: Fig. 1.]

[Illustration: Fig. 2.]

143.--TWO CROSSES FROM ONE.

It will be seen that one cross is cut out entire, as A in Fig. 1, while
the four pieces marked B, C, D and E form the second cross, as in Fig.
2, which will be of exactly the same size as the other. I will leave the
reader the pleasant task of discovering for himself the best way of
finding the direction of the cuts. Note that the Swastika again appears.

The difficult question now presents itself: How are we to cut three
Greek crosses from one in the fewest possible pieces? As a matter of
fact, this problem may be solved in as few as thirteen pieces; but as I
know many of my readers, advanced geometricians, will be glad to have
something to work on of which they are not shown the solution, I leave
the mystery for the present undisclosed.


144.--THE CROSS AND THE TRIANGLE.

The line A B in the following diagram represents the side of a square
having the same area as the cross. I have shown elsewhere, as stated,
how to make a square and equilateral triangle of equal area. I need not
go, therefore, into the preliminary question of finding the dimensions
of the triangle that is to equal our cross. We will assume that we have
already found this, and the question then becomes, How are we to cut up
one of these into pieces that will form the other?

First draw the line A B where A and B are midway between the extremities
of the two side arms. Next make the lines D C and E F equal in length to
half the side of the triangle. Now from E and F describe with the same
radius the intersecting arcs at G and draw F G. Finally make I K equal
to H C and L B equal to A D. If we now draw I L, it should be parallel
to F G, and all the six pieces are marked out. These fit together and
form a perfect equilateral triangle, as shown in the second diagram. Or
we might have first found the direction of the line M N in our triangle,
then placed the point O over the point E in the cross and turned round
the triangle over the cross until the line M N was parallel to A B. The
piece 5 can then be marked off and the other pieces in succession.

[Illustration]

I have seen many attempts at a solution involving the assumption that
the height of the triangle is exactly the same as the height of the
cross. This is a fallacy: the cross will always be higher than the
triangle of equal area.


145.--THE FOLDED CROSS.

[Illustration: FIG. 1., FIG 2.]

First fold the cross along the dotted line A B in Fig. 1. You then have
it in the form shown in Fig. 2. Next fold it along the dotted line C D
(where D is, of course, the centre of the cross), and you get the form
shown in Fig. 3. Now take your scissors and cut from G to F, and the
four pieces, all of the same size and shape, will fit together and form
a square, as shown in Fig. 4.

[Illustration: FIG. 3., FIG. 4.]


146.--AN EASY DISSECTION PUZZLE.

[Illustration

    +===========+===========+-
    | ·         |         · : \
    |   ·       |       ·   :   \
    |     ·     |     ·     :     \
    |       ·   |   ·       :       \
    |         · | ·         :         \
    +-----------+===========+===========+
    |         / : ·         |         · : \
    |       /   :   ·       |       ·   :   \
    |     /     :     ·     |     ·     :     \
    |   /       :       ·   |   ·       :       \
    | /         :         · | ·         :         \
    +===========+===========+===========+===========+

]

The solution to this puzzle is shown in the illustration. Divide the
figure up into twelve equal triangles, and it is easy to discover the
directions of the cuts, as indicated by the dark lines.


147.--AN EASY SQUARE PUZZLE.

[Illustration

    +-----------------------------------------+
    |  .                                     /|
    |      .                                / |
    |          .                           /  |
    |                /                    /   |
    |               /    .               /    |
    |              /         .          /     |
    |             /              .     /      |
    |            /                   ./       |
    |           +--------------------+        |
    |          /                    /         |
    |         /                    /          |
    |        /                    /           |
    |       / .                  /            |
    |      /      .             /             |
    |     /           .        /              |
    |    /                .   /               |
    |   /                     .               |
    |  /                          .           |
    | /                               .       |
    |/                                    .   |
    +-----------------------------------------+

]

The diagram explains itself, one of the five pieces having been cut in
two to form a square.


148.--THE BUN PUZZLE.

[Illustration

                      .   .
               .               .
          _ .                     .
      .    |\              A        .
           |    \
    .   C  |         \               |
           |              \
     .     |                  \     /
         . |______________________\/
           |                       |
            .                     .
             .         B         .
                .              .
                   .        .
                        -


                        _
                  .     |     .
               .        |        .
             .          |          .
                        |
                        |
           |      D     |     E      |
                        |
                        |
             .          |          .
               .        |        .
                  .     |     .
                        _

                        _
                  .     |     .
               .       -+-       .
             .   .             .   .
               -                 -
                                  .
           | G|           F       |  |

               -                 -
             .   .             .   .
               .   -   _ _  -    .
                  .     |     .
                        -

                       -+-
                 .             .
               -                 -
                                  .
              |          H        |

               -                 -
                 .             .
                   -   _ _  -


]

The secret of the bun puzzle lies in the fact that, with the relative
dimensions of the circles as given, the three diameters will form a
right-angled triangle, as shown by A, B, C. It follows that the two
smaller buns are exactly equal to the large bun. Therefore, if we give
David and Edgar the two halves marked D and E, they will have their fair
shares--one quarter of the confectionery each. Then if we place the
small bun, H, on the top of the remaining one and trace its
circumference in the manner shown, Fred's piece, F, will exactly equal
Harry's small bun, H, with the addition of the piece marked G--half the
rim of the other. Thus each boy gets an exactly equal share, and there
are only five pieces necessary.


149.--THE CHOCOLATE SQUARES.

[Illustration]

Square A is left entire; the two pieces marked B fit together and make a
second square; the two pieces C make a third square; and the four pieces
marked D will form the fourth square.


150.--DISSECTING A MITRE.

The diagram on the next page shows how to cut into five pieces to form a
square. The dotted lines are intended to show how to find the points C
and F--the only difficulty. A B is half B D, and A E is parallel to B H.
With the point of the compasses at B describe the arc H E, and A E will
be the distance of C from B. Then F G equals B C less A B.

This puzzle--with the added condition that it shall be cut into four
parts of the same size and shape--I have not been able to trace to an
earlier date than 1835. Strictly speaking, it is, in that form,
impossible of solution; but I give the answer that is always presented,
and that seems to satisfy most people.

[Illustration]

We are asked to assume that the two portions containing the same
letter--AA, BB, CC, DD--are joined by "a mere hair," and are, therefore,
only one piece. To the geometrician this is absurd, and the four shares
are not equal in area unless they consist of two pieces each. If you
make them equal in area, they will not be exactly alike in shape.

[Illustration]


151.--THE JOINER'S PROBLEM.

[Illustration]

Nothing could be easier than the solution of this puzzle--when you know
how to do it. And yet it is apt to perplex the novice a good deal if he
wants to do it in the fewest possible pieces--three. All you have to do
is to find the point A, midway between B and C, and then cut from A to D
and from A to E. The three pieces then form a square in the manner
shown. Of course, the proportions of the original figure must be
correct; thus the triangle BEF is just a quarter of the square BCDF.
Draw lines from B to D and from C to F and this will be clear.


152.--ANOTHER JOINER'S PROBLEM.

[Illustration]

THE point was to find a general rule for forming a perfect square out of
another square combined with a "right-angled isosceles triangle." The
triangle to which geometricians give this high-sounding name is, of
course, nothing more or less than half a square that has been divided
from corner to corner.

The precise relative proportions of the square and triangle are of no
consequence whatever. It is only necessary to cut the wood or material
into five pieces.

Suppose our original square to be ACLF in the above diagram and our
triangle to be the shaded portion CED. Now, we first find half the
length of the long side of the triangle (CD) and measure off this length
at AB. Then we place the triangle in its present position against the
square and make two cuts--one from B to F, and the other from B to E.
Strange as it may seem, that is all that is necessary! If we now remove
the pieces G, H, and M to their new places, as shown in the diagram, we
get the perfect square BEKF.

Take any two square pieces of paper, of different sizes but perfect
squares, and cut the smaller one in half from corner to corner. Now
proceed in the manner shown, and you will find that the two pieces may
be combined to form a larger square by making these two simple cuts, and
that no piece will be required to be turned over.

The remark that the triangle might be "a little larger or a good deal
smaller in proportion" was intended to bar cases where area of triangle
is greater than area of square. In such cases six pieces are necessary,
and if triangle and square are of equal area there is an obvious
solution in three pieces, by simply cutting the square in half
diagonally.


153.--A CUTTING-OUT PUZZLE.

[Illustration]

The illustration shows how to cut the four pieces and form with them a
square. First find the side of the square (the mean proportional between
the length and height of the rectangle), and the method is obvious. If
our strip is exactly in the proportions 9 × 1, or 16 × 1, or 25 × 1, we
can clearly cut it in 3, 4, or 5 rectangular pieces respectively to form
a square. Excluding these special cases, the general law is that for a
strip in length more than n² times the breadth, and not more than (n+1)²
times the breadth, it may be cut in n+2 pieces to form a square, and
there will be n-1 rectangular pieces like piece 4 in the diagram. Thus,
for example, with a strip 24 × 1, the length is more than 16 and less
than 25 times the breadth. Therefore it can be done in 6 pieces (n here
being 4), 3 of which will be rectangular. In the case where n equals 1,
the rectangle disappears and we get a solution in three pieces. Within
these limits, of course, the sides need not be rational: the solution is
purely geometrical.


154.--MRS. HOBSON'S HEARTHRUG.

[Illustration]

As I gave full measurements of the mutilated rug, it was quite an easy
matter to find the precise dimensions for the square. The two pieces cut
off would, if placed together, make an oblong piece 12 × 6, giving an
area of 72 (inches or yards, as we please), and as the original complete
rug measured 36 × 27, it had an area of 972. If, therefore, we deduct
the pieces that have been cut away, we find that our new rug will
contain 972 less 72, or 900; and as 900 is the square of 30, we know
that the new rug must measure 30 × 30 to be a perfect square. This is a
great help towards the solution, because we may safely conclude that the
two horizontal sides measuring 30 each may be left intact.

There is a very easy way of solving the puzzle in four pieces, and also
a way in three pieces that can scarcely be called difficult, but the
correct answer is in only two pieces.

It will be seen that if, after the cuts are made, we insert the teeth of
the piece B one tooth lower down, the two portions will fit together and
form a square.


155.--THE PENTAGON AND SQUARE.

A regular pentagon may be cut into as few as six pieces that will fit
together without any turning over and form a square, as I shall show
below. Hitherto the best answer has been in seven pieces--the solution
produced some years ago by a foreign mathematician, Paul Busschop. We
first form a parallelogram, and from that the square. The process will
be seen in the diagram on the next page.

The pentagon is ABCDE. By the cut AC and the cut FM (F being the middle
point between A and C, and M being the same distance from A as F) we get
two pieces that may be placed in position at GHEA and form the
parallelogram GHDC. We then find the mean proportional between the
length HD and the _height_ of the parallelogram. This distance we mark
off from C at K, then draw CK, and from G drop the line GL,
perpendicular to KC. The rest is easy and rather obvious. It will be
seen that the six pieces will form either the pentagon or the square.

I have received what purported to be a solution in five pieces, but the
method was based on the rather subtle fallacy that half the diagonal
plus half the side of a pentagon equals the side of a square of the same
area. I say subtle, because it is an extremely close approximation that
will deceive the eye, and is quite difficult to prove inexact. I am not
aware that attention has before been drawn to this curious
approximation.

[Illustration]

Another correspondent made the side of his square 1¼ of the side of
the pentagon. As a matter of fact, the ratio is irrational. I calculate
that if the side of the pentagon is 1--inch, foot, or anything else--the
side of the square of equal area is 1.3117 nearly, or say roughly
1+3/10. So we can only hope to solve the puzzle by geometrical methods.


156.--THE DISSECTED TRIANGLE.

Diagram A is our original triangle. We will say it measures 5 inches (or
5 feet) on each side. If we take off a slice at the bottom of any
equilateral triangle by a cut parallel with the base, the portion that
remains will always be an equilateral triangle; so we first cut off
piece 1 and get a triangle 3 inches on every side. The manner of finding
directions of the other cuts in A is obvious from the diagram.

Now, if we want two triangles, 1 will be one of them, and 2, 3, 4, and 5
will fit together, as in B, to form the other. If we want three
equilateral triangles, 1 will be one, 4 and 5 will form the second, as
in C, and 2 and 3 will form the third, as in D. In B and C the piece 5
is turned over; but there can be no objection to this, as it is not
forbidden, and is in no way opposed to the nature of the puzzle.

[Illustration]


157.--THE TABLE-TOP AND STOOLS.

[Illustration]

One object that I had in view when presenting this little puzzle was to
point out the uncertainty of the meaning conveyed by the word "oval."
Though originally derived from the Latin word _ovum_, an egg, yet what
we understand as the egg-shape (with one end smaller than the other) is
only one of many forms of the oval; while some eggs are spherical in
shape, and a sphere or circle is most certainly not an oval. If we speak
of an ellipse--a conical ellipse--we are on safer ground, but here we
must be careful of error. I recollect a Liverpool town councillor, many
years ago, whose ignorance of the poultry-yard led him to substitute the
word "hen" for "fowl," remarking, "We must remember, gentlemen, that
although every cock is a hen, every hen is not a cock!" Similarly, we
must always note that although every ellipse is an oval, every oval is
not an ellipse. It is correct to say that an oval is an oblong
curvilinear figure, having two unequal diameters, and bounded by a curve
line returning into itself; and this includes the ellipse, but all other
figures which in any way approach towards the form of an oval without
necessarily having the properties above described are included in the
term "oval." Thus the following solution that I give to our puzzle
involves the pointed "oval," known among architects as the "vesica
piscis."

[Illustration: THE TWO STOOLS.]

The dotted lines in the table are given for greater clearness, the cuts
being made along the other lines. It will be seen that the eight pieces
form two stools of exactly the same size and shape with similar
hand-holes. These holes are a trifle longer than those in the
schoolmaster's stools, but they are much narrower and of considerably
smaller area. Of course 5 and 6 can be cut out in one piece--also 7 and
8--making only _six pieces_ in all. But I wished to keep the same number
as in the original story.

When I first gave the above puzzle in a London newspaper, in
competition, no correct solution was received, but an ingenious and
neatly executed attempt by a man lying in a London infirmary was
accompanied by the following note: "Having no compasses here, I was
compelled to improvise a pair with the aid of a small penknife, a bit of
firewood from a bundle, a piece of tin from a toy engine, a tin tack,
and two portions of a hairpin, for points. They are a fairly serviceable
pair of compasses, and I shall keep them as a memento of your puzzle."


158.--THE GREAT MONAD.

The areas of circles are to each other as the squares of their
diameters. If you have a circle 2 in. in diameter and another 4 in. in
diameter, then one circle will be four times as great in area as the
other, because the square of 4 is four times as great as the square of
2. Now, if we refer to Diagram 1, we see how two equal squares may be
cut into four pieces that will form one larger square; from which it is
self-evident that any square has just half the area of the square of its
diagonal. In Diagram 2 I have introduced a square as it often occurs in
ancient drawings of the Monad; which was my reason for believing that
the symbol had mathematical meanings, since it will be found to
demonstrate the fact that the area of the outer ring or annulus is
exactly equal to the area of the inner circle. Compare Diagram 2 with
Diagram 1, and you will see that as the square of the diameter CD is
double the square of the diameter of the inner circle, or CE, therefore
the area of the larger circle is double the area of the smaller one, and
consequently the area of the annulus is exactly equal to that of the
inner circle. This answers our first question.

[Illustration: 1 2 3 4]

In Diagram 3 I show the simple solution to the second question. It is
obviously correct, and may be proved by the cutting and superposition of
parts. The dotted lines will also serve to make it evident. The third
question is solved by the cut CD in Diagram 2, but it remains to be
proved that the piece F is really one-half of the Yin or the Yan. This
we will do in Diagram 4. The circle K has one-quarter the area of the
circle containing Yin and Yan, because its diameter is just one-half the
length. Also L in Diagram 3 is, we know, one-quarter the area. It is
therefore evident that G is exactly equal to H, and therefore half G is
equal to half H. So that what F loses from L it gains from K, and F must
be half of Yin or Yan.


159.--THE SQUARE OF VENEER.

[Illustration:

    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    |_ _|___|__:|___|___|:__|___|__:|___|___|:__|__||___|
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    |   |   |  :|   |   |:  |   |  :|   |   |:  |  ||   |
    +---+---+---+---+---+---+---+---+---+---+===+===+---+
    |   |   |  :|   |   |:  |   |  :|   |  ||:  |   |   |
    |   |   |  :|   |   |:  |   |  :|   |  ||:  |   |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |¯¯¯|¯¯¯|¯¯:|¯¯¯|¯¯¯|:¯¯|¯¯¯|¯¯:|¯¯¯|¯¯||:¯¯|¯¯¯|¯¯¯|
    |   |   |  :|   |   |:  |   |  :|   |  ||:  |   |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |  ||:  |   |   |
    |   |   |  :|   |   |:  | B |  :|   |  ||:  | C |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :|   |  ||:  |   |   |
    |_ _|___|__:|___|___|:__|___|__:|___|__||:__|___|___|
    +---+---+---+---+---+---+---+---+===+===+===+===+===+
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    |   |   |  :|   |   |:  |   |  :||  |   |:  | A |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |¯¯¯|¯¯¯|¯¯:|¯¯¯|¯¯¯|:¯¯|¯¯¯|¯¯:||¯¯|¯¯¯|:¯¯|¯¯¯|¯¯¯|
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    +---+---+---+---+---+---+---+---+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    +===+===+===+===+===+===+===+===+---+---+---+---+---+
    |   |   |  :|   |   |:  |   |  :||  |   |:  |   |   |
    |   |   |  :|   | D |:  |   |  :||  |   |:  |   |   |
    +---+---+---+---+---+---+---+---+===+===+===+===+===+

]

Any square number may be expressed as the sum of two squares in an
infinite number of different ways. The solution of the present puzzle
forms a simple demonstration of this rule. It is a condition that we
give actual dimensions.

In this puzzle I ignore the known dimensions of our square and work on
the assumption that it is 13n by 13n. The value of n we can afterwards
determine. Divide the square as shown (where the dotted lines indicate
the original markings) into 169 squares. As 169 is the sum of the two
squares 144 and 25, we will proceed to divide the veneer into two
squares, measuring respectively 12 × 12 and 5 × 5; and as we know that
two squares may be formed from one square by dissection in four pieces,
we seek a solution in this number. The dark lines in the diagram show
where the cuts are to be made. The square 5 × 5 is cut out whole, and
the larger square is formed from the remaining three pieces, B, C, and
D, which the reader can easily fit together.

Now, n is clearly 5/13 of an inch. Consequently our larger square must
be 60/13 in. × 60/13 in., and our smaller square 25/13 in. × 25/13 in.
The square of 60/13 added to the square of 25/13 is 25. The square is
thus divided into as few as four pieces that form two squares of known
dimensions, and all the sixteen nails are avoided.

Here is a general formula for finding two squares whose sum shall equal
a given square, say a². In the case of the solution of our puzzle p = 3,
q = 2, and a = 5.

                         ________________________
       2pqa           \/ a²( p² + q²)² - (2pqa)²
     --------- = x;   --------------------------- = y
      p² + q²                  p² + q²

    Here x² + y² = a².


160.--THE TWO HORSESHOES.

The puzzle was to cut the two shoes (including the hoof contained within
the outlines) into four pieces, two pieces each, that would fit together
and form a perfect circle. It was also stipulated that all four pieces
should be different in shape. As a matter of fact, it is a puzzle based
on the principle contained in that curious Chinese symbol the Monad.
(See No. 158.)

[Illustration]

The above diagrams give the correct solution to the problem. It will be
noticed that 1 and 2 are cut into the required four pieces, all
different in shape, that fit together and form the perfect circle shown
in Diagram 3. It will further be observed that the two pieces A and B of
one shoe and the two pieces C and D of the other form two exactly
similar halves of the circle--the Yin and the Yan of the great Monad. It
will be seen that the shape of the horseshoe is more easily determined
from the circle than the dimensions of the circle from the horseshoe,
though the latter presents no difficulty when you know that the curve of
the long side of the shoe is part of the circumference of your circle.
The difference between B and D is instructive, and the idea is useful in
all such cases where it is a condition that the pieces must be different
in shape. In forming D we simply add on a symmetrical piece, a
curvilinear square, to the piece B. Therefore, in giving either B or D a
quarter turn before placing in the new position, a precisely similar
effect must be produced.


161.--THE BETSY ROSS PUZZLE.

Fold the circular piece of paper in half along the dotted line shown in
Fig. 1, and divide the upper half into five equal parts as indicated.
Now fold the paper along the lines, and it will have the appearance
shown in Fig. 2. If you want a star like Fig. 3, cut from A to B; if you
wish one like Fig. 4, cut from A to C. Thus, the nearer you cut to the
point at the bottom the longer will be the points of the star, and the
farther off from the point that you cut the shorter will be the points
of the star.

[Illustration]


162.--THE CARDBOARD CHAIN.

The reader will probably feel rewarded for any care and patience that
he may bestow on cutting out the cardboard chain. We will suppose that
he has a piece of cardboard measuring 8 in. by 2½ in., though the
dimensions are of no importance. Yet if you want a long chain you
must, of course, take a long strip of cardboard. First rule pencil
lines B B and C C, half an inch from the edges, and also the short
perpendicular lines half an inch apart. (See next page.) Rule lines on
the other side in just the same way, and in order that they shall
coincide it is well to prick through the card with a needle the points
where the short lines end. Now take your penknife and split the card
from A A down to B B, and from D D up to C C. Then cut right through
the card along all the short perpendicular lines, and half through the
card along the short portions of B B and C C that are not dotted. Next
turn the card over and cut half through along the short lines on B B
and C C at the places that are immediately beneath the dotted lines on
the upper side. With a little careful separation of the parts with the
penknife, the cardboard may now be divided into two interlacing
ladder-like portions, as shown in Fig. 2; and if you cut away all the
shaded parts you will get the chain, cut solidly out of the cardboard,
without any join, as shown in the illustrations on page 40.

It is an interesting variant of the puzzle to cut out two keys on a
ring--in the same manner without join.

[Illustration]


164.--THE POTATO PUZZLE.

As many as twenty-two pieces may be obtained by the six cuts. The
illustration shows a pretty symmetrical solution. The rule in such cases
is that every cut shall intersect every other cut and no two
intersections coincide; that is to say, every line passes through every
other line, but more than two lines do not cross at the same point
anywhere. There are other ways of making the cuts, but this rule must
always be observed if we are to get the full number of pieces.

The general formula is that with n cuts we can always produce (n(n +
1) + 1)/2 . One of the problems proposed by the late Sam Loyd was to
produce the maximum number of pieces by n straight cuts through a
solid cheese. Of course, again, the pieces cut off may not be moved or
piled. Here we have to deal with the intersection of planes (instead
of lines), and the general formula is that with n cuts we may produce
((n - 1)n(n + 1))/6 + n + 1 pieces. It is extremely difficult to "see"
the direction and effects of the successive cuts for more than a few
of the lowest values of n.


165.--THE SEVEN PIGS.

The illustration shows the direction for placing the three fences so as
to enclose every pig in a separate sty. The greatest number of spaces
that can be enclosed with three straight lines in a square is seven, as
shown in the last puzzle. Bearing this fact in mind, the puzzle must be
solved by trial.

[Illustration: THE SEVEN PIGS.]


166.--THE LANDOWNER'S FENCES.

Four fences only are necessary, as follows:--

[Illustration]


167.--THE WIZARD'S CATS.

The illustration requires no explanation. It shows clearly how the three
circles may be drawn so that every cat has a separate enclosure, and
cannot approach another cat without crossing a line.

[Illustration: THE WIZARDS' CATS.]


168.--THE CHRISTMAS PUDDING.

The illustration shows how the pudding may be cut into two parts of
exactly the same size and shape. The lines must necessarily pass through
the points A, B, C, D, and E. But, subject to this condition, they may
be varied in an infinite number of ways. For example, at a point midway
between A and the edge, the line may be completed in an unlimited number
of ways (straight or crooked), provided it be exactly reflected from E
to the opposite edge. And similar variations may be introduced at other
places.

[Illustration]


169.--A TANGRAM PARADOX.

The diagrams will show how the figures are constructed--each with the
seven Tangrams. It will be noticed that in both cases the head, hat, and
arm are precisely alike, and the width at the base of the body the
same. But this body contains four pieces in the first case, and in the
second design only three. The first is larger than the second by exactly
that narrow strip indicated by the dotted line between A and B. This
strip is therefore exactly equal in area to the piece forming the foot
in the other design, though when thus distributed along the side of the
body the increased dimension is not easily apparent to the eye.

[Illustration]


170.--THE CUSHION COVERS.

[Illustration]

The two pieces of brocade marked A will fit together and form one
perfect square cushion top, and the two pieces marked B will form the
other.


171.--THE BANNER PUZZLE.

The illustration explains itself. Divide the bunting into 25 squares
(because this number is the sum of two other squares--16 and 9), and
then cut along the thick lines. The two pieces marked A form one square,
and the two pieces marked B form the other.

[Illustration]


172.--MRS. SMILEY'S CHRISTMAS PRESENT.

[Illustration]

[Illustration]

The first step is to find six different square numbers that sum to 196.
For example, 1 + 4 + 25 + 36 + 49 + 81 = 196; 1 + 4 + 9 + 25 + 36 + 121
= 196; 1 + 9 + 16 + 25 + 64 + 81 = 196. The rest calls for individual
judgment and ingenuity, and no definite rules can be given for
procedure. The annexed diagrams will show solutions for the first two
cases stated. Of course the three pieces marked A and those marked B
will fit together and form a square in each case. The assembling of the
parts may be slightly varied, and the reader may be interested in
finding a solution for the third set of squares I have given.


173.--MRS. PERKINS'S QUILT.

The following diagram shows how the quilt should be constructed.

[Illustration]

There is, I believe, practically only one solution to this puzzle. The
fewest separate squares must be eleven. The portions must be of the
sizes given, the three largest pieces must be arranged as shown, and the
remaining group of eight squares may be "reflected," but cannot be
differently arranged.


174.--THE SQUARES OF BROCADE.

[Illustration: Diagram 1]

So far as I have been able to discover, there is only one possible
solution to fulfil the conditions. The pieces fit together as in Diagram
1, Diagrams 2 and 3 showing how the two original squares are to be cut.
It will be seen that the pieces A and C have each twenty chequers, and
are therefore of equal area. Diagram 4 (built up with the dissected
square No. 5) solves the puzzle, except for the small condition
contained in the words, "I cut the _two_ squares in the manner desired."
In this case the smaller square is preserved intact. Still I give it as
an illustration of a feature of the puzzle. It is impossible in a
problem of this kind to give a _quarter-turn_ to any of the pieces if
the pattern is to properly match, but (as in the case of F, in Diagram
4) we may give a symmetrical piece a _half-turn_--that is, turn it
upside down. Whether or not a piece may be given a quarter-turn, a
half-turn, or no turn at all in these chequered problems, depends on the
character of the design, on the material employed, and also on the form
of the piece itself.

[Illustration: Diagram 2]

[Illustration: Diagram 3]

[Illustration: Diagram 4]

[Illustration: Diagram 5]


175.--ANOTHER PATCHWORK PUZZLE.

The lady need only unpick the stitches along the dark lines in the
larger portion of patchwork, when the four pieces will fit together and
form a square, as shown in our illustration.

[Illustration]


176.--LINOLEUM CUTTING.

There is only one solution that will enable us to retain the larger of
the two pieces with as little as possible cut from it. Fig. 1 in the
following diagram shows how the smaller piece is to be cut, and Fig. 2
how we should dissect the larger piece, while in Fig. 3 we have the new
square 10 × 10 formed by the four pieces with all the chequers properly
matched. It will be seen that the piece D contains fifty-two chequers,
and this is the largest piece that it is possible to preserve under the
conditions.

[Illustration]


177.--ANOTHER LINOLEUM PUZZLE.

Cut along the thick lines, and the four pieces will fit together and
form a perfect square in the manner shown in the smaller diagram.

[Illustration: ANOTHER LINOLEUM PUZZLE.]


178.--THE CARDBOARD BOX.

The areas of the top and side multiplied together and divided by the
area of the end give the square of the length. Similarly, the product of
top and end divided by side gives the square of the breadth; and the
product of side and end divided by the top gives the square of the
depth. But we only need one of these operations. Let us take the first.
Thus, 120 × 96 divided by 80 equals 144, the square of 12. Therefore the
length is 12 inches, from which we can, of course, at once get the
breadth and depth--10 in. and 8 in. respectively.


179.--STEALING THE BELL-ROPES.

Whenever we have one side (a) of a right-angled triangle, and know the
difference between the second side and the hypotenuse (which difference
we will call b), then the length of the hypotenuse will be

    a²    b
    --- + -.
    2b    2

In the case of our puzzle this will be

    48 × 48
    ------- + 1½ in. = 32 ft. 1½ in.,
       6

which is the length of the rope.


180-- THE FOUR SONS.

[Illustration]

The diagram shows the most equitable division of the land possible, "so
that each son shall receive land of exactly the same area and exactly
similar in shape," and so that each shall have access to the well in
the centre without trespass on another's land. The conditions do not
require that each son's land shall be in one piece, but it is necessary
that the two portions assigned to an individual should be kept apart, or
two adjoining portions might be held to be one piece, in which case the
condition as to shape would have to be broken. At present there is only
one shape for each piece of land--half a square divided diagonally. And
A, B, C, and D can each reach their land from the outside, and have each
equal access to the well in the centre.


181.--THE THREE RAILWAY STATIONS.

The three stations form a triangle, with sides 13, 14, and 15 miles.
Make the 14 side the base; then the height of the triangle is 12 and the
area 84. Multiply the three sides together and divide by four times the
area. The result is eight miles and one-eighth, the distance required.


182.--THE GARDEN PUZZLE.

Half the sum of the four sides is 144. From this deduct in turn the four
sides, and we get 64, 99, 44, and 81. Multiply these together, and we
have as the result the square of 4,752. Therefore the garden contained
4,752 square yards. Of course the tree being equidistant from the four
corners shows that the garden is a quadrilateral that may be inscribed
in a circle.


183.--DRAWING A SPIRAL.

Make a fold in the paper, as shown by the dotted line in the
illustration. Then, taking any two points, as A and B, describe
semicircles on the line alternately from the centres B and A, being
careful to make the ends join, and the thing is done. Of course this is
not a _true_ spiral, but the puzzle was to produce the _particular_
spiral that was shown, and that was drawn in this simple manner.

[Illustration]


184.--HOW TO DRAW AN OVAL.

If you place your sheet of paper round the surface of a cylindrical
bottle or canister, the oval can be drawn with one sweep of the
compasses.


185.--ST. GEORGE'S BANNER.

As the flag measures 4 ft. by 3 ft., the length of the diagonal (from
corner to corner) is 5 ft. All you need do is to deduct half the
length of this diagonal (2½ ft.) from a quarter of the distance all
round the edge of the flag (3½ ft.)--a quarter of 14 ft. The
difference (1 ft.) is the required width of the arm of the red cross.
The area of the cross will then be the same as that of the white
ground.


186.--THE CLOTHES LINE PUZZLE.

Multiply together, and also add together, the heights of the two poles
and divide one result by the other. That is, if the two heights are a
and b respectively, then ab/(a + b) will give the height of the
intersection. In the particular case of our puzzle, the intersection was
therefore 2 ft. 11 in. from the ground. The distance that the poles are
apart does not affect the answer. The reader who may have imagined that
this was an accidental omission will perhaps be interested in
discovering the reason why the distance between the poles may be
ignored.


187.--THE MILKMAID PUZZLE.

[Illustration:

      A
    |\
    | \
    |  \
    |   \ B       RIVER
    +----+--------------
    |   / \
    |  /   \
    | /     \
    |/       DOOR
     STOOL

]

Draw a straight line, as shown in the diagram, from the milking-stool
perpendicular to the near bank of the river, and continue it to the
point A, which is the same distance from that bank as the stool. If you
now draw the straight line from A to the door of the dairy, it will cut
the river at B. Then the shortest route will be from the stool to B and
thence to the door. Obviously the shortest distance from A to the door
is the straight line, and as the distance from the stool to any point of
the river is the same as from A to that point, the correctness of the
solution will probably appeal to every reader without any acquaintance
with geometry.


188.--THE BALL PROBLEM.

If a round ball is placed on the level ground, six similar balls may be
placed round it (all on the ground), so that they shall all touch the
central ball.

As for the second question, the ratio of the diameter of a circle to its
circumference we call _pi_; and though we cannot express this ratio in
exact numbers, we can get sufficiently near to it for all practical
purposes. However, in this case it is not necessary to know the value of
_pi_ at all. Because, to find the area of the surface of a sphere we
multiply the square of the diameter by _pi_; to find the volume of a
sphere we multiply the cube of the diameter by one-sixth of _pi_.
Therefore we may ignore _pi_, and have merely to seek a number whose
square shall equal one-sixth of its cube. This number is obviously 6.
Therefore the ball was 6 ft. in diameter, for the area of its surface
will be 36 times _pi_ in square feet, and its volume also 36 times _pi_
in cubic feet.


189.--THE YORKSHIRE ESTATES.

The triangular piece of land that was not for sale contains exactly
eleven acres. Of course it is not difficult to find the answer if we
follow the eccentric and tricky tracks of intricate trigonometry; or I
might say that the application of a well-known formula reduces the
problem to finding one-quarter of the square root of (4 × 370 × 116)
-(370 + 116 - 74)²--that is a quarter of the square root of 1936, which
is one-quarter of 44, or 11 acres. But all that the reader really
requires to know is the Pythagorean law on which many puzzles have been
built, that in any right-angled triangle the square of the hypotenuse is
equal to the sum of the squares of the other two sides. I shall dispense
with all "surds" and similar absurdities, notwithstanding the fact that
the sides of our triangle are clearly incommensurate, since we cannot
exactly extract the square roots of the three square areas.

[Illustration:

        A
      |\
      |  \.
      |    \ .
      |5     \  .
      |   7    \   .
    E +--------- +C   .
      |          | ` .   .
      |          |     `.   .
      |4         |4       ` . .
      |   7      |            ` ..
      D----------+----------------- B
                 F

]

In the above diagram ABC represents our triangle. ADB is a right-angled
triangle, AD measuring 9 and BD measuring 17, because the square of 9
added to the square of 17 equals 370, the known area of the square on
AB. Also AEC is a right-angled triangle, and the square of 5 added to
the square of 7 equals 74, the square estate on A C. Similarly, CFB is a
right-angled triangle, for the square of 4 added to the square of 10
equals 116, the square estate on BC. Now, although the sides of our
triangular estate are incommensurate, we have in this diagram all the
exact figures that we need to discover the area with precision.

The area of our triangle ADB is clearly half of 9 × 17, or 76½ acres.
The area of AEC is half of 5 × 7, or 17½ acres; the area of CFB is half
of 4 × 10, or 20 acres; and the area of the oblong EDFC is obviously 4 ×
7, or 28 acres. Now, if we add together 17½, 20, and 28 = 65½, and
deduct this sum from the area of the large triangle ADB (which we have
found to be 76½ acres), what remains must clearly be the area of ABC.
That is to say, the area we want must be 76½ - 65½ = 11 acres exactly.


190.--FARMER WURZEL'S ESTATE.

The area of the complete estate is exactly one hundred acres. To find
this answer I use the following little formula,

      __________________
    \/4ab - (a + b + c)²
    --------------------
              4

where a, b, c represent the three square areas, in any order. The
expression gives the area of the triangle A. This will be found to be 9
acres. It can be easily proved that A, B, C, and D are all equal in
area; so the answer is 26 + 20 + 18 + 9 + 9 + 9 + 9 = 100 acres.

[Illustration]

Here is the proof. If every little dotted square in the diagram
represents an acre, this must be a correct plan of the estate, for the
squares of 5 and 1 together equal 26; the squares of 4 and 2 equal 20;
and the squares of 3 and 3 added together equal 18. Now we see at once
that the area of the triangle E is 2½, F is 4½, and G is 4. These added
together make 11 acres, which we deduct from the area of the rectangle,
20 acres, and we find that the field A contains exactly 9 acres. If you
want to prove that B, C, and D are equal in size to A, divide them in
two by a line from the middle of the longest side to the opposite angle,
and you will find that the two pieces in every case, if cut out, will
exactly fit together and form A.

Or we can get our proof in a still easier way. The complete area of the
squared diagram is 12 × 12 = 144 acres, and the portions 1, 2, 3, 4, not
included in the estate, have the respective areas of 12½, 17½, 9½, and
4½. These added together make 44, which, deducted from 144, leaves 100
as the required area of the complete estate.


191.--THE CRESCENT PUZZLE.

Referring to the original diagram, let AC be x, let CD be x - 9, and let
EC be x - 5. Then x - 5 is a mean proportional between x - 9 and x, from
which we find that x equals 25. Therefore the diameters are 50 in. and
41 in. respectively.


192.--THE PUZZLE WALL.

[Illustration]

The answer given in all the old books is that shown in Fig. 1, where the
curved wall shuts out the cottages from access to the lake. But in
seeking the direction for the "shortest possible" wall most readers
to-day, remembering that the shortest distance between two points is a
straight line, will adopt the method shown in Fig. 2. This is certainly
an improvement, yet the correct answer is really that indicated in Fig.
3. A measurement of the lines will show that there is a considerable
saving of length in this wall.


193.--THE SHEEP-FOLD.

This is the answer that is always given and accepted as correct: Two
more hurdles would be necessary, for the pen was twenty-four by one (as
in Fig. A on next page), and by moving one of the sides and placing an
extra hurdle at each end (as in Fig. B) the area would be doubled. The
diagrams are not to scale. Now there is no condition in the puzzle that
requires the sheep-fold to be of any particular form. But even if we
accept the point that the pen was twenty-four by one, the answer utterly
fails, for two extra hurdles are certainly not at all necessary. For
example, I arrange the fifty hurdles as in Fig. C, and as the area is
increased from twenty-four "square hurdles" to 156, there is now
accommodation for 650 sheep. If it be held that the area must be exactly
double that of the original pen, then I construct it (as in Fig. D) with
twenty-eight hurdles only, and have twenty-two in hand for other
purposes on the farm. Even if it were insisted that all the original
hurdles must be used, then I should construct it as in Fig. E, where I
can get the area as exact as any farmer could possibly require, even if
we have to allow for the fact that the sheep might not be able to graze
at the extreme ends. Thus we see that, from any point of view, the
accepted answer to this ancient little puzzle breaks down. And yet
attention has never before been drawn to the absurdity.

[Illustration

               A                 24
       +--------------------------------+
       |             24                 |1
       +--------------------------------+

               B
       +--------------------------------+
       |             48                 |2
       +--------------------------------+
                                   24
                  C
       +--------------------+          D
       |                    |     +----------+
       |                    |     |          |
       |                    |12   |    48    |6
       |       156          |     |          |
       |                    |     +----------+
       |                    |           8
       |                    |
       |                    |
       +--------------------+
           13


          12         .   E    13
            .    '        '   .
    .   '                         '   .
        '   .                 .   '
          12     '   .    '    13

]


194.--THE GARDEN WALLS.

The puzzle was to divide the circular field into four equal parts by
three walls, each wall being of exactly the same length. There are two
essential difficulties in this problem. These are: (1) the thickness of
the walls, and (2) the condition that these walls are three in number.
As to the first point, since we are told that the walls are brick walls,
we clearly cannot ignore their thickness, while we have to find a
solution that will equally work, whether the walls be of a thickness of
one, two, three, or more bricks.

[Illustration]

The second point requires a little more consideration. How are we to
distinguish between a wall and walls? A straight wall without any bend
in it, no matter how long, cannot ever become "walls," if it is neither
broken nor intersected in any way. Also our circular field is clearly
enclosed by one wall. But if it had happened to be a square or a
triangular enclosure, would there be respectively four and three walls
or only one enclosing wall in each case? It is true that we speak of
"the four walls" of a square building or garden, but this is only a
conventional way of saying "the four sides." If you were speaking of the
actual brickwork, you would say, "I am going to enclose this square
garden with a wall." Angles clearly do not affect the question, for we
may have a zigzag wall just as well as a straight one, and the Great
Wall of China is a good example of a wall with plenty of angles. Now, if
you look at Diagrams 1, 2, and 3, you may be puzzled to declare whether
there are in each case two or four new walls; but you cannot call them
three, as required in our puzzle. The intersection either affects the
question or it does not affect it.

If you tie two pieces of string firmly together, or splice them in a
nautical manner, they become "one piece of string." If you simply let
them lie across one another or overlap, they remain "two pieces of
string." It is all a question of joining and welding. It may similarly
be held that if two walls be built into one another--I might almost say,
if they be made homogeneous--they become one wall, in which case
Diagrams 1, 2, and 3 might each be said to show one wall or two, if it
be indicated that the four ends only touch, and are not really built
into, the outer circular wall.

The objection to Diagram 4 is that although it shows the three required
walls (assuming the ends are not built into the outer circular wall),
yet it is only absolutely correct when we assume the walls to have no
thickness. A brick has thickness, and therefore the fact throws the
whole method out and renders it only approximately correct.

Diagram 5 shows, perhaps, the only correct and perfectly satisfactory
solution. It will be noticed that, in addition to the circular wall,
there are three new walls, which touch (and so enclose) but are not
built into one another. This solution may be adapted to any desired
thickness of wall, and its correctness as to area and length of wall
space is so obvious that it is unnecessary to explain it. I will,
however, just say that the semicircular piece of ground that each tenant
gives to his neighbour is exactly equal to the semicircular piece that
his neighbour gives to him, while any section of wall space found in one
garden is precisely repeated in all the others. Of course there is an
infinite number of ways in which this solution may be correctly varied.


195.--LADY BELINDA'S GARDEN.

All that Lady Belinda need do was this: She should measure from A to B,
fold her tape in four and mark off the point E, which is thus one
quarter of the side. Then, in the same way, mark off the point F,
one-fourth of the side AD Now, if she makes EG equal to AF, and GH equal
to EF, then AH is the required width for the path in order that the bed
shall be exactly half the area of the garden. An exact numerical
measurement can only be obtained when the sum of the squares of the two
sides is a square number. Thus, if the garden measured 12 poles by 5
poles (where the squares of 12 and 5, 144 and 25, sum to 169, the square
of 13), then 12 added to 5, less 13, would equal four, and a quarter of
this, 1 pole, would be the width of the path.


196.--THE TETHERED GOAT.

[Illustration]

This problem is quite simple if properly attacked. Let us suppose the
triangle ABC to represent our half-acre field, and the shaded portion to
be the quarter-acre over which the goat will graze when tethered to the
corner C. Now, as six equal equilateral triangles placed together will
form a regular hexagon, as shown, it is evident that the shaded pasture
is just one-sixth of the complete area of a circle. Therefore all we
require is the radius (CD) of a circle containing six quarter-acres or
1½ acres, which is equal to 9,408,960 square inches. As we only want
our answer "to the nearest inch," it is sufficiently exact for our
purpose if we assume that as 1 is to 3.1416, so is the diameter of a
circle to its circumference. If, therefore, we divide the last number I
gave by 3.1416, and extract the square root, we find that 1,731 inches,
or 48 yards 3 inches, is the required length of the tether "to the
nearest inch."


197.--THE COMPASSES PUZZLE.

Let AB in the following diagram be the given straight line. With the
centres A and B and radius AB describe the two circles. Mark off DE and
EF equal to AD. With the centres A and F and radius DF describe arcs
intersecting at G. With the centres A and B and distance BG describe
arcs GHK and N. Make HK equal to AB and HL equal to HB. Then with
centres K and L and radius AB describe arcs intersecting at I. Make BM
equal to BI. Finally, with the centre M and radius MB cut the line in C,
and the point C is the required middle of the line AB. For greater
exactitude you can mark off R from A (as you did M from B), and from R
describe another arc at C. This also solves the problem, to find a point
midway between two given points without the straight line.

[Illustration]

I will put the young geometer in the way of a rigid proof. First prove
that twice the square of the line AB equals the square of the distance
BG, from which it follows that HABN are the four corners of a square. To
prove that I is the centre of this square, draw a line from H to P
through QIB and continue the arc HK to P. Then, conceiving the necessary
lines to be drawn, the angle HKP, being in a semicircle, is a right
angle. Let fall the perpendicular KQ, and by similar triangles, and from
the fact that HKI is an isosceles triangle by the construction, it can
be proved that HI is half of HB. We can similarly prove that C is the
centre of the square of which AIB are three corners.

I am aware that this is not the simplest possible solution.


198.--THE EIGHT STICKS.

The first diagram is the answer that nearly every one will give to this
puzzle, and at first sight it seems quite satisfactory. But consider the
conditions. We have to lay "every one of the sticks on the table." Now,
if a ladder be placed against a wall with only one end on the ground, it
can hardly be said that it is "laid on the ground." And if we place the
sticks in the above manner, it is only possible to make one end of two
of them touch the table: to say that every one lies on the table would
not be correct. To obtain a solution it is only necessary to have our
sticks of proper dimensions. Say the long sticks are each 2 ft. in
length and the short ones 1 ft. Then the sticks must be 3 in. thick,
when the three equal squares may be enclosed, as shown in the second
diagram. If I had said "matches" instead of "sticks," the puzzle would
be impossible, because an ordinary match is about twenty-one times as
long as it is broad, and the enclosed rectangles would not be squares.

[Illustration]


199.--PAPA'S PUZZLE.

I have found that a large number of people imagine that the following is
a correct solution of the problem. Using the letters in the diagram
below, they argue that if you make the distance BA one-third of BC, and
therefore the area of the rectangle ABE equal to that of the triangular
remainder, the card must hang with the long side horizontal. Readers
will remember the jest of Charles II., who induced the Royal Society to
meet and discuss the reason why the water in a vessel will not rise if
you put a live fish in it; but in the middle of the proceedings one of
the least distinguished among them quietly slipped out and made the
experiment, when he found that the water _did_ rise! If my
correspondents had similarly made the experiment with a piece of
cardboard, they would have found at once their error. Area is one thing,
but gravitation is quite another. The fact of that triangle sticking its
leg out to D has to be compensated for by additional area in the
rectangle. As a matter of fact, the ratio of BA to AC is as 1 is to the
square root of 3, which latter cannot be given in an exact numerical
measure, but is approximately 1.732. Now let us look at the correct
general solution. There are many ways of arriving at the desired result,
but the one I give is, I think, the simplest for beginners.

[Illustration]

Fix your card on a piece of paper and draw the equilateral triangle BCF,
BF and CF being equal to BC. Also mark off the point G so that DG shall
equal DC. Draw the line CG and produce it until it cuts the line BF in
H. If we now make HA parallel to BE, then A is the point from which our
cut must be made to the corner D, as indicated by the dotted line.

A curious point in connection with this problem is the fact that the
position of the point A is independent of the side CD. The reason for
this is more obvious in the solution I have given than in any other
method that I have seen, and (although the problem may be solved with
all the working on the cardboard) that is partly why I have preferred
it. It will be seen at once that however much you may reduce the width
of the card by bringing E nearer to B and D nearer to C, the line CG,
being the diagonal of a square, will always lie in the same direction,
and will cut BF in H. Finally, if you wish to get an approximate measure
for the distance BA, all you have to do is to multiply the length of the
card by the decimal .366. Thus, if the card were 7 inches long, we get 7
× .366 = 2.562, or a little more than 2½ inches, for the distance from B
to A.

But the real joke of the puzzle is this: We have seen that the position
of the point A is independent of the width of the card, and depends
entirely on the length. Now, in the illustration it will be found that
both cards have the same length; consequently all the little maid had to
do was to lay the clipped card on top of the other one and mark off the
point A at precisely the same distance from the top left-hand corner!
So, after all, Pappus' puzzle, as he presented it to his little maid,
was quite an infantile problem, when he was able to show her how to
perform the feat without first introducing her to the elements of
statics and geometry.


200.--A KITE-FLYING PUZZLE.

Solvers of this little puzzle, I have generally found, may be roughly
divided into two classes: those who get within a mile of the correct
answer by means of more or less complex calculations, involving "_pi_,"
and those whose arithmetical kites fly hundreds and thousands of miles
away from the truth. The comparatively easy method that I shall show
does not involve any consideration of the ratio that the diameter of a
circle bears to its circumference. I call it the "hat-box method."

[Illustration]

Supposing we place our ball of wire, A, in a cylindrical hat-box, B,
that exactly fits it, so that it touches the side all round and exactly
touches the top and bottom, as shown in the illustration. Then, by an
invariable law that should be known by everybody, that box contains
exactly half as much again as the ball. Therefore, as the ball is 24 in.
in diameter, a hat-box of the same circumference but two-thirds of the
height (that is, 16 in. high) will have exactly the same contents as the
ball.

Now let us consider that this reduced hat-box is a cylinder of metal
made up of an immense number of little wire cylinders close together
like the hairs in a painter's brush. By the conditions of the puzzle we
are allowed to consider that there are no spaces between the wires. How
many of these cylinders one one-hundredth of an inch thick are equal to
the large cylinder, which is 24 in. thick? Circles are to one another as
the squares of their diameters. The square of 1/100 is 1/100000, and the
square of 24 is 576; therefore the large cylinder contains 5,760,000 of
the little wire cylinders. But we have seen that each of these wires is
16 in. long; hence 16 × 5,760,000 = 92,160,000 inches as the complete
length of the wire. Reduce this to miles, and we get 1,454 miles 2,880
ft. as the length of the wire attached to the professor's kite.

Whether a kite would fly at such a height, or support such a weight, are
questions that do not enter into the problem.


201.--HOW TO MAKE CISTERNS.

Here is a general formula for solving this problem. Call the two sides
of the rectangle a and b. Then

    a + b - (a² + b² - ab)^½
    ---------------------------
                 6

equals the side of the little square pieces to cut away. The
measurements given were 8 ft. by 3 ft., and the above rule gives 8 in.
as the side of the square pieces that have to be cut away. Of course it
will not always come out exact, as in this case (on account of that
square root), but you can get as near as you like with decimals.


202.--THE CONE PUZZLE.

The simple rule is that the cone must be cut at one-third of its
altitude.


203.--CONCERNING WHEELS.

If you mark a point A on the circumference of a wheel that runs on the
surface of a level road, like an ordinary cart-wheel, the curve
described by that point will be a common cycloid, as in Fig. 1. But if
you mark a point B on the circumference of the flange of a
locomotive-wheel, the curve will be a curtate cycloid, as in Fig. 2,
terminating in nodes. Now, if we consider one of these nodes or loops,
we shall see that "at any given moment" certain points at the bottom of
the loop must be moving in the opposite direction to the train. As there
is an infinite number of such points on the flange's circumference,
there must be an infinite number of these loops being described while
the train is in motion. In fact, at any given moment certain points on
the flanges are always moving in a direction opposite to that in which
the train is going.

[Illustration: 1]

[Illustration: 2]

In the case of the two wheels, the wheel that runs round the stationary
one makes two revolutions round its own centre. As both wheels are of
the same size, it is obvious that if at the start we mark a point on the
circumference of the upper wheel, at the very top, this point will be in
contact with the lower wheel at its lowest part when half the journey
has been made. Therefore this point is again at the top of the moving
wheel, and one revolution has been made. Consequently there are two such
revolutions in the complete journey.


204.--A NEW MATCH PUZZLE.

1. The easiest way is to arrange the eighteen matches as in Diagrams 1
and 2, making the length of the perpendicular AB equal to a match and a
half. Then, if the matches are an inch in length, Fig. 1 contains two
square inches and Fig. 2 contains six square inches--4 × 1½. The second
case (2) is a little more difficult to solve. The solution is given in
Figs. 3 and 4. For the purpose of construction, place matches
temporarily on the dotted lines. Then it will be seen that as 3 contains
five equal equilateral triangles and 4 contains fifteen similar
triangles, one figure is three times as large as the other, and exactly
eighteen matches are used.

[Illustration: Figures 1, 2, 3, 4.]


205.--THE SIX SHEEP-PENS.

[Illustration] Place the twelve matches in the manner shown in the
illustration, and you will have six pens of equal size.


206.--THE KING AND THE CASTLES.

There are various ways of building the ten castles so that they shall
form five rows with four castles in every row, but the arrangement in
the next column is the only one that also provides that two castles (the
greatest number possible) shall not be approachable from the outside. It
will be seen that you must cross the walls to reach these two.

[Illustration: The King and the Castles]


207.--CHERRIES AND PLUMS.

There are several ways in which this problem might be solved were it not
for the condition that as few cherries and plums as possible shall be
planted on the north and east sides of the orchard. The best possible
arrangement is that shown in the diagram, where the cherries, plums,
and apples are indicated respectively by the letters C, P, and A. The
dotted lines connect the cherries, and the other lines the plums. It
will be seen that the ten cherry trees and the ten plum trees are so
planted that each fruit forms five lines with four trees of its kind in
line. This is the only arrangement that allows of so few as two cherries
or plums being planted on the north and east outside rows.

[Illustration]


208.--A PLANTATION PUZZLE.

The illustration shows the ten trees that must be left to form five rows
with four trees in every row. The dots represent the positions of the
trees that have been cut down.

[Illustration]


209.--THE TWENTY-ONE TREES.

I give two pleasing arrangements of the trees. In each case there are
twelve straight rows with five trees in every row.

[Illustration: Figure 1, Figure 2.]


210.--THE TEN COINS.

The answer is that there are just 2,400 different ways. Any three coins
may be taken from one side to combine with one coin taken from the other
side. I give four examples on this and the next page. We may thus select
three from the top in ten ways and one from the bottom in five ways,
making fifty. But we may also select three from the bottom and one from
the top in fifty ways. We may thus select the four coins in one hundred
ways, and the four removed may be arranged by permutation in twenty-four
ways. Thus there are 24 × 100 = 2,400 different solutions.

[Illustration]


As all the points and lines puzzles that I have given so far, excepting
the last, are variations of the case of ten points arranged to form five
lines of four, it will be well to consider this particular case
generally. There are six fundamental solutions, and no more, as shown in
the six diagrams. These, for the sake of convenience, I named some years
ago the Star, the Dart, the Compasses, the Funnel, the Scissors, and the
Nail. (See next page.) Readers will understand that any one of these
forms may be distorted in an infinite number of different ways without
destroying its real character.

In "The King and the Castles" we have the Star, and its solution gives
the Compasses. In the "Cherries and Plums" solution we find that the
Cherries represent the Funnel and the Plums the Dart. The solution of
the "Plantation Puzzle" is an example of the Dart distorted. Any
solution to the "Ten Coins" will represent the Scissors. Thus examples
of all have been given except the Nail.

On a reduced chessboard, 7 by 7, we may place the ten pawns in just
three different ways, but they must all represent the Dart. The
"Plantation" shows one way, the Plums show a second way, and the reader
may like to find the third way for himself. On an ordinary chessboard, 8
by 8, we can also get in a beautiful example of the Funnel--symmetrical
in relation to the diagonal of the board. The smallest board that will
take a Star is one 9 by 7. The Nail requires a board 11 by 7, the
Scissors

[Illustration]

11 by 9, and the Compasses 17 by 12. At least these are the best results
recorded in my note-book. They may be beaten, but I do not think so. If
you divide a chessboard into two parts by a diagonal zigzag line, so
that the larger part contains 36 squares and the smaller part 28
squares, you can place three separate schemes on the larger part and one
on the smaller part (all Darts) without their conflicting--that is, they
occupy forty different squares. They can be placed in other ways without
a division of the board. The smallest square board that will contain six
different schemes (not fundamentally different), without any line of one
scheme crossing the line of another, is 14 by 14; and the smallest board
that will contain one scheme entirely enclosed within the lines of a
second scheme, without any of the lines of the one, when drawn from
point to point, crossing a line of the other, is 14 by 12.

[Illustration: STAR DART COMPASSES FUNNEL SCISSORS NAIL]


211.--THE TWELVE MINCE-PIES.

If you ignore the four black pies in our illustration, the remaining
twelve are in their original positions. Now remove the four detached
pies to the places occupied by the black ones, and you will have your
seven straight rows of four, as shown by the dotted lines.

[Illustration: The Twelve Mince Pies.]


212.--THE BURMESE PLANTATION.

The arrangement on the next page is the most symmetrical answer that can
probably be found for twenty-one rows, which is, I believe, the greatest
number of rows possible. There are several ways of doing it.


213.--TURKS AND RUSSIANS.

The main point is to discover the smallest possible number of Russians
that there could have been. As the enemy opened fire from all
directions, it is clearly necessary to find what is the smallest number
of heads that could form sixteen lines with three heads in every line.
Note that I say sixteen, and not thirty-two, because every line taken by
a bullet may be also taken by another bullet fired in exactly the
opposite direction. Now, as few as eleven points, or heads, may be
arranged to form the required sixteen lines of three, but the discovery
of this arrangement is a hard nut. The diagram at the foot of this page
will show exactly how the thing is to be done.

[Illustration]

If, therefore, eleven Russians were in the positions shown by the stars,
and the thirty-two Turks in the positions indicated by the black dots,
it will be seen, by the lines shown, that each Turk may fire exactly
over the heads of three Russians. But as each bullet kills a man, it is
essential that every Turk shall shoot one of his comrades and be shot by
him in turn; otherwise we should have to provide extra Russians to be
shot, which would be destructive of the correct solution of our problem.
As the firing was simultaneous, this point presents no difficulties. The
answer we thus see is that there were at least eleven Russians amongst
whom there was no casualty, and that all the thirty-two Turks were shot
by one another. It was not stated whether the Russians fired any shots,
but it will be evident that even if they did their firing could not have
been effective: for if one of their bullets killed a Turk, then we have
immediately to provide another man for one of the Turkish bullets to
kill; and as the Turks were known to be thirty-two in number, this would
necessitate our introducing another Russian soldier and, of course,
destroying the solution. I repeat that the difficulty of the puzzle
consists in finding how to arrange eleven points so that they shall form
sixteen lines of three. I am told that the possibility of doing this was
first discovered by the Rev. Mr. Wilkinson some twenty years ago.


214.--THE SIX FROGS.

Move the frogs in the following order: 2, 4, 6, 5, 3, 1 (repeat these
moves in the same order twice more), 2, 4, 6. This is a solution in
twenty-one moves--the fewest possible.

If n, the number of frogs, be even, we require (n² + n)/2 moves, of
which (n² - n)/2 will be leaps and n simple moves. If n be odd, we
shall need ((n² + 3n)/2) - 4 moves, of which (n² - n)/2 will be leaps
and 2n - 4 simple moves.

In the even cases write, for the moves, all the even numbers in
ascending order and the odd numbers in descending order. This series
must be repeated ½n times and followed by the even numbers in
ascending order once only. Thus the solution for 14 frogs will be (2, 4,
6, 8, 10, 12, 14, 13, 11, 9, 7, 5, 3, 1) repeated 7 times and followed
by 2, 4, 6, 8, 10, 12, 14 = 105 moves.

In the odd cases, write the even numbers in ascending order and the odd
numbers in descending order, repeat this series ½(n - 1) times, follow
with the even numbers in ascending order (omitting n - 1), the odd
numbers in descending order (omitting 1), and conclude with all the
numbers (odd and even) in their natural order (omitting 1 and n). Thus
for 11 frogs: (2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 1) repeated 5 times, 2,
4, 6, 8, 11, 9, 7, 5, 3, and 2, 3, 4, 5, 6, 7, 8, 9, 10 = 73 moves.

This complete general solution is published here for the first time.


215.--THE GRASSHOPPER PUZZLE.

Move the counters in the following order. The moves in brackets are to
be made four times in succession. 12, 1, 3, 2, 12, 11, 1, 3, 2 (5, 7, 9,
10, 8, 6, 4), 3, 2, 12, 11, 2, 1, 2. The grasshoppers will then be
reversed in forty-four moves.

The general solution of this problem is very difficult. Of course it can
always be solved by the method given in the solution of the last puzzle,
if we have no desire to use the fewest possible moves. But to employ a
full economy of moves we have two main points to consider. There are
always what I call a lower movement (L) and an upper movement (U). L
consists in exchanging certain of the highest numbers, such as 12, 11,
10 in our "Grasshopper Puzzle," with certain of the lower numbers, 1, 2,
3; the former moving in a clockwise direction, the latter in a
non-clockwise direction. U consists in reversing the intermediate
counters. In the above solution for 12, it will be seen that 12, 11, and
1, 2, 3 are engaged in the L movement, and 4, 5, 6, 7, 8, 9, 10 in the
U movement. The L movement needs 16 moves and U 28, making together 44.
We might also involve 10 in the L movement, which would result in L 23,
U 21, making also together 44 moves. These I call the first and second
methods. But any other scheme will entail an increase of moves. You
always get these two methods (of equal economy) for odd or even
counters, but the point is to determine just how many to involve in L
and how many in U. Here is the solution in table form. But first note,
in giving values to n, that 2, 3, and 4 counters are special cases,
requiring respectively 3, 3, and 6 moves, and that 5 and 6 counters do
not give a minimum solution by the second method--only by the first.

                                  FIRST METHOD.
 +----------+---------------------------+-----------------------+-----------+
 | Total No.|         L MOVEMENT.       |       U MOVEMENT.     |           |
 |    of    +-------------+-------------+----------+------------+ Total No. |
 | Counters.|   No. of    |   No. of    | No. of   |   No. of   | of Moves. |
 |          |  Counters.  |   Moves.    |Counters. |   Moves.   |           |
 +----------+-------------+-------------+----------+------------+-----------+
 |    4n    | n-1 and n   |2(n-1)²+5n-7 | 2n+1     |2n²+3n+1    |4(n²+n-1)  |
 |   4n-2   | n-1  "  n   |2(n-1)²+5n-7 | 2n-1     |2(n-1)²+3n-2|4n²-5      |
 |   4n+1   | n    "  n+1 |2n²+5n-2     | 2n       |2n²+3n-4    |2(2n²+4n-3)|
 |   4n-1   | n-1  "  n   |2(n-1)²+5n-7 | 2n       |2n²+3n-4    |4n²+4n-9   |
 +----------+-------------+-------------+----------+------------+-----------+

                                SECOND METHOD.
 +---------+--------------------------+-------------------------+-----------+
 |Total No.|         L MOVEMENT.      |        U MOVEMENT.      |           |
 |   of    +-------------+------------+----------+--------------+ Total No. |
 |Counters.|   No. of    |   No. of   |  No. of  |    No. of    | of Moves. |
 |         |  Counters.  |   Moves.   | Counters.|    Moves.    |           |
 +---------+-------------+------------+----------+--------------+-----------+
 |   4n    | n   and n   |2n²+3n-4    |  2n      | 2(n-1)²+5n-2 |4(n²+n-1)  |
 |  4n-2   | n-1  "  n-1 |2(n-1)²+3n-7|  2n      | 2(n-1)²+5n-2 |4n²-5      |
 |  4n+1   | n    "  n   |2n²+3n-4    |  2n+1    | 2n²+5n-2     |2(2n²+4n-3)|
 |  4n-1   | n    "  n   |2n²+3n-4    |  2n-1    | 2(n-1)²+5n-7 |4n²+4n-9   |
 +---------+-------------+------------+----------+--------------+-----------+

More generally we may say that with m counters, where m is even and
greater than 4, we require (m² + 4m - 16)/4 moves; and where m is odd
and greater than 3, (m² + 6m - 31)/4 moves. I have thus shown the
reader how to find the minimum number of moves for any case, and the
character and direction of the moves. I will leave him to discover for
himself how the actual order of moves is to be determined. This is a
hard nut, and requires careful adjustment of the L and the U
movements, so that they may be mutually accommodating.


216.--THE EDUCATED FROGS.

The following leaps solve the puzzle in ten moves: 2 to 1, 5 to 2, 3 to
5, 6 to 3, 7 to 6, 4 to 7, 1 to 4, 3 to 1, 6 to 3, 7 to 6.


217.--THE TWICKENHAM PUZZLE.

Play the counters in the following order: K C E K W T C E H M K W T A N
C E H M I K C E H M T, and there you are, at Twickenham. The position
itself will always determine whether you are to make a leap or a simple
move.


218.--THE VICTORIA CROSS PUZZLE.

In solving this puzzle there were two things to be achieved: first, so
to manipulate the counters that the word VICTORIA should read round the
cross in the same direction, only with the V on one of the dark arms;
and secondly, to perform the feat in the fewest possible moves. Now, as
a matter of fact, it would be impossible to perform the first part in
any way whatever if all the letters of the word were different; but as
there are two I's, it can be done by making these letters change
places--that is, the first I changes from the 2nd place to the 7th, and
the second I from the 7th place to the 2nd. But the point I referred to,
when introducing the puzzle, as a little remarkable is this: that a
solution in twenty-two moves is obtainable by moving the letters in the
order of the following words: "A VICTOR! A VICTOR! A VICTOR I!"

There are, however, just six solutions in eighteen moves, and the
following is one of them: I (1), V, A, I (2), R, O, T, I (1), I (2), A,
V, I (2), I (1), C, I (2), V, A, I (1). The first and second I in the
word are distinguished by the numbers 1 and 2.

It will be noticed that in the first solution given above one of the I's
never moves, though the movements of the other letters cause it to
change its relative position. There is another peculiarity I may point
out--that there is a solution in twenty-eight moves requiring no letter
to move to the central division except the I's. I may also mention that,
in each of the solutions in eighteen moves, the letters C, T, O, R move
once only, while the second I always moves four times, the V always
being transferred to the right arm of the cross.


219.--THE LETTER BLOCK PUZZLE.

This puzzle can be solved in 23 moves--the fewest possible. Move the
blocks in the following order: A, B, F, E, C, A, B, F, E, C, A, B, D, H,
G, A, B, D, H, G, D, E, F.


220.--A LODGING-HOUSE DIFFICULTY.

The shortest possible way is to move the articles in the following
order: Piano, bookcase, wardrobe, piano, cabinet, chest of drawers,
piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers,
wardrobe, cabinet, bookcase, piano. Thus seventeen removals are
necessary. The landlady could then move chest of drawers, wardrobe, and
cabinet. Mr. Dobson did not mind the wardrobe and chest of drawers
changing rooms so long as he secured the piano.


221.--THE EIGHT ENGINES.

The solution to the Eight Engines Puzzle is as follows: The engine that
has had its fire drawn and therefore cannot move is No. 5. Move the
other engines in the following order: 7, 6, 3, 7, 6, 1, 2, 4, 1, 3, 8,
1, 3, 2, 4, 3, 2, seventeen moves in all, leaving the eight engines in
the required order.

There are two other slightly different solutions.


222.--A RAILWAY PUZZLE.

This little puzzle may be solved in as few as nine moves. Play the
engines as follows: From 9 to 10, from 6 to 9, from 5 to 6, from 2 to 5,
from 1 to 2, from 7 to 1, from 8 to 7, from 9 to 8, and from 10 to 9.
You will then have engines A, B, and C on each of the three circles and
on each of the three straight lines. This is the shortest solution that
is possible.


223.--A RAILWAY MUDDLE.

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

[Illustration: 5]

[Illustration: 6]

Only six reversals are necessary. The white train (from A to D) is
divided into three sections, engine and 7 wagons, 8 wagons, and 1 wagon.
The black train (D to A) never uncouples anything throughout. Fig. 1 is
original position with 8 and 1 uncoupled. The black train proceeds to
position in Fig. 2 (no reversal). The engine and 7 proceed towards D,
and black train backs, leaves 8 on loop, and takes up position in Fig. 3
(first reversal). Black train goes to position in Fig. 4 to fetch single
wagon (second reversal). Black train pushes 8 off loop and leaves single
wagon there, proceeding on its journey, as in Fig. 5 (third and fourth
reversals). White train now backs on to loop to pick up single car and
goes right away to D (fifth and sixth reversals).


224.--THE MOTOR-GARAGE PUZZLE.

The exchange of cars can be made in forty-three moves, as follows: 6-G,
2-B, 1-E, 3-H, 4-I, 3-L, 6-K, 4-G, 1-I, 2-J, 5-H, 4-A, 7-F, 8-E, 4-D,
8-C, 7-A, 8-G, 5-C, 2-B, 1-E, 8-I, 1-G, 2-J, 7-H, 1-A, 7-G, 2-B, 6-E,
3-H, 8-L, 3-I, 7-K, 3-G, 6-I, 2-J, 5-H, 3-C, 5-G, 2-B, 6-E, 5-I, 6-J. Of
course, "6-G" means that the car numbered "6" moves to the point "G."
There are other ways in forty-three moves.


225.--THE TEN PRISONERS.

[Illustration]

It will be seen in the illustration how the prisoners may be arranged so
as to produce as many as sixteen even rows. There are 4 such vertical
rows, 4 horizontal rows, 5 diagonal rows in one direction, and 3
diagonal rows in the other direction. The arrows here show the movements
of the four prisoners, and it will be seen that the infirm man in the
bottom corner has not been moved.


226.--ROUND THE COAST.

In order to place words round the circle under the conditions, it is
necessary to select words in which letters are repeated in certain
relative positions. Thus, the word that solves our puzzle is "Swansea,"
in which the first and fifth letters are the same, and the third and
seventh the same. We make out jumps as follows, taking the letters of
the word in their proper order: 2-5, 7-2, 4-7, 1-4, 6-1, 3-6, 8-3. Or we
could place a word like "Tarapur" (in which the second and fourth
letters, and the third and seventh, are alike) with these moves: 6-1,
7-4, 2-7, 5--2, 8-5, 3-6, 8-3. But "Swansea" is the only word,
apparently, that will fulfil the conditions of the puzzle.

This puzzle should be compared with Sharp's Puzzle, referred to in my
solution to No. 341, "The Four Frogs." The condition "touch and jump
over two" is identical with "touch and move along a line."


227.--CENTRAL SOLITAIRE.

Here is a solution in nineteen moves; the moves enclosed in brackets
count as one move only: 19-17, 16-18, (29-17, 17-19), 30-18, 27-25,
(22-24, 24-26), 31-23, (4-16, 16-28), 7-9, 10-8, 12-10, 3-11, 18-6,
(1-3, 3-11), (13-27, 27-25), (21-7, 7-9), (33-31, 31-23), (10-8, 8-22,
22-24, 24-26, 26-12, 12-10), 5-17. All the counters are now removed
except one, which is left in the central hole. The solution needs
judgment, as one is tempted to make several jumps in one move, where it
would be the reverse of good play. For example, after playing the first
3-11 above, one is inclined to increase the length of the move by
continuing with 11-25, 25-27, or with 11-9, 9-7.

I do not think the number of moves can be reduced.


228.--THE TEN APPLES.

Number the plates (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13, 14,
15, 16) in successive rows from the top to the bottom. Then transfer the
apple from 8 to 10 and play as follows, always removing the apple jumped
over: 9-11, 1-9, 13-5, 16-8, 4-12, 12-10, 3-1, 1-9, 9-11.


229.--THE NINE ALMONDS.

This puzzle may be solved in as few as four moves, in the following
manner: Move 5 over 8, 9, 3, 1. Move 7 over 4. Move 6 over 2 and 7. Move
5 over 6, and all the counters are removed except 5, which is left in
the central square that it originally occupied.


230.--THE TWELVE PENNIES.

Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to 1,
9 to 5, 11 to 2.


231.--PLATES AND COINS.

Number the plates from 1 to 12 in the order that the boy is seen to be
going in the illustration. Starting from 1, proceed as follows, where "1
to 4" means that you take the coin from plate No. 1 and transfer it to
plate No. 4: 1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and
complete the last revolution to 1, making three revolutions in all. Or
you can proceed this way: 4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10
to 1. It is easy to solve in four revolutions, but the solutions in
three are more difficult to discover.

This is "The Riddle of the Fishpond" (No. 41, _Canterbury Puzzles_) in a
different dress.


232.--CATCHING THE MICE.

In order that the cat should eat every thirteenth mouse, and the white
mouse last of all, it is necessary that the count should begin at the
seventh mouse (calling the white one the first)--that is, at the one
nearest the tip of the cat's tail. In this case it is not at all
necessary to try starting at all the mice in turn until you come to the
right one, for you can just start anywhere and note how far distant the
last one eaten is from the starting point. You will find it to be the
eighth, and therefore must start at the eighth, counting backwards from
the white mouse. This is the one I have indicated.

In the case of the second puzzle, where you have to find the smallest
number with which the cat may start at the white mouse and eat this one
last of all, unless you have mastered the general solution of the
problem, which is very difficult, there is no better course open to you
than to try every number in succession until you come to one that works
correctly. The smallest number is twenty-one. If you have to proceed by
trial, you will shorten your labour a great deal by only counting out
the remainders when the number is divided successively by 13, 12, 11,
10, etc. Thus, in the case of 21, we have the remainders 8, 9, 10, 1, 3,
5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the remainders of 7, 3,
and 1 as nought, but as 7, 3, and 1. Now, count round each of these
numbers in turn, and you will find that the white mouse is killed last
of all. Of course, if we wanted simply any number, not the smallest, the
solution is very easy, for we merely take the least common multiple of
13, 12, 11, 10, etc. down to 2. This is 360360, and you will find that
the first count kills the thirteenth mouse, the next the twelfth, the
next the eleventh, and so on down to the first. But the most
arithmetically inclined cat could not be expected to take such a big
number when a small one like twenty-one would equally serve its purpose.

In the third case, the smallest number is 100. The number 1,000 would
also do, and there are just seventy-two other numbers between these that
the cat might employ with equal success.


233.--THE ECCENTRIC CHEESEMONGER.

To leave the three piles at the extreme ends of the rows, the cheeses
may be moved as follows--the numbers refer to the cheeses and not to
their positions in the row: 7-2, 8-7, 9-8, 10-15, 6-10, 5-6, 14-16,
13-14, 12-13, 3-1, 4-3, 11-4. This is probably the easiest solution of
all to find. To get three of the piles on cheeses 13, 14, and 15, play
thus: 9-4, 10-9, 11-10, 6-14, 5-6, 12-15, 8-12, 7-8, 16-5, 3-13, 2-3,
1-2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8-3,
9-14, 16-12, 1-5, 10-9, 7-10, 11-8, 2-1, 4-16, 13-2, 6-11, 15-4.


234.--THE EXCHANGE PUZZLE.

Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F,
I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K. It will be found
that, although the white counters can be moved to their proper places in
11 moves, if we omit all consideration of exchanges, yet the black
cannot be so moved in fewer than 17 moves. So we have to introduce waste
moves with the white counters to equal the minimum required by the
black. Thus fewer than 17 moves must be impossible. Some of the moves
are, of course, interchangeable.


235.--TORPEDO PRACTICE.

[Illustration:

      10  6 7
       \  |/
    4   u u   2
     \   u   /
    3-u u u u
       u   u
      u u u u -----9---
     /   u
    8   u u
       /   \
      1     5

]

If the enemy's fleet be anchored in the formation shown in the
illustration, it will be seen that as many as ten out of the sixteen
ships may be blown up by discharging the torpedoes in the order
indicated by the numbers and in the directions indicated by the arrows.
As each torpedo in succession passes under three ships and sinks the
fourth, strike out each vessel with the pencil as it is sunk.


236.--THE HAT PUZZLE.

[Illustration:

      1  2  3  4  5  6  7  8  9 10 11 12
    +--+--+--+--+--+--+--+--+--+--+--+--+
    | *| o| *| O| *| O| *| O| *| O|  |  |
    +--+--+--+--+--+--+--+--+--+--+--+--+
    | *|  |  | O| *| O| *| O| *| O| O| *|
    +--+--+--+--+--+--+--+--+--+--+--+--+
    | *| *| O| O| *| O|  |  | *| O| O| *|
    +--+--+--+--+--+--+--+--+--+--+--+--+
    | *| *| O|  |  | O| O| *| *| O| O| *|
    +--+--+--+--+--+--+--+--+--+--+--+--+
    | *| *| O| O| O| O| O| *| *|  |  | *|
    +--+--+--+--+--+--+--+--+--+--+--+--+
    |  |  | O| O| O| O| O| *| *| *| *| *|
    +--+--+--+--+--+--+--+--+--+--+--+--+

]

I suggested that the reader should try this puzzle with counters, so I
give my solution in that form. The silk hats are represented by black
counters and the felt hats by white counters. The first row shows the
hats in their original positions, and then each successive row shows how
they appear after one of the five manipulations. It will thus be seen
that we first move hats 2 and 3, then 7 and 8, then 4 and 5, then 10 and
11, and, finally, 1 and 2, leaving the four silk hats together, the four
felt hats together, and the two vacant pegs at one end of the row. The
first three pairs moved are dissimilar hats, the last two pairs being
similar. There are other ways of solving the puzzle.


237.--BOYS AND GIRLS.

There are a good many different solutions to this puzzle. Any contiguous
pair, except 7-8, may be moved first, and after the first move there are
variations. The following solution shows the position from the start
right through each successive move to the end:--

    . . 1 2 3 4 5 6 7 8
    4 3 1 2 . . 5 6 7 8
    4 3 1 2 7 6 5 . . 8
    4 3 1 2 7 . . 5 6 8
    4 . . 2 7 1 3 5 6 8
    4 8 6 2 7 1 3 5 . .


238.--ARRANGING THE JAM POTS.

Two of the pots, 13 and 19, were in their proper places. As every
interchange may result in a pot being put in its place, it is clear that
twenty-two interchanges will get them all in order. But this number of
moves is not the fewest possible, the correct answer being seventeen.
Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17),
(24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14,
18-9). When you have made the interchanges within any pair of brackets,
all numbers within those brackets are in their places. There are five
pairs of brackets, and 5 from 22 gives the number of changes
required--17.


239.--A JUVENILE PUZZLE.

[Illustration:

    +-----------------+
    |     C     E     |
    |     |     |     |
    |     D     F     |
    +---------------B |
             G        |
    A        |        |
    |        H        |
    +-----------------+

]

As the conditions are generally understood, this puzzle is incapable of
solution. This can be demonstrated quite easily. So we have to look for
some catch or quibble in the statement of what we are asked to do. Now
if you fold the paper and then push the point of your pencil down
between the fold, you can with one stroke make the two lines CD and EF
in our diagram. Then start at A, and describe the line ending at B.
Finally put in the last line GH, and the thing is done strictly within
the conditions, since folding the paper is not actually forbidden. Of
course the lines are here left unjoined for the purpose of clearness.

In the rubbing out form of the puzzle, first rub out A to B with a
single finger in one stroke. Then rub out the line GH with one finger.
Finally, rub out the remaining two vertical lines with two fingers at
once! That is the old trick.


240.--THE UNION JACK.

[Illustration:

    +-------+    +-----
    A  B    |    |    /
        \   |    |   /
    |\   \  |    |  /   /|
    | \   \ |    | /   / |
    |  \   \|    |/   /  |
    |   \   |    /   /   |
    |    \  |\  /|  /    |
    +-----\-|-\/-|-/-----+
           \| /\ |/
            |/  \/
            |\  /\
           /| \/ |\
    +-----/-|-/\-|-\-----+
    |    /   /  \|  \    |
    |   /   |    \   \   |
    |  /   /|    |\   \  |
    | /   / |    | \   \ |
    |/   /  |    |  \   \|
        /   |    |   \
       /    |    |    \
       -----+    +-----

]

There are just sixteen points (all on the outside) where three roads may
be said to join. These are called by mathematicians "odd nodes." There
is a rule that tells us that in the case of a drawing like the present
one, where there are sixteen odd nodes, it requires eight separate
strokes or routes (that is, half as many as there are odd nodes) to
complete it. As we have to produce as much as possible with only one of
these eight strokes, it is clearly necessary to contrive that the seven
strokes from odd node to odd node shall be as short as possible. Start
at A and end at B, or go the reverse way.


241.--THE DISSECTED CIRCLE.

[Illustration:

           /---------------\
          /                 \
         /  /------B         \
        /  /       |     /^\  \
       /  /  |\    |    /   \  \
      /  /   | \   |   /     \  \
     /  /    |  \  |  /  A    \  \
    /  /     |   \ | /   |     \  \
    | /      |    \|/    |      \ |
    | | -----+-----*-----+----- | |
    | | \    |    /|\    |    / | |
    | |  \   |   / | \   |   /  | |
    | |   \  |  /  |  \  |  /   | |
    | |    \ | /   |   \ | /    | |
    | |     \|/    |    \|/     | |
    D-+------*-----+-----*----E | |
      |     /|\    |    /|\     | |
      |    / | \   |   / | \    | |
      |   /  |  \  |  /  |  \   | |
      |  /   |   \ | /   |   \  | |
      | /    |    \|/    |    \ | |
      | -----+-----*-----+----- | |
      \      |    /|\    |      / |
       \     |   / | \   |     /  /
        \    |  /  |  \  |    /  /
         \   | /   |   \ |   /  /
          \  |/    |    \|  /  /
           \       |       /  /
            \------+------/  /
                   |        /
                   C-------/

]

It can be done in twelve continuous strokes, thus: Start at A in the
illustration, and eight strokes, forming the star, will bring you back
to A; then one stroke round the circle to B, one stroke to C, one round
the circle to D, and one final stroke to E--twelve in all. Of course, in
practice the second circular stroke will be over the first one; it is
separated in the diagram, and the points of the star not joined to the
circle, to make the solution clear to the eye.

242.--THE TUBE INSPECTOR'S PUZZLE.

The inspector need only travel nineteen miles if he starts at B and
takes the following route: BADGDEFIFCBEHKLIHGJK. Thus the only portions
of line travelled over twice are the two sections D to G and F to I. Of
course, the route may be varied, but it cannot be shortened.

243.--VISITING THE TOWNS.

Note that there are six towns, from which only two roads issue. Thus 1
must lie between 9 and 12 in the circular route. Mark these two roads as
settled. Similarly mark 9, 5, 14, and 4, 8, 14, and 10, 6, 15, and 10,
2, 13, and 3, 7, 13. All these roads must be taken. Then you will find
that he must go from 4 to 15, as 13 is closed, and that he is compelled
to take 3, 11, 16, and also 16, 12. Thus, there is only one route, as
follows: 1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16, 12, 1, or
its reverse--reading the line the other way. Seven roads are not used.

244.--THE FIFTEEN TURNINGS.

[Illustration]

It will be seen from the illustration (where the roads not used are
omitted) that the traveller can go as far as seventy miles in fifteen
turnings. The turnings are all numbered in the order in which they are
taken. It will be seen that he never visits nineteen of the towns. He
might visit them all in fifteen turnings, never entering any town twice,
and end at the black town from which he starts (see "The Rook's Tour,"
No. 320), but such a tour would only take him sixty-four miles.

245.--THE FLY ON THE OCTAHEDRON.

[Illustration]

Though we cannot really see all the sides of the octahedron at once, we
can make a projection of it that suits our purpose just as well. In the
diagram the six points represent the six angles of the octahedron, and
four lines proceed from every point under exactly the same conditions as
the twelve edges of the solid. Therefore if we start at the point A and
go over all the lines once, we must always end our route at A. And the
number of different routes is just 1,488, counting the reverse way of
any route as different. It would take too much space to show how I make
the count. It can be done in about five minutes, but an explanation of
the method is difficult. The reader is therefore asked to accept my
answer as correct.

246.--THE ICOSAHEDRON PUZZLE.

[Illustration]

There are thirty edges, of which eighteen were visible in the original
illustration, represented in the following diagram by the hexagon
NAESGD. By this projection of the solid we get an imaginary view of the
remaining twelve edges, and are able to see at once their direction and
the twelve points at which all the edges meet. The difference in the
length of the lines is of no importance; all we want is to present their
direction in a graphic manner. But in case the novice should be puzzled
at only finding nineteen triangles instead of the required twenty, I
will point out that the apparently missing triangle is the outline HIK.

In this case there are twelve odd nodes; therefore six distinct and
disconnected routes will be needful if we are not to go over any lines
twice. Let us therefore find the greatest distance that we may so travel
in one route.

It will be noticed that I have struck out with little cross strokes five
lines or edges in the diagram. These five lines may be struck out
anywhere so long as they do not join one another, and so long as one of
them does not connect with N, the North Pole, from which we are to
start. It will be seen that the result of striking out these five lines
is that all the nodes are now even except N and S. Consequently if we
begin at N and stop at S we may go over all the lines, except the five
crossed out, without traversing any line twice. There are many ways of
doing this. Here is one route: N to H, I, K, S, I, E, S, G, K, D, H, A,
N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making five of the routes
as short as is possible--simply from one node to the next--we are able
to get the greatest possible length for our sixth line. A greater
distance in one route, without going over the same ground twice, it is
not possible to get.

It is now readily seen that those five erased lines must be gone over
twice, and they may be "picked up," so to speak, at any points of our
route. Thus, whenever the traveller happens to be at I he can run up to
A and back before proceeding on his route, or he may wait until he is at
A and then run down to I and back to A. And so with the other lines that
have to be traced twice. It is, therefore, clear that he can go over 25
of the lines once only (25 × 10,000 miles = 250,000 miles) and 5 of the
lines twice (5 × 20,000 miles = 100,000 miles), the total, 350,000 miles,
being the length of his travels and the shortest distance that is
possible in visiting the whole body.

It will be noticed that I have made him end his travels at S, the South
Pole, but this is not imperative. I might have made him finish at any of
the other nodes, except the one from which he started. Suppose it had
been required to bring him home again to N at the end of his travels.
Then instead of suppressing the line AI we might leave that open and
close IS. This would enable him to complete his 350,000 miles tour at A,
and another 10,000 miles would take him to his own fireside. There are a
great many different routes, but as the lengths of the edges are all
alike, one course is as good as another. To make the complete 350,000
miles tour from N to S absolutely clear to everybody, I will give it
entire: N to H, I, A, I, K, H, K, S, I, E, S, G, F, G, K, D, C, D, H, A,
N, B, E, B, A, E, F, B, C, G, D, N, C, F, S--that is, thirty-five lines
of 10,000 miles each.


247.--INSPECTING A MINE.

Starting from A, the inspector need only travel 36 furlongs if he takes
the following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R,
M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K, P, Q. He thus
passes between A and B twice, between C and D twice, between F and K
twice, between J and O twice, and between R and S twice--five
repetitions. Therefore 31 passages plus 5 repeated equal 36 furlongs.
The little pitfall in this puzzle lies in the fact that we start from an
even node. Otherwise we need only travel 35 furlongs.


248.--THE CYCLIST'S TOUR.

When Mr. Maggs replied, "No way, I'm sure," he was not saying that the
thing was impossible, but was really giving the actual route by which
the problem can be solved. Starting from the star, if you visit the
towns in the order, NO WAY, I'M SURE, you will visit every town once,
and only once, and end at E. So both men were correct. This was the
little joke of the puzzle, which is not by any means difficult.


249.--THE SAILOR'S PUZZLE.

[Illustration]

There are only four different routes (or eight, if we count the reverse
ways) by which the sailor can start at the island marked A, visit all
the islands once, and once only, and return again to A. Here they are:--

A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D K
M B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E U G
N S K M B Q D C F R H A

Now, if the sailor takes the first route he will make C his 12th island
(counting A as 1); by the second route he will make C his 13th island;
by the third route, his 16th island; and by the fourth route, his 17th
island. If he goes the reverse way, C will be respectively his 10th,
9th, 6th, and 5th island. As these are the only possible routes, it is
evident that if the sailor puts off his visit to C as long as possible,
he must take the last route reading from left to right. This route I
show by the dark lines in the diagram, and it is the correct answer to
the puzzle.

The map may be greatly simplified by the "buttons and string" method,
explained in the solution to No. 341, "The Four Frogs."

250.--THE GRAND TOUR.

The first thing to do in trying to solve a puzzle like this is to
attempt to simplify it. If you look at Fig. 1, you will see that it is a
simplified version of the map. Imagine the circular towns to be buttons
and the railways to be connecting strings. (See solution to No. 341.)
Then, it will be seen, we have simply "straightened out" the previous
diagram without affecting the conditions. Now we can further simplify by
converting Fig. 1 into Fig. 2, which is a portion of a chessboard. Here
the directions of the railways will resemble the moves of a rook in
chess--that is, we may move in any direction parallel to the sides of
the diagram, but not diagonally. Therefore the first town (or square)
visited must be a black one; the second must be a white; the third must
be a black; and so on. Every odd square visited will thus be black and
every even one white. Now, we have 23 squares to visit (an odd number),
so the last square visited must be black. But Z happens to be white, so
the puzzle would seem to be impossible of solution.

[Illustration: Fig. 1.]

[Illustration: Fig. 2.]

As we were told that the man "succeeded" in carrying put his plan, we
must try to find some loophole in the conditions. He was to "enter every
town once and only once," and we find no prohibition against his
entering once the town A after leaving it, especially as he has never
left it since he was born, and would thus be "entering" it for the first
time in his life. But he must return at once from the first town he
visits, and then he will have only 22 towns to visit, and as 22 is an
even number, there is no reason why he should not end on the white
square Z. A possible route for him is indicated by the dotted line from
A to Z. This route is repeated by the dark lines in Fig. 1, and the
reader will now have no difficulty in applying; it to the original map.
We have thus proved that the puzzle can only be solved by a return to A
immediately after leaving it.

251.--WATER, GAS, AND ELECTRICITY.

[Illustration]

According to the conditions, in the strict sense in which one at first
understands them, there is no possible solution to this puzzle. In such
a dilemma one always has to look for some verbal quibble or trick. If
the owner of house A will allow the water company to run their pipe for
house C through his property (and we are not bound to assume that he
would object), then the difficulty is got over, as shown in our
illustration. It will be seen that the dotted line from W to C passes
through house A, but no pipe ever crosses another pipe.


252.--A PUZZLE FOR MOTORISTS.

[Illustration]

The routes taken by the eight drivers are shown in the illustration,
where the dotted line roads are omitted to make the paths clearer to the
eye.


253.--A BANK HOLIDAY PUZZLE.

The simplest way is to write in the number of routes to all the towns in
this manner. Put a 1 on all the towns in the top row and in the first
column. Then the number of routes to any town will be the sum of the
routes to the town immediately above and to the town immediately to the
left. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc.,
in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with the other
rows. It will then be seen that the only town to which there are exactly
1,365 different routes is the twelfth town in the fifth row--the one
immediately over the letter E. This town was therefore the cyclist's
destination.

The general formula for the number of routes from one corner to the
corner diagonally opposite on any such rectangular reticulated
arrangement, under the conditions as to direction, is (m+n)!/m!n!,
where m is the number of towns on one side, less one, and n the number
on the other side, less one. Our solution involves the case where
there are 12 towns by 5. Therefore m = 11 and n = 4. Then the formula
gives us the answer 1,365 as above.


254.-- THE MOTOR-CAR TOUR.

First of all I will ask the reader to compare the original square
diagram with the circular one shown in Figs. 1, 2, and 3 below. If for
the moment we ignore the shading (the purpose of which I shall proceed
to explain), we find that the circular diagram in each case is merely a
simplification of the original square one--that is, the roads from A
lead to B, E, and M in both cases, the roads from L (London) lead to I,
K, and S, and so on. The form below, being circular and symmetrical,
answers my purpose better in applying a mechanical solution, and I
therefore adopt it without altering in any way the conditions of the
puzzle. If such a question as distances from town to town came into the
problem, the new diagrams might require the addition of numbers to
indicate these distances, or they might conceivably not be at all
practicable.

[Illustration: Figs. 1, 2, and 3]

Now, I draw the three circular diagrams, as shown, on a sheet of paper
and then cut out three pieces of cardboard of the forms indicated by the
shaded parts of these diagrams. It can be shown that every route, if
marked out with a red pencil, will form one or other of the designs
indicated by the edges of the cards, or a reflection thereof. Let us
direct our attention to Fig. 1. Here the card is so placed that the star
is at the town T; it therefore gives us (by following the edge of the
card) one of the circular routes from London: L, S, R, T, M, A, E, P, O,
J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we should
get L, I, F, H, K, Q, etc., but these reverse routes were not to be
counted. When we have written out this first route we revolve the card
until the star is at M, when we get another different route, at A a
third route, at E a fourth route, and at P a fifth route. We have thus
obtained five different routes by revolving the card as it lies. But it
is evident that if we now take up the card and replace it with the other
side uppermost, we shall in the same manner get five other routes by
revolution.

We therefore see how, by using the revolving card in Fig. 1, we may,
without any difficulty, at once write out ten routes. And if we employ
the cards in Figs. 2 and 3, we similarly obtain in each case ten other
routes. These thirty routes are all that are possible. I do not give the
actual proof that the three cards exhaust all the possible cases, but
leave the reader to reason that out for himself. If he works out any
route at haphazard, he will certainly find that it falls into one or
other of the three categories.


255.--THE LEVEL PUZZLE.

Let us confine our attention to the L in the top left-hand corner.
Suppose we go by way of the E on the right: we must then go straight on
to the V, from which letter the word may be completed in four ways, for
there are four E's available through which we may reach an L. There are
therefore four ways of reading through the right-hand E. It is also
clear that there must be the same number of ways through the E that is
immediately below our starting point. That makes eight. If, however, we
take the third route through the E on the diagonal, we then have the
option of any one of the three V's, by means of each of which we may
complete the word in four ways. We can therefore spell LEVEL in twelve
ways through the diagonal E. Twelve added to eight gives twenty
readings, all emanating from the L in the top left-hand corner; and as
the four corners are equal, the answer must be four times twenty, or
eighty different ways.


256.--THE DIAMOND PUZZLE.

There are 252 different ways. The general formula is that, for words of
n letters (not palindromes, as in the case of the next puzzle), when
grouped in this manner, there are always 2^(n+1) - 4 different readings.
This does not allow diagonal readings, such as you would get if you used
instead such a word as DIGGING, where it would be possible to pass from
one G to another G by a diagonal step.


257.--THE DEIFIED PUZZLE.

The correct answer is 1,992 different ways. Every F is either a corner F
or a side F--standing next to a corner in its own square of F's. Now,
FIED may be read _from_ a corner F in 16 ways; therefore DEIF may be
read _into_ a corner F also in 16 ways; hence DEIFIED may be read
_through_ a corner F in 16 × 16 = 256 ways. Consequently, the four
corner F's give 4 × 256 = 1,024 ways. Then FIED may be read from a side
F in 11 ways, and DEIFIED therefore in 121 ways. But there are eight
side F's; consequently these give together 8 × 121 = 968 ways. Add 968
to 1,024 and we get the answer, 1,992.

In this form the solution will depend on whether the number of letters
in the palindrome be odd or even. For example, if you apply the word NUN
in precisely the same manner, you will get 64 different readings; but if
you use the word NOON, you will only get 56, because you cannot use the
same letter twice in immediate succession (since you must "always pass
from one letter to another") or diagonal readings, and every reading
must involve the use of the central N.

The reader may like to find for himself the general formula in this
case, which is complex and difficult. I will merely add that for such a
case as MADAM, dealt with in the same way as DEIFIED, the number of
readings is 400.


258.-- THE VOTERS' PUZZLE.

THE number of readings here is 63,504, as in the case of "WAS IT A RAT I
SAW" (No. 30, _Canterbury Puzzles_). The general formula is that for
palindromic sentences containing 2n + 1 letters there are (4(2^n -1))²
readings.


259.-- HANNAH'S PUZZLE.

Starting from any one of the N's, there are 17 different readings of
NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways
of spelling HAN. If we were allowed to use the same N twice in a
spelling, the answer would be 68 times 68, or 4,624 ways. But the
conditions were, "always passing from one letter to another." Therefore,
for every one of the 17 ways of spelling HAN with a particular N, there
would be 51 ways (3 times 17) of completing the NAH, or 867 (17 times
51) ways for the complete word. Hence, as there are four N's to use in
HAN, the correct solution of the puzzle is 3,468 (4 times 867) different
ways.


260.--THE HONEYCOMB PUZZLE.

The required proverb is, "There is many a slip 'twixt the cup and the
lip." Start at the T on the outside at the bottom right-hand corner,
pass to the H above it, and the rest is easy.


261.-- THE MONK AND THE BRIDGES.

[Illustration]

The problem of the Bridges may be reduced to the simple diagram shown
in illustration. The point M represents the Monk, the point I the
Island, and the point Y the Monastery. Now the only direct ways from M
to I are by the bridges a and b; the only direct ways from I to Y are
by the bridges c and d; and there is a direct way from M to Y by the
bridge e. Now, what we have to do is to count all the routes that will
lead from M to Y, passing over all the bridges, a, b, c, d, and e once
and once only. With the simple diagram under the eye it is quite easy,
without any elaborate rule, to count these routes methodically. Thus,
starting from a, b, we find there are only two ways of completing the
route; with _a, c_, there are only two routes; with a, d, only two
routes; and so on. It will be found that there are sixteen such routes
in all, as in the following list:--

    a b e c d                      b c d a e
    a b e d c                      b c e a d
    a c d b e                      b d c a e
    a c e b d                      b d e a c
    a d e b c                      e c a b d
    a d c b e                      e c b a d
    b a e c d                      e d a b c
    b a e d c                      e d b a c

If the reader will transfer the letters indicating the bridges from the
diagram to the corresponding bridges in the original illustration,
everything will be quite obvious.


262.--THOSE FIFTEEN SHEEP.

If we read the exact words of the writer in the cyclopædia, we find that
we are not told that the pens were all necessarily empty! In fact, if
the reader will refer back to the illustration, he will see that one
sheep is already in one of the pens. It was just at this point that the
wily farmer said to me, "_Now_ I'm going to start placing the fifteen
sheep." He thereupon proceeded to drive three from his flock into the
already occupied pen, and then placed four sheep in each of the other
three pens. "There," says he, "you have seen me place fifteen sheep in
four pens so that there shall be the same number of sheep in every pen."
I was, of course, forced to admit that he was perfectly correct,
according to the exact wording of the question.


263.--KING ARTHUR'S KNIGHTS.

On the second evening King Arthur arranged the knights and himself in
the following order round the table: A, F, B, D, G, E, C. On the third
evening they sat thus, A, E, B, G, C, F, D. He thus had B next but one
to him on both occasions (the nearest possible), and G was the third
from him at both sittings (the furthest position possible). No other way
of sitting the knights would have been so satisfactory.


264.--THE CITY LUNCHEONS.

The men may be grouped as follows, where each line represents a day and
each column a table:--

    AB  CD  EF  GH  IJ  KL
    AE  DL  GK  FI  CB  HJ
    AG  LJ  FH  KC  DE  IB
    AF  JB  KI  HD  LG  CE
    AK  BE  HC  IL  JF  DG
    AH  EG  ID  CJ  BK  LF
    AI  GF  CL  DB  EH  JK
    AC  FK  DJ  LE  GI  BH
    AD  KH  LB  JG  FC  EI
    AL  HI  JE  BF  KD  GC
    AJ  IC  BG  EK  HL  FD

Note that in every column (except in the case of the A's) all the
letters descend cyclically in the same order, B, E, G, F, up to J, which
is followed by B.


265.--A PUZZLE FOR CARD-PLAYERS.

In the following solution each of the eleven lines represents a sitting,
each column a table, and each pair of letters a pair of partners.

    A B -- I L  |  E J -- G K  |  F H -- C D
    A C -- J B  |  F K -- H L  |  G I -- D E
    A D -- K C  |  G L -- I B  |  H J -- E F
    A E -- L D  |  H B -- J C  |  I K -- F G
    A F -- B E  |  I C -- K D  |  J L -- G H
    A G -- C F  |  J D -- L E  |  K B -- H I
    A H -- D G  |  K E -- B F  |  L C -- I J
    A I -- E H  |  L F -- C G  |  B D -- J K
    A J -- F I  |  B G -- D H  |  C E -- K L
    A K -- G J  |  C H -- E I  |  D F -- L B
    A L -- H K  |  D I -- F J  |  E G -- B C

It will be seen that the letters B, C, D ...L descend cyclically. The
solution given above is absolutely perfect in all respects. It will be
found that every player has every other player once as his partner and
twice as his opponent.

266.--A TENNIS TOURNAMENT.

Call the men A, B, D, E, and their wives a, b, d, e. Then they may play
as follows without any person ever playing twice with or against any
other person:--

                First Court.     Second Court.
    1st Day | A d against B e | D a against E b
    2nd Day | A e    "    D b | E a    "    B d
    3rd Day | A b    "    E d | B a    "    D e

It will be seen that no man ever plays with or against his own wife--an
ideal arrangement. If the reader wants a hard puzzle, let him try to
arrange eight married couples (in four courts on seven days) under
exactly similar conditions. It can be done, but I leave the reader in
this case the pleasure of seeking the answer and the general solution.


267.--THE WRONG HATS.

The number of different ways in which eight persons, with eight hats,
can each take the wrong hat, is 14,833.

Here are the successive solutions for any number of persons from one to
eight:--

    1 = 0
    2 = 1
    3 = 2
    4 = 9
    5 = 44
    6 = 265
    7 = 1,854
    8 = 14,833

To get these numbers, multiply successively by 2, 3, 4, 5, etc. When the
multiplier is even, add 1; when odd, deduct 1. Thus, 3 × 1 - 1 = 2; 4 ×
2 + 1 = 9; 5 × 9 - 1 = 44; and so on. Or you can multiply the sum of the
number of ways for n - 1 and n - 2 persons by n - 1, and so get the
solution for n persons. Thus, 4(2 + 9) = 44; 5(9 + 44) = 265; and so on.


268.--THE PEAL OF BELLS.

The bells should be rung as follows:--

    1 2 3 4
    2 1 4 3
    2 4 1 3
    4 2 3 1
    4 3 2 1
    3 4 1 2
    3 1 4 2
    1 3 2 4
    3 1 2 4
    1 3 4 2
    1 4 3 2
    4 1 2 3
    4 2 1 3
    2 4 3 1
    2 3 4 1
    3 2 1 4
    2 3 1 4
    3 2 4 1
    3 4 2 1
    4 3 1 2
    4 1 3 2
    1 4 2 3
    1 2 4 3
    2 1 3 4

I have constructed peals for five and six bells respectively, and a
solution is possible for any number of bells under the conditions
previously stated.


269.--THREE MEN IN A BOAT.

If there were no conditions whatever, except that the men were all to go
out together, in threes, they could row in an immense number of
different ways. If the reader wishes to know how many, the number is
455^7. And with the condition that no two may ever be together more than
once, there are no fewer than 15,567,552,000 different solutions--that
is, different ways of arranging the men. With one solution before him,
the reader will realize why this must be, for although, as an example, A
must go out once with B and once with C, it does not necessarily follow
that he must go out with C on the same occasion that he goes with B. He
might take any other letter with him on that occasion, though the fact
of his taking other than B would have its effect on the arrangement of
the other triplets.

Of course only a certain number of all these arrangements are available
when we have that other condition of using the smallest possible number
of boats. As a matter of fact we need employ only ten different boats.
Here is one the arrangements:--

              1     2     3     4     5
    1st Day (ABC) (DBF) (GHI) (JKL) (MNO)
              8     6     7     9    10
    2nd Day (ADG) (BKN) (COL) (JEI) (MHF)
              3     5     4     1     2
    3rd Day (AJM) (BEH) (CFI) (DKO) (GNL)
              7     6     8     9     1
    4th Day (AEK) (CGM) (BOI) (DHL) (JNF)
              4     5     3    10     2
    5th Day (AHN) (CDJ) (BFL) (GEO) (MKI)
              6     7     8    10     1
    6th Day (AFO) (BGJ) (CKH) (DNI) (MEL)
              5     4     3     9     2
    7th Day (AIL) (BDM) (CEN) (GKF) (JHO)

It will be found that no two men ever go out twice together, and that no
man ever goes out twice in the same boat.

This is an extension of the well-known problem of the "Fifteen
Schoolgirls," by Kirkman. The original conditions were simply that
fifteen girls walked out on seven days in triplets without any girl ever
walking twice in a triplet with another girl. Attempts at a general
solution of this puzzle had exercised the ingenuity of mathematicians
since 1850, when the question was first propounded, until recently. In
1908 and the two following years I indicated (see _Educational Times
Reprints_, Vols. XIV., XV., and XVII.) that all our trouble had arisen
from a failure to discover that 15 is a special case (too small to enter
into the general law for all higher numbers of girls of the form 6n+3),
and showed what that general law is and how the groups should be posed
for any number of girls. I gave actual arrangements for numbers that had
previously baffled all attempts to manipulate, and the problem may now
be considered generally solved. Readers will find an excellent full
account of the puzzle in W.W. Rouse Ball's _Mathematical Recreations_,
5th edition.


270.--THE GLASS BALLS.

There are, in all, sixteen balls to be broken, or sixteen places in the
order of breaking. Call the four strings A, B, C, and D--order is here
of no importance. The breaking of the balls on A may occupy any 4 out of
these 16 places--that is, the combinations of 16 things, taken 4
together, will be

    13 × 14 × 15 × 16
    ----------------- = 1,820
     1 ×  2 ×  3 ×  4

ways for A. In every one of these cases B may occupy any 4 out of the
remaining 12 places, making

     9 × 10 × 11 × 12
    ----------------- = 495
     1 ×  2 ×  3 ×  4

ways. Thus 1,820 × 495 = 900,900 different placings are open to A and B.
But for every one of these cases C may occupy

    5 × 6 × 7 × 8
    ------------- = 70
    1 × 2 × 3 × 4

different places; so that 900,900 × 70 = 63,063,000 different placings
are open to A, B, and C. In every one of these cases, D has no choice
but to take the four places that remain. Therefore the correct answer is
that the balls may be broken in 63,063,000 different ways under the
conditions. Readers should compare this problem with No. 345, "The Two
Pawns," which they will then know how to solve for cases where there are
three, four, or more pawns on the board.


271.--FIFTEEN LETTER PUZZLE.

The following will be found to comply with the conditions of grouping:--

    ALE  MET  MOP  BLM
    BAG  CAP  YOU  CLT
    IRE  OIL  LUG  LNR
    NAY  BIT  BUN  BPR
    AIM  BEY  RUM  GMY
    OAR  GIN  PLY  CGR
    PEG  ICY  TRY  CMN
    CUE  COB  TAU  PNT
    ONE  GOT  PIU

The fifteen letters used are A, E, I, O, U, Y, and B, C, G, L, M, N, P,
R, T. The number of words is 27, and these are all shown in the first
three columns. The last word, PIU, is a musical term in common use; but
although it has crept into some of our dictionaries, it is Italian,
meaning "a little; slightly." The remaining twenty-six are good words.
Of course a TAU-cross is a T-shaped cross, also called the cross of St.
Anthony, and borne on a badge in the Bishop's Palace at Exeter. It is
also a name for the toad-fish.

We thus have twenty-six good words and one doubtful, obtained under the
required conditions, and I do not think it will be easy to improve on
this answer. Of course we are not bound by dictionaries but by common
usage. If we went by the dictionary only in a case of this kind, we
should find ourselves involved in prefixes, contractions, and such
absurdities as I.O.U., which Nuttall actually gives as a word.


272.--THE NINE SCHOOLBOYS.

The boys can walk out as follows:--

    1st Day.  2nd Day.  3rd Day.
     A B C     B F H     F A G
     D E F     E I A     I D B
     G H I     C G D     H C E

    4th Day.  5th Day.  6th Day.
     A D H     G B I     D C A
     B E G     C F D     E H B
     F I C     H A E     I G F

Every boy will then have walked by the side of every other boy once and
once only.

Dealing with the problem generally, 12n+9 boys may walk out in triplets
under the conditions on 9n+6 days, where n may be nought or any integer.
Every possible pair will occur once. Call the number of boys m. Then
every boy will pair m-1 times, of which (m-1)/4 times he will be in the
middle of a triplet and (m-1)/2 times on the outside. Thus, if we refer
to the solution above, we find that every boy is in the middle twice
(making 4 pairs) and four times on the outside (making the remaining 4
pairs of his 8). The reader may now like to try his hand at solving the
two next cases of 21 boys on 15 days, and 33 boys on 24 days. It is,
perhaps, interesting to note that a school of 489 boys could thus walk
out daily in one leap year, but it would take 731 girls (referred to in
the solution to No. 269) to perform their particular feat by a daily
walk in a year of 365 days.


273.--THE ROUND TABLE.

The history of this problem will be found in _The Canterbury Puzzles_
(No. 90). Since the publication of that book in 1907, so far as I know,
nobody has succeeded in solving the case for that unlucky number of
persons, 13, seated at a table on 66 occasions. A solution is possible
for any number of persons, and I have recorded schedules for every
number up to 25 persons inclusive and for 33. But as I know a good many
mathematicians are still considering the case of 13, I will not at this
stage rob them of the pleasure of solving it by showing the answer. But
I will now display the solutions for all the cases up to 12 persons
inclusive. Some of these solutions are now published for the first time,
and they may afford useful clues to investigators.

The solution for the case of 3 persons seated on 1 occasion needs no
remark.

A solution for the case of 4 persons on 3 occasions is as follows:--

    1 2 3 4
    1 3 4 2
    1 4 2 3

Each line represents the order for a sitting, and the person represented
by the last number in a line must, of course, be regarded as sitting
next to the first person in the same line, when placed at the round
table.

The case of 5 persons on 6 occasions may be solved as follows:--

    1 2 3 4 5
    1 2 4 5 3
    1 2 5 3 4
    ---------
    1 3 2 5 4
    1 4 2 3 5
    1 5 2 4 3

The case for 6 persons on 10 occasions is solved thus:--

    1 2 3 6 4 5
    1 3 4 2 5 6
    1 4 5 3 6 2
    1 5 6 4 2 3
    1 6 2 5 3 4
    -----------
    1 2 4 5 6 3
    1 3 5 6 2 4
    1 4 6 2 3 5
    1 5 2 3 4 6
    1 6 3 4 5 2

It will now no longer be necessary to give the solutions in full, for
reasons that I will explain. It will be seen in the examples above that
the 1 (and, in the case of 5 persons, also the 2) is repeated down the
column. Such a number I call a "repeater." The other numbers descend in
cyclical order. Thus, for 6 persons we get the cycle, 2, 3, 4, 5, 6, 2,
and so on, in every column. So it is only necessary to give the two
lines 1 2 3 6 4 5 and 1 2 4 5 6 3, and denote the cycle and repeaters,
to enable any one to write out the full solution straight away. The
reader may wonder why I do not start the last solution with the numbers
in their natural order, 1 2 3 4 5 6. If I did so the numbers in the
descending cycle would not be in their natural order, and it is more
convenient to have a regular cycle than to consider the order in the
first line.

The difficult case of 7 persons on 15 occasions is solved as follows,
and was given by me in _The Canterbury Puzzles_:--

    1 2 3 4 5 7 6
    1 6 2 7 5 3 4
    1 3 5 2 6 7 4
    1 5 7 4 3 6 2
    1 5 2 7 3 4 6

In this case the 1 is a repeater, and there are _two_ separate cycles,
2, 3, 4, 2, and 5, 6, 7, 5. We thus get five groups of three lines each,
for a fourth line in any group will merely repeat the first line.

A solution for 8 persons on 21 occasions is as follows:--

    1 8 6 3 4 5 2 7
    1 8 4 5 7 2 3 6
    1 8 2 7 3 6 4 5

The 1 is here a repeater, and the cycle 2, 3, 4, 5, 6, 7, 8. Every one
of the 3 groups will give 7 lines.

Here is my solution for 9 persons on 28 occasions:--

    2 1 9 7 4 5 6 3 8
    2 9 5 1 6 8 3 4 7
    2 9 3 1 8 4 7 5 6
    2 9 1 5 6 4 7 8 3

There are here two repeaters, 1 and 2, and the cycle is 3, 4, 5, 6, 7,
8, 9. We thus get 4 groups of 7 lines each.

The case of 10 persons on 36 occasions is solved as follows:--

    1 10 8 3 6 5 4 7 2 9
    1 10 6 5 2 9 7 4 3 8
    1 10 2 9 3 8 6 5 7 4
    1 10 7 4 8 3 2 9 5 6

The repeater is 1, and the cycle, 2, 3, 4, 5, 6, 7, 8, 9, 10. We here
have 4 groups of 9 lines each.

My solution for 11 persons on 45 occasions is as follows:--

    2 11  9  4 7 6  5 1 8 3 10
    2  1 11  7 6 3 10 8 5 4  9
    2 11 10  3 9 4  8 5 1 7  6
    2 11  5  8 1 3 10 6 7 9  4
    2 11  1 10 3 4  9 6 7 5  8

There are two repeaters, 1 and 2, and the cycle is, 3, 4, 5,... 11. We
thus get 5 groups of 9 lines each.

The case of 12 persons on 55 occasions is solved thus:--

    1 2 3 12  4 11  5 10  6  9  7  8
    1 2 4 11  6  9  8  7 10  5 12  3
    1 2 5 10  8  7 11  4  3 12  6  9
    1 2 6  9 10  5  3 12  7  8 11  4
    1 2 7  8 12  3  6  9 11  4  5 10

Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12. We thus get 5
groups of 11 lines each.


274.--THE MOUSE-TRAP PUZZLE.

If we interchange cards 6 and 13 and begin our count at 14, we may take
up all the twenty-one cards--that is, make twenty-one "catches"--in the
following order: 6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7, 21, 12, 15, 20,
9, 16, 18, 17, 19. We may also exchange 10 and 14 and start at 16, or
exchange 6 and 8 and start at 19.


275.--THE SIXTEEN SHEEP.

The six diagrams on next page show solutions for the cases where we
replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the
hurdles that have been replaced. There are, of course, other ways of
making the removals.


276.--THE EIGHT VILLAS.

There are several ways of solving the puzzle, but there is very little
difference between them. The solver should, however, first of all bear
in mind that in making his calculations he need only consider the four
villas that stand at the corners, because the intermediate villas can
never vary when the corners are known. One way is to place the numbers
nought to 9 one at a time in the top left-hand corner, and then consider
each case in turn.

Now, if we place 9 in the corner as shown in the Diagram A, two of the
corners cannot be occupied, while the corner that is diagonally opposite
may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We thus see
that there are 10

[Illustration:

    +---+---+  +-+-----+  +---+---+
    |O OHO O|  |OHO O O|  |O OHO O|
    |   H   |  | +     |  |   +=+ |
    |O OHO O|  |OHO O O|  |O OHOHO|
    +-+ +-+-+  +-+-----+  +---+ + |
    |O|O O|O|  |O|O O O|  |O O O|O|
    | +---+ |  | +-+-+ |  |   +-+ |
    |O O O O|  |O O OHO|  |O O|O O|
    +-------+  +-------+  +-------+
        2          3          4

    +-----+-+  +-+-----+  +-------+
    |O O OHO|  |OHO O O|  |O O O O|
    |   +=+ |  | +=+   |  | +=+=+=+
    |O OHO O|  |OHOHO O|  |OHOHO O|
    | +-+-+ +  | + +-+ |  + + +   |
    |O|O O|O|  |O|O O|O|  |O|OHO O|
    +=+   +=+  | +   +=+  +=+ +   |
    |O O O O|  |OHO O O|  |O O|O O|
    +-------+  +-+-----+  +---+---+
        5          6          7
           THE SIXTEEN SHEEP

]

solutions with a 9 in the corner. If, however, we substitute 8, the two
corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1,
or 1, 0. In the case of B, ten different selections may be made for the
fourth corner; but in each of the cases C, D, and E, only nine
selections are possible, because we cannot use the 9. Therefore with 8
in the top left-hand corner there are 10 + (3 × 9) = 37 different
solutions. If we then try 7 in the corner, the result will be 10 + 27 +
40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 +
27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 55 = 235 ; with
3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 =
332; with 1, the same as with 2, + 34 = 366, and with nought in the top
left-hand corner the number of solutions will be found to be 10 + 27 +
40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385. As there is no other number
to be placed in the top left-hand corner, we have now only to add these
totals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 +
385 = 2,035. We therefore find that the total number of ways in which
tenants may occupy some or all of the eight villas so that there shall
be always nine persons living along each side of the square is 2,035. Of
course, this method must obviously cover all the reversals and
reflections, since each corner in turn is occupied by every number in
all possible combinations with the other two corners that are in line
with it.

[Illustration:

       A         B         C         D         E
    +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+
    |9| |0|   |8| |0|   |8| |1|   |8| |0|   |8| |1|
    +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+
    | |*| |   | |*| |   | |*| |   | |*| |   | |*| |
    +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+
    |0| | |   |0| | |   |1| | |   |1| | |   |0| | |
    +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+   +-+-+-+

]

Here is a general formula for solving the puzzle: (n² + 3n + 2)(n² +
3n + 3)/6. Whatever may be the stipulated number of residents along
each of the sides (which number is represented by n), the total number
of different arrangements may be thus ascertained. In our particular
case the number of residents was nine. Therefore (81 + 27 + 2) × (81 +
27 + 3) and the product, divided by 6, gives 2,035. If the number of
residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total
arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365
respectively.


277.--COUNTER CROSSES.

Let us first deal with the Greek Cross. There are just eighteen forms in
which the numbers may be paired for the two arms. Here they are:--

    12978  13968  14958
    34956  24957  23967

    23958  13769  14759
    14967  24758  23768

    12589  23759  13579
    34567  14768  24568

    14569  23569  14379
    23578  14578  25368

    15369  24369  23189
    24378  15378  45167

    24179  25169  34169
    35168  34178  25178

Of course, the number in the middle is common to both arms. The first
pair is the one I gave as an example. I will suppose that we have
written out all these crosses, always placing the first row of a pair in
the upright and the second row in the horizontal arm. Now, if we leave
the central figure fixed, there are 24 ways in which the numbers in the
upright may be varied, for the four counters may be changed in 1 × 2 × 3
× 4 = 24 ways. And as the four in the horizontal may also be changed in
24 ways for every arrangement on the other arm, we find that there are
24 × 24 = 576 variations for every form; therefore, as there are 18
forms, we get 18 × 576 = 10,368 ways. But this will include half the
four reversals and half the four reflections that we barred, so we must
divide this by 4 to obtain the correct answer to the Greek Cross, which
is thus 2,592 different ways. The division is by 4 and not by 8, because
we provided against half the reversals and reflections by always
reserving one number for the upright and the other for the horizontal.

In the case of the Latin Cross, it is obvious that we have to deal with
the same 18 forms of pairing. The total number of different ways in this
case is the full number, 18 × 576. Owing to the fact that the upper and
lower arms are unequal in length, permutations will repeat by
reflection, but not by reversal, for we cannot reverse. Therefore this
fact only entails division by 2. But in every pair we may exchange the
figures in the upright with those in the horizontal (which we could not
do in the case of the Greek Cross, as the arms are there all alike);
consequently we must multiply by 2. This multiplication by 2 and
division by 2 cancel one another. Hence 10,368 is here the correct
answer.


278.--A DORMITORY PUZZLE.

[Illustration:

         MON.           TUES.            WED.
    +---+---+---+   +---+---+---+   +---+---+---+
    | 1 | 2 | 1 |   | 1 | 3 | 1 |   | 1 | 4 | 1 |
    +---+---+---+   +---+---+---+   +---+---+---+
    | 2 |   | 2 |   | 1 |   | 1 |   | 1 |   | 1 |
    +---+---+---+   +---+---+---+   +---+---+---+
    | 1 | 22| 1 |   | 3 | 19| 3 |   | 4 | 16| 4 |
    +---+---+---+   +---+---+---+   +---+---+---+

        THURS.           FRI.            SAT.
    +---+---+---+   +---+---+---+   +---+---+---+
    | 1 | 5 | 1 |   | 2 | 6 | 2 |   | 4 | 4 | 4 |
    +---+---+---+   +---+---+---+   +---+---+---+
    | 2 |   | 2 |   | 1 |   | 1 |   | 4 |   | 4 |
    +---+---+---+   +---+---+---+   +---+---+---+
    | 4 | 13| 4 |   | 7 | 6 | 7 |   | 4 | 4 | 4 |
    +---+---+---+   +---+---+---+   +---+---+---+

]

Arrange the nuns from day to day as shown in the six diagrams. The
smallest possible number of nuns would be thirty-two, and the
arrangements on the last three days admit of variation.


279.--THE BARRELS OF BALSAM.

This is quite easy to solve for any number of barrels--if you know how.
This is the way to do it. There are five barrels in each row Multiply
the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10
together. Divide one result by the other, and we get the number of
different combinations or selections of ten things taken five at a time.
This is here 252. Now, if we divide this by 6 (1 more than the number in
the row) we get 42, which is the correct answer to the puzzle, for there
are 42 different ways of arranging the barrels. Try this method of
solution in the case of six barrels, three in each row, and you will
find the answer is 5 ways. If you check this by trial, you will discover
the five arrangements with 123, 124, 125, 134, 135 respectively in the
top row, and you will find no others.

The general solution to the problem is, in fact, this:

      n
     C
      2n
    -----
    n + 1

where 2n equals the number of barrels. The symbol C, of course, implies
that we have to find how many combinations, or selections, we can make
of 2n things, taken n at a time.


280.--BUILDING THE TETRAHEDRON.

Take your constructed pyramid and hold it so that one stick only lies on
the table. Now, four sticks must branch off from it in different
directions--two at each end. Any one of the five sticks may be left out
of this connection; therefore the four may be selected in 5 different
ways. But these four matches may be placed in 24 different orders. And
as any match may be joined at either of its ends, they may further be
varied (after their situations are settled for any particular
arrangement) in 16 different ways. In every arrangement the sixth stick
may be added in 2 different ways. Now multiply these results together,
and we get 5 × 24 × 16 × 2 = 3,840 as the exact number of ways in which
the pyramid may be constructed. This method excludes all possibility of
error.

A common cause of error is this. If you calculate your combinations by
working upwards from a basic triangle lying on the table, you will get
half the correct number of ways, because you overlook the fact that an
equal number of pyramids may be built on that triangle downwards, so to
speak, through the table. They are, in fact, reflections of the others,
and examples from the two sets of pyramids cannot be set up to resemble
one another--except under fourth dimensional conditions!


281.--PAINTING A PYRAMID.

It will be convenient to imagine that we are painting our pyramids on
the flat cardboard, as in the diagrams, before folding up. Now, if we
take any _four_ colours (say red, blue, green, and yellow), they may be
applied in only 2 distinctive ways, as shown in Figs, 1 and 2. Any other
way will only result in one of these when the pyramids are folded up. If
we take any _three_ colours, they may be applied in the 3 ways shown in
Figs. 3, 4, and 5. If we take any _two_ colours, they may be applied in
the 3 ways shown in Figs. 6, 7, and 8. If we take any _single_ colour,
it may obviously be applied in only 1 way. But four colours may be
selected in 35 ways out of seven; three in 35 ways; two in 21 ways; and
one colour in 7 ways. Therefore 35 applied in 2 ways = 70; 35 in 3 ways
= 105; 21 in 3 ways = 63; and 7 in 1 way = 7. Consequently the pyramid
may be painted in 245 different ways (70 + 105 + 63 + 7), using the
seven colours of the solar spectrum in accordance with the conditions of
the puzzle.

[Illustration:

                      1                   2
              +---------------+   +---------------+
               \  R  / \  B  /     \  B  / \  R  /
                \   /   \   /       \   /   \   /
                 \ /  G  \ /         \ /  G  \ /
                  \-------/           \-------/
                   \     /             \     /
                    \ Y /               \ Y /
                     \ /                 \ /
                      '                   '

            3                   4                   5
    +---------------+   +---------------+   +---------------+
     \  R  / \  R  /     \  R  / \  G  /     \  Y  / \  R  /
      \   /   \   /       \   /   \   /       \   /   \   /
       \ /  G  \ /         \ /  G  \ /         \ /  G  \ /
        \-------/           \-------/           \-------/
         \     /             \     /             \     /
          \ Y /               \ Y /               \ Y /
           \ /                 \ /                 \ /
            '                   '                   '

            6                   7                   8
    +---------------+   +---------------+   +---------------+
     \  G  / \  Y  /     \  Y  / \  Y  /     \  G  / \  G  /
      \   /   \   /       \   /   \   /       \   /   \   /
       \ /  G  \ /         \ /  G  \ /         \ /  G  \ /
        \-------/           \-------/           \-------/
         \     /             \     /             \     /
          \ Y /               \ Y /               \ Y /
           \ /                 \ /                 \ /
            '                   '                   '

]


282.--THE ANTIQUARY'S CHAIN.

[Illustration]

THE number of ways in which nine things may be arranged in a row without
any restrictions is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362,880. But we
are told that the two circular rings must never be together; therefore
we must deduct the number of times that this would occur. The number is
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40,320 × 2 = 80,640, because if we
consider the two circular links to be inseparably joined together they
become as one link, and eight links are capable of 40,320 arrangements;
but as these two links may always be put on in the orders AB or BA, we
have to double this number, it being a question of arrangement and not
of design. The deduction required reduces our total to 282,240. Then one
of our links is of a peculiar form, like an 8. We have therefore the
option of joining on either one end or the other on every occasion, so
we must double the last result. This brings up our total to 564,480.

We now come to the point to which I directed the reader's
attention--that every link may be put on in one of two ways. If we join
the first finger and thumb of our left hand horizontally, and then link
the first finger and thumb of the right hand, we see that the right
thumb may be either above or below. But in the case of our chain we must
remember that although that 8-shaped link has two independent _ends_ it
is like every other link in having only two _sides_--that is, you cannot
turn over one end without turning the other at the same time.

We will, for convenience, assume that each link has a black side and a
side painted white. Now, if it were stipulated that (with the chain
lying on the table, and every successive link falling over its
predecessor in the same way, as in the diagram) only the white sides
should be uppermost as in A, then the answer would be 564,480, as
above--ignoring for the present all reversals of the completed chain.
If, however, the first link were allowed to be placed either side up,
then we could have either A or B, and the answer would be 2 × 564,480 =
1,128,960; if two links might be placed either way up, the answer would
be 4 × 564,480; if three links, then 8 × 564,480, and so on. Since,
therefore, every link may be placed either side up, the number will be
564,480 multiplied by 2^9, or by 512. This raises our total to
289,013,760.

But there is still one more point to be considered. We have not yet
allowed for the fact that with any given arrangement three of the other
arrangements may be obtained by simply turning the chain over through
its entire length and by reversing the ends. Thus C is really the same
as A, and if we turn this page upside down, then A and C give two other
arrangements that are still really identical. Thus to get the correct
answer to the puzzle we must divide our last total by 4, when we find
that there are just 72,253,440 different ways in which the smith might
have put those links together. In other words, if the nine links had
originally formed a piece of chain, and it was known that the two
circular links were separated, then it would be 72,253,439 chances to 1
that the smith would not have put the links together again precisely as
they were arranged before!


283.--THE FIFTEEN DOMINOES.

The reader may have noticed that at each end of the line I give is a
four, so that, if we like, we can form a ring instead of a line. It can
easily be proved that this must always be so. Every line arrangement
will make a circular arrangement if we like to join the ends. Now,
curious as it may at first appear, the following diagram exactly
represents the conditions when we leave the doubles out of the question
and devote our attention to forming circular arrangements. Each number,
or half domino, is in line with every other number, so that if we start
at any one of the five numbers and go over all the lines of the pentagon
once and once only we shall come back to the starting place, and the
order of our route will give us one of the circular arrangements for the
ten dominoes. Take your pencil and follow out the following route,
starting at the 4: 41304210234. You have been over all the lines once
only, and by repeating all these figures in this way,
41--13--30--04--42--21--10--02--23--34, you get an arrangement of the
dominoes (without the doubles) which will be perfectly clear. Take other
routes and you will get other arrangements. If, therefore, we can
ascertain just how many of these circular routes are obtainable from
the pentagon, then the rest is very easy.

Well, the number of different circular routes over the pentagon is 264.
How I arrive at these figures I will not at present explain, because it
would take a lot of space. The dominoes may, therefore, be arranged in a
circle in just 264 different ways, leaving out the doubles. Now, in any
one of these circles the five doubles may be inserted in 2^5 = 32
different ways. Therefore when we include the doubles there are 264 × 32
= 8,448 different circular arrangements. But each of those circles may
be broken (so as to form our straight line) in any one of 15 different
places. Consequently, 8,448 × 15 gives 126,720 different ways as the
correct answer to the puzzle.

[Illustration:

                -----
               |     |
            /  |     |  \
           /    -----    \
          /     .  .      \
     -----     .    .       -----
    |     |   .      .     | o o |
    |  o  | -.--------.--- |     |
    |     | .          . . | o o |
     ----- .  .        ..   -----
        \ .     .    .   .  /
       -----      ..     -----
      |  o  |    .  .   |o    |
      |     | --------- |  o  |
      |  o  |.         .|    o|
       -----             -----

]

I purposely refrained from asking the reader to discover in just how
many different ways the full set of twenty-eight dominoes may be
arranged in a straight line in accordance with the ordinary rules of the
game, left to right and right to left of any arrangement counting as
different ways. It is an exceedingly difficult problem, but the correct
answer is 7,959,229,931,520 ways. The method of solving is very complex.


284.--THE CROSS TARGET.

[Illustration:

             --  --
            (CD)(  )
             --  --
            (AE)(A )
     --  --  --  --  --  --
    (CE)(E )(A )(AB)(C )(D )
     --  --  --  --  --  --
    (D )(  )(B )(E )(EB)(  )
     --  --  --  --  --  --
            (C )(B )
             --  --
            (  )(ED)
             --  --


]

Twenty-one different squares may be selected. Of these nine will be of
the size shown by the four A's in the diagram, four of the size shown by
the B's, four of the size shown by the C's, two of the size shown by the
D's, and two of the size indicated by the upper single A, the upper
single E, the lower single C, and the EB. It is an interesting fact that
you cannot form any one of these twenty-one squares without using at
least one of the six circles marked E.


285.--THE FOUR POSTAGE STAMPS.

Referring to the original diagram, the four stamps may be given in the
shape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways;
in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, in
twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways;
in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, in
fourteen ways. Thus there are sixty-five ways in all.


286.--PAINTING THE DIE.

The 1 can be marked on any one of six different sides. For every side
occupied by 1 we have a selection of four sides for the 2. For every
situation of the 2 we have two places for the 3. (The 6, 5, and 4 need
not be considered, as their positions are determined by the 1, 2, and
3.) Therefore 6, 4, and 2 multiplied together make 48 different
ways--the correct answer.


287.--AN ACROSTIC PUZZLE.

There are twenty-six letters in the alphabet, giving 325 different
pairs. Every one of these pairs may be reversed, making 650 ways. But
every initial letter may be repeated as the final, producing 26 other
ways. The total is therefore 676 different pairs. In other words, the
answer is the square of the number of letters in the alphabet.


288.--CHEQUERED BOARD DIVISIONS.

There are 255 different ways of cutting the board into two pieces of
exactly the same size and shape. Every way must involve one of the five
cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by
reversal and reflection, we need only consider cuts that enter at the
points a, b, and c. But the exit must always be at a point in a straight
line from the entry through the centre. This is the most important
condition to remember. In case B you cannot enter at a, or you will get
the cut provided for in E. Similarly in C or D, you must not enter the
key-line in the same direction as itself, or you will get A or B. If you
are working on A or C and entering at a, you must consider joins at one
end only of the key-line, or you will get repetitions. In other cases
you must consider joins at both ends of the key; but after leaving a in
case D, turn always either to right or left--use one direction only.
Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and
6 come under C;

[Illustration]

and 7 is a pretty example of D. Of course, E is a peculiar type, and
obviously admits of only one way of cutting, for you clearly cannot
enter at b or c.

Here is a table of the results:--


         a    b    c   Ways.
    A =  8 + 17 + 21 =  46
    B =  0 + 17 + 21 =  38
    C = 15 + 31 + 39 =  85
    D = 17 + 29 + 39 =  85
    E =  1 +  0 +  0 =   1
        --   --   --   ---
        41  94   120   255

I have not attempted the task of enumerating the ways of dividing a
board 8 × 8--that is, an ordinary chessboard. Whatever the method
adopted, the solution would entail considerable labour.


289.--LIONS AND CROWNS.

[Illustration]

Here is the solution. It will be seen that each of the four pieces
(after making the cuts along the thick lines) is of exactly the same
size and shape, and that each piece contains a lion and a crown. Two of
the pieces are shaded so as to make the solution quite clear to the eye.


290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

There are fifteen different ways of cutting the 5 × 5 board (with the
central square removed) into two pieces of the same size and shape.
Limitations of space will not allow me to give diagrams of all these,
but I will enable the reader to draw them all out for himself without
the slightest difficulty. At whatever point on the edge your cut enters,
it must always end at a point on the edge, exactly opposite in a line
through the centre of the square. Thus, if you enter at point 1 (see
Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and
2 are the only two really different points of entry; if we use any
others they will simply produce similar solutions. The directions of the
cuts in the following fifteen

[Illustration: Fig. 1. Fig. 2.]

solutions are indicated by the numbers on the diagram. The duplication
of the numbers can lead to no confusion, since every successive number
is contiguous to the previous one. But whichever direction you take from
the top downwards you must repeat from the bottom upwards, one direction
being an exact reflection of the other.

    1, 4, 8.
    1, 4, 3, 7, 8.
    1, 4, 3, 7, 10, 9.
    1, 4, 3, 7, 10, 6, 5, 9.
    1, 4, 5, 9.
    1, 4, 5, 6, 10, 9.
    1, 4, 5, 6, 10, 7, 8.
    2, 3, 4, 8.
    2, 3, 4, 5, 9.
    2, 3, 4, 5, 6, 10, 9.
    2, 3, 4, 5, 6, 10, 7, 8.
    2, 3, 7, 8.
    2, 3, 7, 10, 9.
    2, 3, 7, 10, 6, 5, 9.
    2, 3, 7, 10, 6, 5, 4, 8.

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9)
produces the solution shown in Fig. 2. The thirteenth produces the
solution given in propounding the puzzle, where the cut entered at the
side instead of at the top. The pieces, however, will be of the same
shape if turned over, which, as it was stated in the conditions, would
not constitute a different solution.


291.--THE GRAND LAMA'S PROBLEM.

The method of dividing the chessboard so that each of the four parts
shall be of exactly the same size and shape, and contain one of the
gems, is shown in the diagram. The method of shading the squares is
adopted to make the shape of the pieces clear to the eye. Two of the
pieces are shaded and two left white.

The reader may find it interesting to compare this puzzle with that of
the "Weaver" (No. 14, _Canterbury Puzzles_).

[Illustration: THE GRAND LAMA'S PROBLEM.

    +===+===+===+===+===+===+===+===+
    |:o:|   :   :   :   :   :   :   :
    I...I...+===+===+===+===+===+===+
    |:::| o |:::::::::::::::::::::::|
    I...I...I...+===+===+===+===+...I
    |:::|   |:o:|   :   :   :   |:::|
    I...I...I...I...I===+===+...I...I
    |:::|   |:::| o |:::::::|   |:::|
    I...I...I...+===I===+...I...I...I
    |:::|   |:::::::|   |:::|   |:::|
    I...I...+===+===+...+...I...I...I
    |:::|   :   :   :   |:::|   |:::|
    I...+===+===+===+===I...I...I...I
    |:::::::::::::::::::::::|   |:::|
    +===+===+===+===+===+===+...I...I
    |   :   :   :   :   :   :   |:::|
    +===+===+===+===+===+===+===+===+

]


292.--THE ABBOT'S WINDOW.

THE man who was "learned in strange mysteries" pointed out to Father
John that the orders of the Lord Abbot of St. Edmondsbury might be
easily carried out by blocking up twelve of the lights in the window as
shown by the dark squares in the following sketch:--


[Illustration:

    +===+===+===+===+===+===+===+===+
    |   :   :   :   :   :   :   :   |
    I...+===+...+...+...+...+===+...I
    |   IIIII   :   :   :   IIIII   |
    I...+===+===+...+...+===+===+...I
    |   :   IIIII   :   IIIII   :   |
    I...+...+===+===+===+===+...+...I
    |   :   :   IIIIIIIII   :   :   |
    I...+...+...+===+===+...+...+...I
    |   :   :   IIIIIIIII   :   :   |
    I...+...+===+===+===+===+...+...I
    |   :   IIIII   :   IIIII   :   |
    I...+===+===+...+...+===+===+...I
    |   IIIII   :   :   :   IIIII   |
    I...+===+...+...+...+...+===+...I
    |   :   :   :   :   :   :   :   |
    +===+===+===+===+===+===+===+===+

]

Father John held that the four corners should also be darkened, but the
sage explained that it was desired to obstruct no more light than was
absolutely necessary, and he said, anticipating Lord Dundreary, "A
single pane can no more be in a _line_ with itself than one bird can go
into a corner and flock in solitude. The Abbot's condition was that no
diagonal _lines_ should contain an odd number of lights."

Now, when the holy man saw what had been done he was well pleased, and
said, "Truly, Father John, thou art a man of deep wisdom, in that thou
hast done that which seemed impossible, and yet withal adorned our
window with a device of the cross of St. Andrew, whose name I received
from my godfathers and godmothers." Thereafter he slept well and arose
refreshed. The window might be seen intact to-day in the monastery of
St. Edmondsbury, if it existed, which, alas! the window does not.


293.--THE CHINESE CHESSBOARD.

    +===I===+===+===+===I===+===+===+
    |   |:::: 2 ::::| 3 |:::| 5 |:6:|
    I...+===+...+===+...I...I...+===I
    |:::: 1 |:::|   ::::| 4 |:::| 7 |
    I...+===+===+...I===I...I===+===I
    |   |::::   |:::|   ::::| 9 |:::|
    I===I...I===============I...I...I
    |:::: 11|::::   ::::: 10|:::| 8 |
    I=======I===I===========I...I...I
    |   ::::: 12|:::: 13::::|   |:::|
    I=======+...I...+===+===|===+===I
    |:::: 14|:::|   |:::| 16::::| 17|
    I...+...I===I===+...+...+===+...I
    |   ::::|   ::::: 15|:::|   ::::|
    I=======+===========+===+=======I
    |::::   ::::: 18:::::   :::::   |
    +===+===+===+===+===+===+===+===+

    +===+===I===I===+===I===+===+===+
    |   ::::|   |::::   |:::|   ::::|
    I...+===I...I=======I...I===+...I
    |:::|   |::::   |::::   |:::|   |
    I...I===I===============I===I...I
    |   |::::   ::::|   :::::   |:::|
    I===I=======I=======I=======I===I
    |:::|   ::::|   ::::|   ::::|   |
    I...I===+...I...+...I...+===+...I
    |   ::::|   |::::   |:::|   ::::|
    I...+===I...+===I===+...I===+...I
    |:::|   |::::   |::::   |:::|   |
    I===I...+=======I=======+...I===I
    |   |::::   ::::|   :::::   |:::|
    I...+=======+...I...+=======+...I
    |::::   ::::|   |:::|   :::::   |
    +===+===+===+===+===+===+===+===+


Eighteen is the maximum number of pieces. I give two solutions. The
numbered diagram is so cut that the eighteenth piece has the largest
area--eight squares--that is possible under the conditions. The second
diagram was prepared under the added condition that no piece should
contain more than five squares.

No. 74 in _The Canterbury Puzzles_ shows how to cut the board into
twelve pieces, all different, each containing five squares, with one
square piece of four squares.


294.--THE CHESSBOARD SENTENCE.

    +===I===I===I===I=======I=======+
    |   |:::|   |:::|   ::::|   ::::|
    I===I...I===I...I...+===I...+===I
    |:::|   :::::   |:::|   :::::   |
    |...|...+===I...I...+===+...+===I
    |   |:::|   |:::|   ::::|   ::::|
    |...+===+...+===I===I===I=======I
    |::::   :::::   |:::|   :::::   |
    I===========I===I...I===I===+...|
    |   :::::   |:::|   |:::|   |:::|
    |...+===+...|...|...|...I===+...|
    |:::|   |:::|   |:::|   |::::   |
    |...|...|...|...I===+...+===+...|
    |   |:::|   |:::|   :::::   |:::|
    I===+...+===I...+=======I===+...|
    |::::   ::::|   :::::   |::::   |
    +===========I===================+

The pieces may be fitted together, as shown in the illustration, to form
a perfect chessboard.


295.--THE EIGHT ROOKS.

Obviously there must be a rook in every row and every column. Starting
with the top row, it is clear that we may put our first rook on any one
of eight different squares. Wherever it is placed, we have the option of
seven squares for the second rook in the second row. Then we have six
squares from which to select the third row, five in the fourth, and so
on. Therefore the number of our different ways must be 8 × 7 × 6 × 5 × 4
× 3 × 2 × 1 = 40,320 (that is 8!), which is the correct answer.

How many ways there are if mere reversals and reflections are not
counted as different has not yet been determined; it is a difficult
problem. But this point, on a smaller square, is considered in the next
puzzle.


296.--THE FOUR LIONS.


There are only seven different ways under the conditions. They are as
follows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1 3.
Taking the last example, this notation means that we place a lion in the
second square of first row, fourth square of second row, first square of
third row, and third square of fourth row. The first example is, of
course, the one we gave when setting the puzzle.


297.--BISHOPS--UNGUARDED.

    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+...+...+
    ::B:: B ::B:: B ::B:: B ::B:: B :
    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+...+...+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+...+...+


This cannot be done with fewer bishops than eight, and the simplest
solution is to place the bishops in line along the fourth or fifth row
of the board (see diagram). But it will be noticed that no bishop is
here guarded by another, so we consider that point in the next puzzle.


298.--BISHOPS--GUARDED.

    +...+...+...+...+.......+.......+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+...+...+.......+
    ::::: B ::B:: B ::::: B ::B::   :
    +...........+...+...+...+...+...+
    :   ::B:: B ::B::   ::B:: B :::::
    +...+...+...+...+...+...+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+...+...+...+...+
    :   :::::   :::::   :::::   :::::
    +...+...+...+...+.......+...+...+
    :::::   :::::   :::::   :::::   :
    +...+...+...+...+.......+...+...+

This puzzle is quite easy if you first of all give it a little thought.
You need only consider squares of one colour, for whatever can be done
in the case of the white squares can always be repeated on the black,
and they are here quite independent of one another. This equality, of
course, is in consequence of the fact that the number of squares on an
ordinary chessboard, sixty-four, is an even number. If a square
chequered board has an odd number of squares, then there will always be
one more square of one colour than of the other.

Ten bishops are necessary in order that every square shall be attacked
and every bishop guarded by another bishop. I give one way of arranging
them in the diagram. It will be noticed that the two central bishops in
the group of six on the left-hand side of the board serve no purpose,
except to protect those bishops that are on adjoining squares. Another
solution would therefore be obtained by simply raising the upper one of
these one square and placing the other a square lower down.


299.--BISHOPS IN CONVOCATION.

The fourteen bishops may be placed in 256 different ways. But every
bishop must always be placed on one of the sides of the board--that
is, somewhere on a row or file on the extreme edge. The puzzle,
therefore, consists in counting the number of different ways that we
can arrange the fourteen round the edge of the board without attack.
This is not a difficult matter. On a chessboard of n² squares 2n - 2
bishops (the maximum number) may always be placed in 2^n ways without
attacking. On an ordinary chessboard n would be 8; therefore 14
bishops may be placed in 256 different ways. It is rather curious that
the general result should come out in so simple a form.

[Illustration]


300.--THE EIGHT QUEENS.

[Illustration]

The solution to this puzzle is shown in the diagram. It will be found
that no queen attacks another, and also that no three queens are in a
straight line in any oblique direction. This is the only arrangement out
of the twelve fundamentally different ways of placing eight queens
without attack that fulfils the last condition.


301.--THE EIGHT STARS.

The solution of this puzzle is shown in the first diagram. It is the
only possible solution within the conditions stated. But if one of the
eight stars had not already been placed as shown, there would then have
been eight ways of arranging the stars according to this scheme, if we
count reversals and reflections as different. If you turn this page
round so that each side is in turn at the bottom, you will get the four
reversals; and if you reflect each of these in a mirror, you will get
the four reflections. These are, therefore, merely eight aspects of one
"fundamental solution." But without that first star being so placed,
there is another fundamental solution, as shown in the second diagram.
But this arrangement being in a way symmetrical, only produces four
different aspects by reversal and reflection.

[Illustration]


302.--A PROBLEM IN MOSAICS.

[Illustration]

The diagram shows how the tiles may be rearranged. As before, one yellow
and one purple tile are dispensed with. I will here point out that in
the previous arrangement the yellow and purple tiles in the seventh row
might have changed places, but no other arrangement was possible.


303.--UNDER THE VEIL.

Some schemes give more diagonal readings of four letters than others,
and we are at first tempted to favour these; but this is a false scent,
because what you appear to gain in this direction you lose in others. Of
course it immediately occurs to the solver that every LIVE or EVIL is
worth twice as much as any other word, since it reads both ways and
always counts as 2. This is an important consideration, though sometimes
those arrangements that contain most readings of these two words are
fruitless in other words, and we lose in the general count.

[Illustration:

     _ _ I V E L _ _
     E V L _ _ I _ _
     L _ _ I _ _ V E
     I _ V E _ _ _ L
     _ E _ _ L V _ I
     _ L I _ _ I _ E V
    /V _ E L _ _ I _
    _ I _ _ V E L _\

]

The above diagram is in accordance with the conditions requiring no
letter to be in line with another similar letter, and it gives twenty
readings of the five words--six horizontally, six vertically, four in
the diagonals indicated by the arrows on the left, and four in the
diagonals indicated by the arrows on the right. This is the maximum.

Four sets of eight letters may be placed on the board of sixty-four
squares in as many as 604 different ways, without any letter ever being
in line with a similar one. This does not count reversals and
reflections as different, and it does not take into consideration the
actual permutations of the letters among themselves; that is, for
example, making the L's change places with the E's. Now it is a singular
fact that not only do the twenty word-readings that I have given prove
to be the real maximum, but there is actually only that one arrangement
from which this maximum may be obtained. But if you make the V's change
places with the I's, and the L's with the E's, in the solution given,
you still get twenty readings--the same number as before in every
direction. Therefore there are two ways of getting the maximum from the
same arrangement. The minimum number of readings is zero--that is, the
letters can be so arranged that no word can be read in any of the
directions.


304.--BACHET'S SQUARE.

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

Let us use the letters A, K, Q, J, to denote ace, king, queen, jack; and
D, S, H, C, to denote diamonds, spades, hearts, clubs. In Diagrams 1
and 2 we have the two available ways of arranging either group of
letters so that no two similar letters shall be in line--though a
quarter-turn of 1 will give us the arrangement in 2. If we superimpose
or combine these two squares, we get the arrangement of Diagram 3, which
is one solution. But in each square we may put the letters in the top
line in twenty-four different ways without altering the scheme of
arrangement. Thus, in Diagram 4 the S's are similarly placed to the D's
in 2, the H's to the S's, the C's to the H's, and the D's to the C's. It
clearly follows that there must be 24×24 = 576 ways of combining the two
primitive arrangements. But the error that Labosne fell into was that of
assuming that the A, K, Q, J must be arranged in the form 1, and the D,
S, H, C in the form 2. He thus included reflections and half-turns, but
not quarter-turns. They may obviously be interchanged. So that the
correct answer is 2 × 576 = 1,152, counting reflections and reversals as
different. Put in another manner, the pairs in the top row may be
written in 16 × 9 × 4 × 1 = 576 different ways, and the square then
completed in 2 ways, making 1,152 ways in all.


305.--THE THIRTY-SIX LETTER BLOCKS.

I pointed out that it was impossible to get all the letters into the box
under the conditions, but the puzzle was to place as many as possible.

This requires a little judgment and careful investigation, or we are
liable to jump to the hasty conclusion that the proper way to solve the
puzzle must be first to place all six of one letter, then all six of
another letter, and so on. As there is only one scheme (with its
reversals) for placing six similar letters so that no two shall be in a
line in any direction, the reader will find that after he has placed
four different kinds of letters, six times each, every place is occupied
except those twelve that form the two long diagonals. He is, therefore,
unable to place more than two each of his last two letters, and there
are eight blanks left. I give such an arrangement in Diagram 1.

[Illustration: 1]

[Illustration: 2]

The secret, however, consists in not trying thus to place all six of
each letter. It will be found that if we content ourselves with placing
only five of each letter, this number (thirty in all) may be got into
the box, and there will be only six blanks. But the correct solution is
to place six of each of two letters and five of each of the remaining
four. An examination of Diagram 2 will show that there are six each of C
and D, and five each of A, B, E, and F. There are, therefore, only four
blanks left, and no letter is in line with a similar letter in any
direction.


306.--THE CROWDED CHESSBOARD.

[Illustration]

Here is the solution. Only 8 queens or 8 rooks can be placed on the
board without attack, while the greatest number of bishops is 14, and of
knights 32. But as all these knights must be placed on squares of the
same colour, while the queens occupy four of each colour and the bishops
7 of each colour, it follows that only 21 knights can be placed on the
same colour in this puzzle. More than 21 knights can be placed alone on
the board if we use both colours, but I have not succeeded in placing
more than 21 on the "crowded chessboard." I believe the above solution
contains the maximum number of pieces, but possibly some ingenious
reader may succeed in getting in another knight.


307.--THE COLOURED COUNTERS.

The counters may be arranged in this order:--

    R1,  B2,  Y3,  O4,  GS.
    Y4,  O5,  G1,  R2,  B3.
    G2,  R3,  B4,  Y5,  O1.
    B5,  Y1,  O2,  G3,  R4.
    O3,  G4,  R5,  B1,  Y2.


308.--THE GENTLE ART OF STAMP-LICKING.

The following arrangement shows how sixteen stamps may be stuck on the
card, under the conditions, of a total value of fifty pence, or 4s.
2d.:--

[Illustration]

If, after placing the four 5d. stamps, the reader is tempted to place
four 4d. stamps also, he can afterwards only place two of each of the
three other denominations, thus losing two spaces and counting no more
than forty-eight pence, or 4s. This is the pitfall that was hinted at.
(Compare with No. 43, _Canterbury Puzzles_.)


309.--THE FORTY-NINE COUNTERS.

The counters may be arranged in this order:--

    A1,  B2,  C3,  D4,  E5,  F6,  G7.
    F4,  G5,  A6,  B7,  C1,  D2,  E3.
    D7,  E1,  F2,  G3,  A4,  B5,  C6.
    B3,  C4,  D5,  E6,  F7,  G1,  A2.
    G6,  A7,  B1,  C2,  D3,  E4,  F5.
    E2,  F3,  G4,  A5,  B6,  C7,  D1.
    C5,  D6,  E7,  F1,  G2,  A3,  B4.


310.--THE THREE SHEEP.

The number of different ways in which the three sheep may be placed so
that every pen shall always be either occupied or in line with at least
one sheep is forty-seven.

The following table, if used with the key in Diagram 1, will enable the
reader to place them in all these ways:--

    +------------+---------------------------+----------+
    |            |                           |  No. of  |
    | Two Sheep. |       Third Sheep.        |   Ways.  |
    +------------+---------------------------+----------+
    | A and B    | C, E, G, K, L, N, or P    |    7     |
    | A and C    | I, J, K, or O             |    4     |
    | A and D    | M, N, or J                |    3     |
    | A and F    | J, K, L, or P             |    4     |
    | A and G    | H, J, K, N, O, or P       |    6     |
    | A and H    | K, L, N, or O             |    4     |
    | A and O    | K or L                    |    2     |
    | B and C    | N                         |    1     |
    | B and E    | F, H, K, or L             |    4     |
    | B and F    | G, J, N, or O             |    4     |
    | B and G    | K, L, or N                |    3     |
    | B and H    | J or N                    |    2     |
    | B and J    | K or L                    |    2     |
    | F and G    | J                         |    1     |
    |            |                           |   ----   |
    |            |                           |    47    |
    +------------+---------------------------+----------+

This, of course, means that if you place sheep in the pens marked A and
B, then there are seven different pens in which you may place the third
sheep, giving seven different solutions. It was understood that
reversals and reflections do not count as different.

If one pen at least is to be _not_ in line with a sheep, there would be
thirty solutions to that problem. If we counted all the reversals and
reflections of these 47 and 30 cases respectively as different, their
total would be 560, which is the number of different ways in which the
sheep may be placed in three pens without any conditions. I will remark
that there are three ways in which two sheep may be placed so that every
pen is occupied or in line, as in Diagrams 2, 3, and 4, but in every
case each sheep is in line with its companion. There are only two ways
in which three sheep may be so placed that every pen shall be occupied
or in line, but no sheep in line with another. These I show in Diagrams
5 and 6. Finally, there is only one way in which three sheep may be
placed so that at least one pen shall not be in line with a sheep and
yet no sheep in line with another. Place the sheep in C, E, L. This is
practically all there is to be said on this pleasant pastoral subject.

[Illustration]


311.--THE FIVE DOGS PUZZLE.

The diagrams show four fundamentally different solutions. In the case of
A we can reverse the order, so that the single dog is in the bottom row
and the other four shifted up two squares. Also we may use the next
column to the right and both of the two central horizontal rows. Thus A
gives 8 solutions. Then B may be reversed and placed in either diagonal,
giving 4 solutions. Similarly C will give 4 solutions. The line in D
being symmetrical, its reversal will not be different, but it may be
disposed in 4 different directions. We thus have in all 20 different
solutions.

[Illustration]


312.--THE FIVE CRESCENTS OF BYZANTIUM.

[Illustration]

If that ancient architect had arranged his five crescent tiles in the
manner shown in the following diagram, every tile would have been
watched over by, or in a line with, at least one crescent, and space
would have been reserved for a perfectly square carpet equal in area to
exactly half of the pavement. It is a very curious fact that, although
there are two or three solutions allowing a carpet to be laid down
within the conditions so as to cover an area of nearly twenty-nine of
the tiles, this is the only possible solution giving exactly half the
area of the pavement, which is the largest space obtainable.


313.--QUEENS AND BISHOP PUZZLE.

[Illustration: FIG. 1.]

[Illustration: FIG. 2.]

The bishop is on the square originally occupied by the rook, and the
four queens are so placed that every square is either occupied or
attacked by a piece. (Fig. 1.)

I pointed out in 1899 that if four queens are placed as shown in the
diagram (Fig. 2), then the fifth queen may be placed on any one of the
twelve squares marked a, b, c, d, and e; or a rook on the two squares,
c; or a bishop on the eight squares, a, b, and e; or a pawn on the
square b; or a king on the four squares, b, c, and e. The only known
arrangement for four queens and a knight is that given by Mr. J. Wallis
in _The Strand Magazine_ for August 1908, here reproduced. (Fig. 3.)

[Illustration: FIG. 3.]

I have recorded a large number of solutions with four queens and a rook,
or bishop, but the only arrangement, I believe, with three queens and
two rooks in which all the pieces are guarded is that of which I give an
illustration (Fig. 4), first published by Dr. C. Planck. But I have
since found the accompanying solution with three queens, a rook, and a
bishop, though the pieces do not protect one another. (Fig. 5.)

[Illustration: FIG. 4.]

[Illustration: FIG. 5.]


314.--THE SOUTHERN CROSS.

My readers have been so familiarized with the fact that it requires at
least five planets to attack every one of a square arrangement of
sixty-four stars that many of them have, perhaps, got to believe that a
larger square arrangement of stars must need an increase of planets. It
was to correct this possible error of reasoning, and so warn readers
against another of those numerous little pitfalls in the world of
puzzledom, that I devised this new stellar problem. Let me then state at
once that, in the case of a square arrangement of eighty one stars,
there are several ways of placing five planets so that every star shall
be in line with at least one planet vertically, horizontally, or
diagonally. Here is the solution to the "Southern Cross": --

It will be remembered that I said that the five planets in their new
positions "will, of course, obscure five other stars in place of those
at present covered." This was to exclude an easier solution in which
only four planets need be moved.


315.--THE HAT-PEG PUZZLE.

The moves will be made quite clear by a reference to the diagrams, which
show the position on the board after each of the four moves. The darts
indicate the successive removals that have been made. It will be seen
that at every stage all the squares are either attacked or occupied, and
that after the fourth move no queen attacks any other. In the case of
the last move the queen in the top row might also have been moved one
square farther to the left. This is, I believe, the only solution to the
puzzle.

[Illustration: 1]

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]


316.--THE AMAZONS.

It will be seen that only three queens have been removed from their
positions on the edge of the board, and that, as a consequence, eleven
squares (indicated by the black dots) are left unattacked by any queen.
I will hazard the statement that eight queens cannot be placed on the
chessboard so as to leave more than eleven squares unattacked. It is
true that we have no rigid proof of this yet, but I have entirely
convinced myself of the truth of the statement. There are at least five
different ways of arranging the queens so as to leave eleven squares
unattacked.

[Illustration]


317.--A PUZZLE WITH PAWNS.

Sixteen pawns may be placed so that no three shall be in a straight line
in any possible direction, as in the diagram. We regard, as the
conditions required, the pawns as mere points on a plane.

[Illustration]

318.--LION-HUNTING.

There are 6,480 ways of placing the man and the lion, if there are no
restrictions whatever except that they must be on different spots. This
is obvious, because the man may be placed on any one of the 81 spots,
and in every case there are 80 spots remaining for the lion; therefore
81 × 80 = 6,480. Now, if we deduct the number of ways in which the lion
and the man may be placed on the same path, the result must be the
number of ways in which they will not be on the same path. The number of
ways in which they may be in line is found without much difficulty to be
816. Consequently, 6,480 - 816 = 5,664, the required answer.

The general solution is this: 1/3n(n - 1)(3n² - n + 2). This is, of
course, equivalent to saying that if we call the number of squares on
the side of a "chessboard" n, then the formula shows the number of
ways in which two bishops may be placed without attacking one another.
Only in this case we must divide by two, because the two bishops have no
distinct individuality, and cannot produce a different solution by mere
exchange of places.


319.--THE KNIGHT-GUARDS.

[Illustration: DIAGRAM 1.]

[Illustration: DIAGRAM 2.]

The smallest possible number of knights with which this puzzle can be
solved is fourteen.

It has sometimes been assumed that there are a great many different
solutions. As a matter of fact, there are only three arrangements--not
counting mere reversals and reflections as different. Curiously enough,
nobody seems ever to have hit on the following simple proof, or to have
thought of dealing with the black and the white squares separately.

[Illustration: DIAGRAM 3.]

[Illustration: DIAGRAM 4.]

[Illustration: DIAGRAM 5.]

Seven knights can be placed on the board on white squares so as to
attack every black square in two ways only. These are shown in Diagrams
1 and 2. Note that three knights occupy the same position in both
arrangements. It is therefore clear that if we turn the board so that a
black square shall be in the top left-hand corner instead of a white,
and place the knights in exactly the same positions, we shall have two
similar ways of attacking all the white squares. I will assume the
reader has made the two last described diagrams on transparent paper,
and marked them _1a_ and _2a_. Now, by placing the transparent Diagram
_1a_ over 1 you will be able to obtain the solution in Diagram 3, by
placing _2a_ over 2 you will get Diagram 4, and by placing _2a_ over 1
you will get Diagram 5. You may now try all possible combinations of
those two pairs of diagrams, but you will only get the three
arrangements I have given, or their reversals and reflections. Therefore
these three solutions are all that exist.


320.--THE ROOK'S TOUR.

[Illustration]

The only possible minimum solutions are shown in the two diagrams, where
it will be seen that only sixteen moves are required to perform the
feat. Most people find it difficult to reduce the number of moves below
seventeen*.

[Illustration: THE ROOK'S TOUR.]


321.--THE ROOK'S JOURNEY.

[Illustration]

I show the route in the diagram. It will be seen that the tenth move
lands us at the square marked "10," and that the last move, the
twenty-first, brings us to a halt on square "21."


322.--THE LANGUISHING MAIDEN.

The dotted line shows the route in twenty-two straight paths by which
the knight may rescue the maiden. It is necessary, after entering the
first cell, immediately to return before entering another. Otherwise a
solution would not be possible. (See "The Grand Tour," p. 200.)


323.--A DUNGEON PUZZLE.

If the prisoner takes the route shown in the diagram--where for
clearness the doorways are omitted--he will succeed in visiting every
cell once, and only once, in as many as fifty-seven straight lines. No
rook's path over the chessboard can exceed this number of moves.

[Illustration: THE LANGUISHING MAIDEN]

[Illustration: A DUNGEON PUZZLE.]


324.--THE LION AND THE MAN.

First of all, the fewest possible straight lines in each case are
twenty-two, and in order that no cell may be visited twice it is
absolutely necessary that each should pass into one cell and then
immediately "visit" the one from which he started, afterwards proceeding
by way of the second available cell. In the following diagram the man's
route is indicated by the unbroken lines, and the lion's by the dotted
lines. It will be found, if the two routes are followed cell by cell
with two pencil points, that the lion and the man never meet. But there
was one little point that ought not to be overlooked--"they occasionally
got glimpses of one another." Now, if we take one route for the man and
merely reverse it for the lion, we invariably find that, going at the
same speed, they never get a glimpse of one another. But in our diagram
it will be found that the man and the lion are in the cells marked A at
the same moment, and may see one another through the open doorways;
while the same happens when they are in the two cells marked B, the
upper letters indicating the man and the lower the lion. In the first
case the lion goes straight for the man, while the man appears to
attempt to get in the rear of the lion; in the second case it looks
suspiciously like running away from one another!

[Illustration]


325.--AN EPISCOPAL VISITATION.

[Illustration]

In the diagram I show how the bishop may be made to visit every one of
his white parishes in seventeen moves. It is obvious that we must start
from one corner square and end at the one that is diagonally opposite to
it. The puzzle cannot be solved in fewer than seventeen moves.


326.--A NEW COUNTER PUZZLE.

Play as follows: 2--3, 9--4, 10--7, 3--8, 4--2, 7--5, 8--6, 5--10, 6--9,
2--5, 1--6, 6--4, 5--3, 10--8, 4--7, 3--2, 8--1, 7--10. The white
counters have now changed places with the red ones, in eighteen moves,
without breaking the conditions.


327.--A NEW BISHOP'S PUZZLE.

[Illustration: A]

[Illustration: B]

Play as follows, using the notation indicated by the numbered squares in
Diagram A:--

      White.   |   Black.   |   White.    |    Black.
    1. 18--15  |  1. 3--6   | 10. 20--10  |  10. 1--11
    2. 17--8   |  2. 4--13  | 11. 3--9    |  11. 18--12
    3. 19--14  |  3. 2--7   | 12. 10--13  |  12. 11--8
    4. 15--5   |  4. 6--16  | 13. 19--16  |  13. 2--5
    5. 8--3    |  5. 13-18  | 14. 16--1   |  14. 5--20
    6. 14--9   |  6. 7--12  | 15. 9--6    |  15. 12--15
    7. 5--10   |  7. 16-11  | 16. 13-7    |  16. 8--14
    8. 9--19   |  8. 12--2  | 17. 6--3    |  17. 15-18
    9. 10--4   |  9. 11-17  | 18. 7--2    |  18. 14--19

Diagram B shows the position after the ninth move. Bishops at 1 and 20
have not yet moved, but 2 and 19 have sallied forth and returned. In the
end, 1 and 19, 2 and 20, 3 and 17, and 4 and 18 will have exchanged
places. Note the position after the thirteenth move.

328.--THE QUEEN'S TOUR.

[Illustration]

The annexed diagram shows a second way of performing the Queen's Tour.
If you break the line at the point J and erase the shorter portion of
that line, you will have the required path solution for any J square. If
you break the line at I, you will have a non-re-entrant solution
starting from any I square. And if you break the line at G, you will
have a solution for any G square. The Queen's Tour previously given may
be similarly broken at three different places, but I seized the
opportunity of exhibiting a second tour.


329.--THE STAR PUZZLE.

The illustration explains itself. The stars are all struck out in
fourteen straight strokes, starting and ending at a white star.

[Illustration]


330.--THE YACHT RACE.

The diagram explains itself. The numbers will show the direction of the
lines in their proper order, and it will be seen that the seventh course
ends at the flag-buoy, as stipulated.

[Illustration]


331.--THE SCIENTIFIC SKATER.

In this case we go beyond the boundary of the square. Apart from that,
the moves are all queen moves. There are three or four ways in which it
can be done.

Here is one way of performing the feat:--

[Illustration]

It will be seen that the skater strikes out all the stars in one
continuous journey of fourteen straight lines, returning to the point
from which he started. To follow the skater's course in the diagram it
is necessary always to go as far as we can in a straight line before
turning.


332.--THE FORTY-NINE STARS.

The illustration shows how all the stars may be struck out in twelve
straight strokes, beginning and ending at a black star.

[Illustration]


333.--THE QUEEN'S JOURNEY.

The correct solution to this puzzle is shown in the diagram by the dark
line. The five moves indicated will take the queen the greatest distance
that it is possible for her to go in five moves, within the conditions.
The dotted line shows the route that most people suggest, but it is not
quite so long as the other. Let us assume that the distance from the
centre of any square to the centre of the next in the same horizontal or
vertical line is 2 inches, and that the queen travels from the centre of
her original square to the centre of the one at which she rests. Then
the first route will be found to exceed 67.9 inches, while the dotted
route is less than 67.8 inches. The difference is small, but it is
sufficient to settle the point as to the longer route. All other routes
are shorter still than these two.

[Illustration]


334.--ST. GEORGE AND THE DRAGON.

We select for the solution of this puzzle one of the prettiest designs
that can be formed by representing the moves of the knight by lines from
square to square. The chequering of the squares is omitted to give
greater clearness. St. George thus slays the Dragon in strict accordance
with the conditions and in the elegant manner we should expect of him.

[Illustration: St. George and the Dragon.]


335.--FARMER LAWRENCE'S CORNFIELDS.

There are numerous solutions to this little agricultural problem. The
version I give in the next column is rather curious on account of the
long parallel straight lines formed by some of the moves.

[Illustration: Farmer Lawrence's Cornfields.]


336.--THE GREYHOUND PUZZLE.

There are several interesting points involved in this question. In the
first place, if we had made no stipulation as to the positions of the
two ends of the string, it is quite impossible to form any such string
unless we begin and end in the top and bottom row of kennels. We may
begin in the top row and end in the bottom (or, of course, the reverse),
or we may begin in one of these rows and end in the same. But we can
never begin or end in one of the two central rows. Our places of
starting and ending, however, were fixed for us. Yet the first half of
our route must be confined entirely to those squares that are
distinguished in the following diagram by circles, and the second half
will therefore be confined to the squares that are not circled. The
squares reserved for the two half-strings will be seen to be symmetrical
and similar.

The next point is that the first half-string must end in one of the
central rows, and the second half-string must begin in one of these
rows. This is now obvious, because they have to link together to form
the complete string, and every square on an outside row is connected by
a knight's move with similar squares only--that is, circled or
non-circled as the case may be. The half-strings can, therefore, only be
linked in the two central rows.

[Illustration]

Now, there are just eight different first half-strings, and consequently
also eight second half-strings. We shall see that these combine to form
twelve complete strings, which is the total number that exist and the
correct solution of our puzzle. I do not propose to give all the routes
at length, but I will so far indicate them that if the reader has
dropped any he will be able to discover which they are and work them out
for himself without any difficulty. The following numbers apply to those
in the above diagram.

The eight first half-strings are: 1 to 6 (2 routes); 1 to 8 (1 route);
1 to 10 (3 routes); 1 to 12 (1 route); and 1 to 14 (1 route). The eight
second half-strings are: 7 to 20 (1 route); 9 to 20 (1 route); 11 to 20
(3 routes); 13 to 20 (1 route); and 15 to 20 (2 routes). Every different
way in which you can link one half-string to another gives a different
solution. These linkings will be found to be as follows: 6 to 13 (2
cases); 10 to 13 (3 cases); 8 to 11 (3 cases); 8 to 15 (2 cases); 12 to
9 (1 case); and 14 to 7 (1 case). There are, therefore, twelve different
linkings and twelve different answers to the puzzle. The route given in
the illustration with the greyhound will be found to consist of one of
the three half-strings 1 to 10, linked to the half-string 13 to 20. It
should be noted that ten of the solutions are produced by five
distinctive routes and their reversals--that is, if you indicate these
five routes by lines and then turn the diagrams upside down you will get
the five other routes. The remaining two solutions are symmetrical
(these are the cases where 12 to 9 and 14 to 7 are the links), and
consequently they do not produce new solutions by reversal.


337.--THE FOUR KANGAROOS.

[Illustration]

A pretty symmetrical solution to this puzzle is shown in the diagram.
Each of the four kangaroos makes his little excursion and returns to his
corner, without ever entering a square that has been visited by another
kangaroo and without crossing the central line. It will at once occur to
the reader, as a possible improvement of the puzzle, to divide the board
by a central vertical line and make the condition that this also shall
not be crossed. This would mean that each kangaroo had to confine
himself to a square 4 by 4, but it would be quite impossible, as I shall
explain in the next two puzzles.


338.--THE BOARD IN COMPARTMENTS.

[Illustration]

In attempting to solve this problem it is first necessary to take the
two distinctive compartments of twenty and twelve squares respectively
and analyse them with a view to determining where the necessary points
of entry and exit lie. In the case of the larger compartment it will be
found that to complete a tour of it we must begin and end on two of the
outside squares on the long sides. But though you may start at any one
of these ten squares, you are restricted as to those at which you can
end, or (which is the same thing) you may end at whichever of these you
like, provided you begin your tour at certain particular squares. In the
case of the smaller compartment you are compelled to begin and end at
one of the six squares lying at the two narrow ends of the compartments,
but similar restrictions apply as in the other instance. A very little
thought will show that in the case of the two small compartments you
must begin and finish at the ends that lie together, and it then
follows that the tours in the larger compartments must also start and
end on the contiguous sides.

In the diagram given of one of the possible solutions it will be seen
that there are eight places at which we may start this particular tour;
but there is only one route in each case, because we must complete the
compartment in which we find ourself before passing into another. In any
solution we shall find that the squares distinguished by stars must be
entering or exit points, but the law of reversals leaves us the option
of making the other connections either at the diamonds or at the
circles. In the solution worked out the diamonds are used, but other
variations occur in which the circle squares are employed instead. I
think these remarks explain all the essential points in the puzzle,
which is distinctly instructive and interesting.


339.--THE FOUR KNIGHTS' TOURS.

[Illustration]

It will be seen in the illustration how a chessboard may be divided into
four parts, each of the same size and shape, so that a complete
re-entrant knight's tour may be made on each portion. There is only one
possible route for each knight and its reversal.


340.--THE CUBIC KNIGHT'S TOUR.

[Illustration]

If the reader should cut out the above diagram, fold it in the form of a
cube, and stick it together by the strips left for that purpose at the
edges, he would have an interesting little curiosity. Or he can make one
on a larger scale for himself. It will be found that if we imagine the
cube to have a complete chessboard on each of its sides, we may start
with the knight on any one of the 384 squares, and make a complete tour
of the cube, always returning to the starting-point. The method of
passing from one side of the cube to another is easily understood, but,
of course, the difficulty consisted in finding the proper points of
entry and exit on each board, the order in which the different boards
should be taken, and in getting arrangements that would comply with the
required conditions.


341.--THE FOUR FROGS.

The fewest possible moves, counting every move separately, are sixteen.
But the puzzle may be solved in seven plays, as follows, if any number
of successive moves by one frog count as a single play. All the moves
contained within a bracket are a single play; the numbers refer to the
toadstools: (1--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4,
4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1).

This is the familiar old puzzle by Guarini, propounded in 1512, and I
give it here in order to explain my "buttons and string" method of
solving this class of moving-counter problem.

Diagram A shows the old way of presenting Guarini's puzzle, the point
being to make the white knights change places with the black ones. In
"The Four Frogs" presentation of the idea the possible directions of the
moves are indicated by lines, to obviate the necessity of the reader's
understanding the nature of the knight's move in chess. But it will at
once be seen that the two problems are identical. The central square
can, of course, be ignored, since no knight can ever enter it. Now,
regard the toadstools as buttons and the connecting lines as strings, as
in Diagram B. Then by disentangling these strings we can clearly present
the diagram in the form shown in Diagram C, where the relationship
between the buttons is precisely the same as in B. Any solution on C
will be applicable to B, and to A. Place your white knights on 1 and 3
and your black knights on 6 and 8 in the C diagram, and the simplicity
of the solution will be very evident. You have simply to move the
knights round the circle in one direction or the other. Play over the
moves given above, and you will find that every little difficulty has
disappeared.

[Illustrations: A B C D E]

In Diagram D I give another familiar puzzle that first appeared in a
book published in Brussels in 1789, _Les Petites Aventures de Jerome
Sharp_. Place seven counters on seven of the eight points in the
following manner. You must always touch a point that is vacant with a
counter, and then move it along a straight line leading from that point
to the next vacant point (in either direction), where you deposit the
counter. You proceed in the same way until all the counters are placed.
Remember you always touch a vacant place and slide the counter from it
to the next place, which must be also vacant. Now, by the "buttons and
string" method of simplification we can transform the diagram into E.
Then the solution becomes obvious. "Always move _to_ the point that you
last moved _from_." This is not, of course, the only way of placing the
counters, but it is the simplest solution to carry in the mind.

There are several puzzles in this book that the reader will find lend
themselves readily to this method.


342.--THE MANDARIN'S PUZZLE.

The rather perplexing point that the solver has to decide for himself in
attacking this puzzle is whether the shaded numbers (those that are
shown in their right places) are mere dummies or not. Ninety-nine
persons out of a hundred might form the opinion that there can be no
advantage in moving any of them, but if so they would be wrong.

The shortest solution without moving any shaded number is in thirty-two
moves. But the puzzle can be solved in thirty moves. The trick lies in
moving the 6, or the 15, on the second move and replacing it on the
nineteenth move. Here is the solution: 2, 6, 13, 4, 1, 21, 4, 1, 10, 2,
21, 10, 2, 5, 22, 16, 1, 13, 6, 19, 11, 2, 5, 22, 16, 5, 13, 4, 10, 21.
Thirty moves.


343.--EXERCISE FOR PRISONERS.

There are eighty different arrangements of the numbers in the form of a
perfect knight's path, but only forty of these can be reached without
two men ever being in a cell at the same time. Two is the greatest
number of men that can be given a complete rest, and though the knight's
path can be arranged so as to leave either 7 and 13, 8 and 13, 5 and 7,
or 5 and 13 in their original positions, the following four
arrangements, in which 7 and 13 are unmoved, are the only ones that can
be reached under the moving conditions. It therefore resolves itself
into finding the fewest possible moves that will lead up to one of these
positions. This is certainly no easy matter, and no rigid rules can be
laid down for arriving at the correct answer. It is largely a matter for
individual judgment, patient experiment, and a sharp eye for revolutions
and position.

          A
    +--+--+--+--+
    | 6| 1|10|15|
    +--+--+--+--+
    | 9|12| 7| 4|
    +--+--+--+--+
    | 2| 5|14|11|
    +--+--+--+--+
    |13| 8| 3|**|
    +--+--+--+--+

          B
    +--+--+--+--+
    | 6| 1|10|15|
    +--+--+--+--+
    |11|14| 7| 4|
    +--+--+--+--+
    | 2| 5|12| 9|
    +--+--+--+--+
    |13| 8| 3|**|
    +--+--+--+--+

          C
    +--+--+--+--+
    | 6| 9| 4|15|
    +--+--+--+--+
    | 1|12| 7|10|
    +--+--+--+--+
    | 8| 5|14| 3|
    +--+--+--+--+
    |13| 2|11|**|
    +--+--+--+--+

          D
    +--+--+--+--+
    | 6|11| 4|15|
    +--+--+--+--+
    | 1|14| 7|10|
    +--+--+--+--+
    | 8| 5|12| 3|
    +--+--+--+--+
    |13| 2| 9|**|
    +--+--+--+--+

[Illustration: A, B, C, D]

As a matter of fact, the position C can be reached in as few as
sixty-six moves in the following manner: 12, 11, 15, 12, 11, 8, 4, 3, 2,
6, 5, 1, 6, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8,
4, 12, 11, 3, 2, 5, 10, 15, 6, 1, 8, 4, 9, 8, 1, 6, 4, 9, 12, 2, 5, 10,
15, 4, 9, 12, 2, 5, 3, 11, 14, 2, 5, 14, 11 = 66 moves. Though this is
the shortest that I know of, and I do not think it can be beaten, I
cannot state positively that there is not a shorter way yet to be
discovered. The most tempting arrangement is certainly A; but things
are not what they seem, and C is really the easiest to reach.

If the bottom left-hand corner cell might be left vacant, the following
is a solution in forty-five moves by Mr. R. Elrick: 15, 11, 10, 9, 13,
14, 11, 10, 7, 8, 4, 3, 8, 6, 9, 7, 12, 4, 6, 9, 5, 13, 7, 5, 13, 1, 2,
13, 5, 7, 1, 2, 13, 8, 3, 6, 9, 12, 7, 11, 14, 1, 11, 14, 1. But every
man has moved.


344.--THE KENNEL PUZZLE.

The first point is to make a choice of the most promising knight's
string and then consider the question of reaching the arrangement in the
fewest moves. I am strongly of opinion that the best string is the one
represented in the following diagram, in which it will be seen that each
successive number is a knight's move from the preceding one, and that
five of the dogs (1, 5, 10, 15, and 20) never leave their original
kennels.

    +-----+------+------+------+------+
    |1    |2     |3     |4     |5     |
    |     |      |      |      |      |
    |   1 |  18  |   9  |  14  |   5  |
    |     |      |      |      |      |
    +-----+------+------+------+------+
    |6    |7     |8     |9     |10    |
    |     |      |      |      |      |
    |   8 |  13  |   4  |  19  |  10  |
    |     |      |      |      |      |
    +-----+------+------+------+------+
    |11   |12    |13    |14    |15    |
    |     |      |      |      |      |
    |  17 |   2  |  11  |   6  |  15  |
    |     |      |      |      |      |
    +-----+------+------+------+------+
    |16   |17    |18    |19    |20    |
    |     |      |      |      |      |
    |  12 |   7  |  16  |   3  |  20  |
    |     |      |      |      |      |
    +-----+------+------+------+------+
    |21   |22    |23    |24    |25    |
    |     |      |      |      |      |
    |     |      |      |      |      |
    |     |      |      |      |      |
    +-----+------+------+------+------+

[Illustration]

This position may be arrived at in as few as forty-six moves, as
follows: 16--21, 16--22, 16--23, 17--16, 12--17, 12--22, 12--21,7--12,
7--17, 7--22, 11--12, 11--17, 2--7, 2--12, 6--11, 8--7, 8--6, 13--8,
18--13, 11--18, 2--17, 18--12, 18--7, 18--2, 13--7, 3--8, 3--13, 4--3,
4--8, 9--4, 9--3, 14--9, 14--4, 19--14, 19--9, 3--14, 3--19, 6--12,
6--13, 6--14, 17--11, 12--16, 2--12, 7--17, 11--13, 16--18 = 46 moves. I
am, of course, not able to say positively that a solution cannot be
discovered in fewer moves, but I believe it will be found a very hard
task to reduce the number.


345.--THE TWO PAWNS.

Call one pawn A and the other B. Now, owing to that optional first move,
either pawn may make either 5 or 6 moves in reaching the eighth square.
There are, therefore, four cases to be considered: (1) A 6 moves and B 6
moves; (2) A 6 moves and B 5 moves; (3) A 5 moves and B 6 moves; (4) A 5
moves and B 5 moves. In case (1) there are 12 moves, and we may select
any 6 of these for A. Therefore 7×8×9×10×11×12 divided by 1×2×3×4×5×6
gives us the number of variations for this case--that is, 924. Similarly
for case (2), 6 selections out of 11 will be 462; in case (3), 5
selections out of 11 will also be 462; and in case (4), 5 selections out
of 10 will be 252. Add these four numbers together and we get 2,100,
which is the correct number of different ways in which the pawns may
advance under the conditions. (See No. 270, on p. 204.)


346.--SETTING THE BOARD.

The White pawns may be arranged in 40,320 ways, the White rooks in 2
ways, the bishops in 2 ways, and the knights in 2 ways. Multiply these
numbers together, and we find that the White pieces may be placed in
322,560 different ways. The Black pieces may, of course, be placed in
the same number of ways. Therefore the men may be set up in 322,560 ×
322,560 = 104,044,953,600 ways. But the point that nearly everybody
overlooks is that the board may be placed in two different ways for
every arrangement. Therefore the answer is doubled, and is
208,089,907,200 different ways.


347.--COUNTING THE RECTANGLES.

There are 1,296 different rectangles in all, 204 of which are squares,
counting the square board itself as one, and 1,092 rectangles that are
not squares. The general formula is that a board of n² squares
contains ((n² + n)²)/4 rectangles, of which (2n³ + 3n² + n)/6 are
squares and (3n^4 + 2n³ - 3n² - 2n)/12 are rectangles that are not
squares. It is curious and interesting that the total number of
rectangles is always the square of the triangular number whose side is
n.


348.--THE ROOKERY.

The answer involves the little point that in the final position the
numbered rooks must be in numerical order in the direction contrary to
that in which they appear in the original diagram, otherwise it cannot
be solved. Play the rooks in the following order of their numbers. As
there is never more than one square to which a rook can move (except on
the final move), the notation is obvious--5, 6, 7, 5, 6, 4, 3, 6, 4, 7,
5, 4, 7, 3, 6, 7, 3, 5, 4, 3, 1, 8, 3, 4, 5, 6, 7, 1, 8, 2, 1, and rook
takes bishop, checkmate. These are the fewest possible
moves--thirty-two. The Black king's moves are all forced, and need not
be given.


349.--STALEMATE.


Working independently, the same position was arrived at by Messrs. S.
Loyd, E.N. Frankenstein, W.H. Thompson, and myself. So the following may
be accepted as the best solution possible to this curious problem :--

        White.       Black.
     1. P--Q4     1. P--K4
     2. Q--Q3     2. Q--R5
     3. Q--KKt3   3. B--Kt5 ch
     4. Kt--Q2    4. P--QR4
     5. P--R4     5. P--Q3
     6. P--R3     6. B--K3
     7. R--R3     7. P--KB4
     8. Q--R2     8. P--B4
     9. R--KKt3   9. B--Kt6
    10. P--QB4   10. P--B5
    11. P--B3    11. P--K5
    12. P--Q5    12. P--K6

And White is stalemated.

We give a diagram of the curious position arrived at. It will be seen
that not one of White's pieces may be moved.

[Illustration]

    +-+-+-+-+-+-+-+-+
    |r|n| | |k| |n|r|
    +-+-+-+-+-+-+-+-+
    | |p| | | | |p|p|
    +-+-+-+-+-+-+-+-+
    | | | |p| | | | |
    +-+-+-+-+-+-+-+-+
    |p| |p|P| | | | |
    +-+-+-+-+-+-+-+-+
    |P|b|P| | |p| |q|
    +-+-+-+-+-+-+-+-+
    | |b| | |p|P|R|P|
    +-+-+-+-+-+-+-+-+
    | |P| |N|P| |P|Q|
    +-+-+-+-+-+-+-+-+
    | | |B| |K|B|N|R|
    +-+-+-+-+-+-+-+-+


350.--THE FORSAKEN KING.

Play as follows:--

       White.            Black.
    1. P to K 4th     1. Any move
    2. Q to Kt 4th    2. Any move except on KB file (a)
    3. Q to Kt 7th    3. K moves to royal row
    4. B to Kt 5th    4. Any move
    5. Mate in two moves
                   If 3. K other than to royal row
    4. P to Q 4th     4. Any move
    5. Mate in two moves
               (a) If 2. Any move on KB file
    3. Q to Q 7th     3. K moves to royal row
    4. P to Q Kt 3rd  4. Any move
    5. Mate in two moves
                   If 3. K other than to royal row
    4. P to Q 4th     4. Any move
    5. Mate in two moves

Of course, by "royal row" is meant the row on which the king originally
stands at the beginning of a game. Though, if Black plays badly, he may,
in certain positions, be mated in fewer moves, the above provides for
every variation he can possibly bring about.


351.--THE CRUSADER.

        White.                    Black.
     1. Kt to QB 3rd           1. P to Q 4th
     2. Kt takes QP            2. Kt to QB 3rd
     3. Kt takes KP            3. P to KKt 4th
     4. Kt takes B             4. Kt to KB 3rd
     5. Kt takes P             5. Kt to K 5th
     6. Kt takes Kt            6. Kt to B 6th
     7. Kt takes Q             7. R to KKt sq
     8. Kt takes BP            8. R to KKt 3rd
     9. Kt takes P             9. R to K 3rd
    10. Kt takes P            10. Kt to Kt 8th
    11. Kt takes B            11. R to R 6th
    12. Kt takes R            12. P to Kt 4th
    13. Kt takes P (ch)       13. K to B 2nd
    14. Kt takes P            14. K to Kt 3rd
    15. Kt takes R            15. K to R 4th
    16. Kt takes Kt           16. K to R 5th
        White now mates in three moves.
    17. P to Q 4th            17. K to R 4th
    18. Q to Q 3rd            18. K moves
    19. Q to KR 3rd (mate)
                           If 17. K to Kt 5th
    18. P to K 4th (dis. ch)  18. K moves
    19. P to KKt 3rd (mate)

The position after the sixteenth move, with the mate in three moves, was
first given by S. Loyd in _Chess Nuts_.


352.--IMMOVABLE PAWNS.

     1. Kt to KB 3
     2. Kt to KR 4
     3. Kt to Kt 6
     4. Kt takes R
     5. Kt to Kt 6
     6. Kt takes B
     7. K takes Kt
     8. Kt to QB 3
     9. Kt to R 4
    10. Kt to Kt 6
    11. Kt takes R
    12. Kt to Kt 6
    13. Kt takes B
    14. Kt to Q 6
    15. Q to K sq
    16. Kt takes Q
    17. K takes Kt, and the position is reached.

Black plays precisely the same moves as White, and therefore we give one
set of moves only. The above seventeen moves are the fewest possible.


353.--THIRTY-SIX MATES.

Place the remaining eight White pieces thus: K at KB 4th, Q at QKt 6th,
R at Q 6th, R at KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 5th, and Kt
at QB 5th. The following mates can then be given:--

    By discovery from Q            8
    By discovery from R at Q 6th  13
    By discovery from B at R 8th  11
    Given by Kt at R 5th           2
    Given by pawns                 2
                                  --
       Total                      36

Is it possible to construct a position in which more than thirty-six
different mates on the move can be given? So far as I know, nobody has
yet beaten my arrangement.


354.--AN AMAZING DILEMMA.

Mr Black left his king on his queen's knight's 7th, and no matter what
piece White chooses for his pawn, Black cannot be checkmated. As we
said, the Black king takes no notice of checks and never moves. White
may queen his pawn, capture the Black rook, and bring his three pieces
up to the attack, but mate is quite impossible. The Black king cannot be
left on any other square without a checkmate being possible.

The late Sam Loyd first pointed out the peculiarity on which this puzzle
is based.


355.--CHECKMATE!

Remove the White pawn from B 6th to K 4th and place a Black pawn on
Black's KB 2nd. Now, White plays P to K 5th, check, and Black must play
P to B 4th. Then White plays P takes P _en passant_, checkmate. This was
therefore White's last move, and leaves the position given. It is the
only possible solution.


356.--QUEER CHESS.

    +-+-+-+-+-+-+-+-+
    | | | | | | | | |
    +-+-+-+-+-+-+-+-+
    | | |R|k|R|N| | |
    +-+-+-+-+-+-+-+-+
    | | | | | | | | |
    +-+-+-+-+-+-+-+-+

If you place the pieces as follows (where only a portion of the board is
given, to save space), the Black king is in check, with no possible move
open to him. The reader will now see why I avoided the term "checkmate,"
apart from the fact that there is no White king. The position is
impossible in the game of chess, because Black could not be given check
by both rooks at the same time, nor could he have moved into check on
his last move.

I believe the position was first published by the late S. Loyd.


357.--ANCIENT CHINESE PUZZLE.

Play as follows:--

    1. R--Q 6
    2. K--R 7
    3. R (R 6)--B 6 (mate).

Black's moves are forced, so need not be given.


358.--THE SIX PAWNS.

The general formula for six pawns on all squares greater than 2² is
this: Six times the square of the number of combinations of n things
taken three at a time, where n represents the number of squares on the
side of the board. Of course, where n is even the unoccupied squares
in the rows and columns will be even, and where n is odd the number of
squares will be odd. Here n is 8, so the answer is 18,816 different
ways. This is "The Dyer's Puzzle" (_Canterbury Puzzles_, No. 27) in
another form. I repeat it here in order to explain a method of solving
that will be readily grasped by the novice. First of all, it is evident
that if we put a pawn on any line, we must put a second one in that line
in order that the remainder may be even in number. We cannot put four or
six in any row without making it impossible to get an even number in all
the columns interfered with. We have, therefore, to put two pawns in
each of three rows and in each of three columns. Now, there are just six
schemes or arrangements that fulfil these conditions, and these are
shown in Diagrams A to F, inclusive, on next page.

[Illustration]

I will just remark in passing that A and B are the only distinctive
arrangements, because, if you give A a quarter-turn, you get F; and if
you give B three quarter-turns in the direction that a clock hand
moves, you will get successively C, D, and E. No matter how you may
place your six pawns, if you have complied with the conditions of the
puzzle they will fall under one of these arrangements. Of course it
will be understood that mere expansions do not destroy the essential
character of the arrangements. Thus G is only an expansion of form A.
The solution therefore consists in finding the number of these
expansions. Supposing we confine our operations to the first three
rows, as in G, then with the pairs a and b placed in the first and
second columns the pair c may be disposed in any one of the remaining
six columns, and so give six solutions. Now slide pair b into the
third column, and there are five possible positions for c. Slide b
into the fourth column, and c may produce four new solutions. And so
on, until (still leaving a in the first column) you have b in the
seventh column, and there is only one place for c--in the eighth
column. Then you may put a in the second column, b in the third, and c
in the fourth, and start sliding c and b as before for another series
of solutions.

We find thus that, by using form A alone and confining our operations to
the three top rows, we get as many answers as there are combinations of
8 things taken 3 at a time. This is (8 × 7 × 6)/(1 × 2 × 3) = 56. And it
will at once strike the reader that if there are 56 different ways of
electing the columns, there must be for each of these ways just 56 ways
of selecting the rows, for we may simultaneously work that "sliding"
process downwards to the very bottom in exactly the same way as we have
worked from left to right. Therefore the total number of ways in which
form A may be applied is 56 × 6 = 3,136. But there are, as we have seen,
six arrangements, and we have only dealt with one of these, A. We must,
therefore, multiply this result by 6, which gives us 3,136 × 6 = 18,816,
which is the total number of ways, as we have already stated.


359.--COUNTER SOLITAIRE.

Play as follows: 3--11, 9--10, 1--2, 7--15, 8--16, 8--7, 5--13, 1--4,
8--5, 6--14, 3--8, 6--3, 6--12, 1--6, 1--9, and all the counters will
have been removed, with the exception of No. 1, as required by the
conditions.


360.--CHESSBOARD SOLITAIRE.

Play as follows: 7--15, 8--16, 8--7, 2--10, 1--9, 1--2, 5--13, 3--4,
6--3, 11--1, 14--8, 6--12, 5--6, 5--11, 31--23, 32--24, 32--31, 26--18,
25--17, 25--26, 22--32, 14--22, 29--21, 14--29, 27--28, 30--27, 25--14,
30--20, 25--30, 25--5. The two counters left on the board are 25 and
19--both belonging to the same group, as stipulated--and 19 has never
been moved from its original place.

I do not think any solution is possible in which only one counter is
left on the board.


361.--THE MONSTROSITY.

        White          Black,
     1. P to KB 4      P to QB 3
     2. K to B 2       Q to R 4
     3. K to K 3       K to Q sq
     4. P to B 5       K to B 2
     5. Q to K sq      K to Kt 3
     6. Q to Kt 3      Kt to QR 3
     7. Q to Kt 8      P to KR 4
     8. Kt to KB 3     R to R 3
     9. Kt to K 5      R to Kt 3
    10. Q takes B      R to Kt 6, ch
    11. P takes R      K to Kt 4
    12. R to R 4       P to B 3
    13. R to Q 4       P takes Kt
    14. P to QKt 4     P takes R, ch
    15. K to B 4       P to R 5
    16. Q to K 8       P to R 6
    17. Kt to B 3, ch  P takes Kt
    18. B to R 3       P to R 7
    19. R to Kt sq     P to R 8 (Q)
    20. R to Kt 2      P takes R
    21. K to Kt 5      Q to KKt 8
    22. Q to R 5       K to R 5
    23. P to Kt 5      R to B sq
    24. P to Kt 6      R to B 2
    25. P takes R      P to Kt 8 (B)
    26. P to B 8 (R)   Q to B 2
    27. B to Q 6       Kt to Kt 5
    28. K to Kt 6      K to R 6
    29. R to R 8       K to Kt 7
    30. P to R 4       Q (Kt 8) to Kt 3
    31. P to R 5       K to B 8
    32. P takes Q      K to Q 8
    33. P takes Q      K to K 8
    34. K to B 7       Kt to KR 3, ch
    35. K to K 8       B to R 7
    36. P to B 6       B to Kt sq
    37. P to B 7       K takes B
    38. P to B 8 (B)   Kt to Q 4
    39. B to Kt 8      Kt to B 3, ch
    40. K to Q 8       Kt to K sq
    41. P takes Kt (R) Kt to B 2, ch
    42. K to B 7       Kt to Q sq
    43. Q to B 7, ch   K to Kt 8

And the position is reached.

The order of the moves is immaterial, and this order may be greatly
varied. But, although many attempts have been made, nobody has succeeded
in reducing the number of my moves.


362.--THE WASSAIL BOWL.

The division of the twelve pints of ale can be made in eleven
manipulations, as below. The six columns show at a glance the quantity
of ale in the barrel, the five-pint jug, the three-pint jug, and the
tramps X, Y, and Z respectively after each manipulation.

        Barrel. 5-pint. 3-pint.    X.       Y.       Z.

        7   ..   5   ..   0   ..   0   ..   0   ..   0
        7   ..   2   ..   3   ..   0   ..   0   ..   0
        7   ..   0   ..   3   ..   2   ..   0   ..   0
        7   ..   3   ..   0   ..   2   ..   0   ..   0
        4   ..   3   ..   3   ..   2   ..   0   ..   0
        0   ..   3   ..   3   ..   2   ..   4   ..   0
        0   ..   5   ..   1   ..   2   ..   4   ..   0
        0   ..   5   ..   0   ..   2   ..   4   ..   1
        0   ..   2   ..   3   ..   2   ..   4   ..   1
        0   ..   0   ..   3   ..   4   ..   4   ..   1
        0   ..   0   ..   0   ..   4   ..   4   ..   4

And each man has received his four pints of ale.


363.--THE DOCTOR'S QUERY.

The mixture of spirits of wine and water is in the proportion of 40 to
1, just as in the other bottle it was in the proportion of 1 to 40.



364.--THE BARREL PUZZLE.

[Illustration: Figs. 1, 2, and 3]

All that is necessary is to tilt the barrel as in Fig. 1, and if the
edge of the surface of the water exactly touches the lip a at the same
time that it touches the edge of the bottom b, it will be just half
full. To be more exact, if the bottom is an inch or so from the ground,
then we can allow for that, and the thickness of the bottom, at the top.
If when the surface of the water reached the lip a it had risen to the
point c in Fig. 2, then it would be more than half full. If, as in
Fig. 3, some portion of the bottom were visible and the level of the
water fell to the point d, then it would be less than half full.

This method applies to all symmetrically constructed vessels.



365.--NEW MEASURING PUZZLE.

The following solution in eleven manipulations shows the contents of
every vessel at the start and after every manipulation:--

    10-quart. 10-quart. 5-quart.  4-quart.

       10   ..   10   ..   0   ..   0
        5   ..   10   ..   5   ..   0
        5   ..   10   ..   1   ..   4
        9   ..   10   ..   1   ..   0
        9   ..    6   ..   1   ..   4
        9   ..    7   ..   0   ..   4
        9   ..    7   ..   4   ..   0
        9   ..    3   ..   4   ..   4
        9   ..    3   ..   5   ..   3
        9   ..    8   ..   0   ..   3
        4   ..    8   ..   5   ..   3
        4   ..   10   ..   3   ..   3



366.--THE HONEST DAIRYMAN.

Whatever the respective quantities of milk and water, the relative
proportion sent to London would always be three parts of water to one of
milk. But there are one or two points to be observed. There must
originally be more water than milk, or there will be no water in A to
double in the second transaction. And the water must not be more than
three times the quantity of milk, or there will not be enough liquid in
B to effect the second transaction. The third transaction has no effect
on A, as the relative proportions in it must be the same as after the
second transaction. It was introduced to prevent a quibble if the
quantity of milk and water were originally the same; for though double
"nothing" would be "nothing," yet the third transaction in such a case
could not take place.



367.--WINE AND WATER.

The wine in small glass was one-sixth of the total liquid, and the wine
in large glass two-ninths of total. Add these together, and we find that
the wine was seven-eighteenths of total fluid, and therefore the water
eleven-eighteenths.



368.--THE KEG OF WINE.

The capacity of the jug must have been a little less than three gallons.
To be more exact, it was 2.93 gallons.



369.--MIXING THE TEA.

There are three ways of mixing the teas. Taking them in the order of
quality, 2s. 6d., 2s. 3d., 1s. 9p., mix 16 lbs., 1 lb., 3 lbs.; or 14
lbs., 4 lbs., 2 lbs.; or 12 lbs., 7 lbs., 1 lb. In every case the
twenty pounds mixture should be worth 2s. 4½d. per pound; but the last
case requires the smallest quantity of the best tea, therefore it is
the correct answer.


370.--A PACKING PUZZLE.

On the side of the box, 14 by 22+4/5, we can arrange 13 rows containing
alternately 7 and 6 balls, or 85 in all. Above this we can place another
layer consisting of 12 rows of 7 and 6 alternately, or a total of 78. In
the length of 24+9/10 inches 15 such layers may be packed, the alternate
layers containing 85 and 78 balls. Thus 8 times 85 added to 7 times 78
gives us 1,226 for the full contents of the box.


371.--GOLD PACKING IN RUSSIA.

The box should be 100 inches by 100 inches by 11 inches deep, internal
dimensions. We can lay flat at the bottom a row of eight slabs,
lengthways, end to end, which will just fill one side, and nine of these
rows will dispose of seventy-two slabs (all on the bottom), with a space
left over on the bottom measuring 100 inches by 1 inch by 1 inch. Now
make eleven depths of such seventy-two slabs, and we have packed 792,
and have a space 100 inches by 1 inch by 11 inches deep. In this we may
exactly pack the remaining eight slabs on edge, end to end.


372.--THE BARRELS OF HONEY.

The only way in which the barrels could be equally divided among the
three brothers, so that each should receive his 3½ barrels of honey
and his 7 barrels, is as follows:--

      Full.  Half-full. Empty.
    A   3        1        3
    B   2        3        2
    C   2        3        2

There is one other way in which the division could be made, were it not
for the objection that all the brothers made to taking more than four
barrels of the same description. Except for this difficulty, they might
have given B his quantity in exactly the same way as A above, and then
have left C one full barrel, five half-full barrels, and one empty
barrel. It will thus be seen that in any case two brothers would have to
receive their allowance in the same way.


373.--CROSSING THE STREAM.

First, the two sons cross, and one returns Then the man crosses and the
other son returns. Then both sons cross and one returns. Then the lady
crosses and the other son returns Then the two sons cross and one of
them returns for the dog. Eleven crossings in all.

It would appear that no general rule can be given for solving these
river-crossing puzzles. A formula can be found for a particular case
(say on No. 375 or 376) that would apply to any number of individuals
under the restricted conditions; but it is not of much use, for some
little added stipulation will entirely upset it. As in the case of the
measuring puzzles, we generally have to rely on individual ingenuity.


374.--CROSSING THE RIVER AXE.

Here is the solution:--

           | {J 5) | G T8 3
        5  | ( J } | G T8 3
        5  | {G 3) |  JT8
        53 | ( G } |  JT8
        53 | {J T) | G  8
     J  5  | (T 3} | G  8
     J  5  | {G 8) |   T  3
    G  8   | (J 5} |   T
    G  8   | {J T) |     53
     JT8   | ( G } |     53
     JT8   | {G 3) |     5
    G T8 3 | ( J } |     5
    G T8 3 | {J 5) |

G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for £800,
£500, and £300 respectively. The two side columns represent the left
bank and the right bank, and the middle column the river. Thirteen
crossings are necessary, and each line shows the position when the boat
is in mid-stream during a crossing, the point of the bracket indicating
the direction.

It will be found that not only is no person left alone on the land or in
the boat with more than his share of the spoil, but that also no two
persons are left with more than their joint shares, though this last
point was not insisted upon in the conditions.


375.--FIVE JEALOUS HUSBANDS.

It is obvious that there must be an odd number of crossings, and that if
the five husbands had not been jealous of one another the party might
have all got over in nine crossings. But no wife was to be in the
company of a man or men unless her husband was present. This entails two
more crossings, eleven in all.

The following shows how it might have been done. The capital letters
stand for the husbands, and the small letters for their respective
wives. The position of affairs is shown at the start, and after each
crossing between the left bank and the right, and the boat is
represented by the asterisk. So you can see at a glance that a, b, and c
went over at the first crossing, that b and c returned at the second
crossing, and so on.

       ABCDE abcde *|..|
                    |  |
    1. ABCDE    de  |..|*       abc
    2. ABCDE  bcde *|..|        a
    3. ABCDE     e  |..|*       abcd
    4. ABCDE    de *|..|        abc
    5.    DE    de  |,,|* ABC   abc
    6.   CDE   cde *|..|  AB    ab
    7.         cde  |..|* ABCDE ab
    8.        bcde *|..|  ABCDE a
    9.           e  |..|* ABCDE abcd
    10.       bc e *|..|  ABCDE a  d
    11.             |..|* ABCDE abcde

There is a little subtlety concealed in the words "show the _quickest_
way."

Everybody correctly assumes that, as we are told nothing of the rowing
capabilities of the party, we must take it that they all row equally
well. But it is obvious that two such persons should row more quickly
than one.

Therefore in the second and third crossings two of the ladies should
take back the boat to fetch d, not one of them only. This does not
affect the number of landings, so no time is lost on that account. A
similar opportunity occurs in crossings 10 and 11, where the party again
had the option of sending over two ladies or one only.

To those who think they have solved the puzzle in nine crossings I would
say that in every case they will find that they are wrong. No such
jealous husband would, in the circumstances, send his wife over to the
other bank to a man or men, even if she assured him that she was coming
back next time in the boat. If readers will have this fact in mind, they
will at once discover their errors.


376.--THE FOUR ELOPEMENTS.

If there had been only three couples, the island might have been
dispensed with, but with four or more couples it is absolutely necessary
in order to cross under the conditions laid down. It can be done in
seventeen passages from land to land (though French mathematicians have
declared in their books that in such circumstances twenty-four are
needed), and it cannot be done in fewer. I will give one way. A, B, C,
and D are the young men, and a, b, c, and d are the girls to whom they
are respectively engaged. The three columns show the positions of the
different individuals on the lawn, the island, and the opposite shore
before starting and after each passage, while the asterisk indicates the
position of the boat on every occasion.

    Lawn.      | Island. | Shore.
               |         |
    ABCDabcd * |         |
    ABCD  cd   |         |      ab   *
    ABCD bcd * |         |      a
    ABCD   d   |   bc  * |      a
    ABCD  cd * |   b     |      a
      CD  cd   |   b     | AB   a    *
     BCD  cd * |   b     | A    a
     BCD       |   bcd * | A    a
     BCD   d * |   bc    | A    a
       D   d   |   bc    | ABC  a    *
       D   d   |  abc  * | ABC
       D   d   |   b     | ABC  a c  *
     B D   d * |   b     | A C  a c
           d   |   b     | ABCD a c  *
           d   |   bc  * | ABCD a
           d   |         | ABCD abc  *
          cd * |         | ABCD ab
               |         | ABCD abcd *

Having found the fewest possible passages, we should consider two other
points in deciding on the "quickest method": Which persons were the most
expert in handling the oars, and which method entails the fewest
possible delays in getting in and out of the boat? We have no data upon
which to decide the first point, though it is probable that, as the boat
belonged to the girls' household, they would be capable oarswomen. The
other point, however, is important, and in the solution I have given
(where the girls do 8-13ths of the rowing and A and D need not row at
all) there are only sixteen gettings-in and sixteen gettings-out. A man
and a girl are never in the boat together, and no man ever lands on the
island. There are other methods that require several more exchanges of
places.


377.--STEALING THE CASTLE TREASURE.

Here is the best answer, in eleven manipulations:--

    Treasure down.
    Boy down--treasure up.
    Youth down--boy up.
    Treasure down.
    Man down--youth and treasure up.
    Treasure down.
    Boy down--treasure up.
    Treasure down.
    Youth down--boy up.
    Boy down--treasure up.
    Treasure down.


378.--DOMINOES IN PROGRESSION.

There are twenty-three different ways. You may start with any domino,
except the 4--4 and those that bear a 5 or 6, though only certain
initial dominoes may be played either way round. If you are given the
common difference and the first domino is played, you have no option as
to the other dominoes. Therefore all I need do is to give the initial
domino for all the twenty-three ways, and state the common difference.
This I will do as follows:--

With a common difference of 1, the first domino may be either of these:
0--0, 0--1, 1--0, 0--2, 1--1, 2--0, 0--3, 1--2, 2--1, 3--0, 0--4, 1--3,
2--2, 3--1, 1--4, 2--3, 3--2, 2--4, 3--3, 3--4. With a difference of 2,
the first domino may be 0--0, 0--2, or 0--1. Take the last case of all
as an example. Having played the 0--1, and the difference being 2, we
are compelled to continue with 1--2, 2--3, 3--4. 4--5, 5--6. There are
three dominoes that can never be used at all. These are 0--5, 0--6, and
1--6. If we used a box of dominoes extending to 9--9, there would be
forty different ways.


379.--THE FIVE DOMINOES.

There are just ten different ways of arranging the dominoes. Here is one
of them:--

(2--0) (0--0) (0--1) (1--4) (4--0).

I will leave my readers to find the remaining nine for themselves.


380.--THE DOMINO FRAME PUZZLE.


[Illustration:

    +---+-------+-------+-------+-------+-------+-------+-------+
    | 2 | 2 | 5 | 5 | 6 | 6 | 6 | 6 | 1 | 1 |   |   |   |   | 4 |
    | - +-------+-------+-------+-------+-------+-------+---+---+
    | 2 |                                                   | 4 |
    +---+                                                   | - |
    | 2 |                                                   | 3 |
    | - |                                                   +---+
    | 6 |                                                   | 3 |
    +---+                       T H E                       | - |
    | 6 |                                                   | 3 |
    | - |                                                   +---+
    | 3 |                                                   | 3 |
    +---+                                                   | - |
    | 3 |                                                   | 1 |
    | - |              D O M I N O   F R A M E              +---+
    |   |                                                   | 1 |
    +---+                                                   | - |
    |   |                                                   | 1 |
    | - |                                                   +---+
    | 5 |                                                   | 1 |
    +---+                 -S-O-L-U-T-I-O-N-                 | - |
    | 5 |                                                   | 4 |
    | - |                                                   +---+
    | 3 |                                                   | 4 |
    +---+                                                   | - |
    | 3 |                                                   | 6 |
    | - |                                                   +---+
    | 2 |                                                   | 6 |
    +---+---+-------+-------+-------+-------+-------+-------+ - |
    | 2 | 1 | 1 | 5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 | 2 | 2 |   |   |
    +-------+-------+-------+-------+-------+-------+-------+---+

]

The illustration is a solution. It will be found that all four sides of
the frame add up 44. The sum of the pips on all the dominoes is 168, and
if we wish to make the sides sum to 44, we must take care that the four
corners sum to 8, because these corners are counted twice, and 168 added
to 8 will equal 4 times 44, which is necessary. There are many different
solutions. Even in the example given certain interchanges are possible
to produce different arrangements. For example, on the left-hand side
the string of dominoes from 2--2 down to 3--2 may be reversed, or from
2--6 to 3--2, or from 3--0 to 5--3. Also, on the right-hand side we may
reverse from 4--3 to 1--4. These changes will not affect the correctness
of the solution.


381.--THE CARD FRAME PUZZLE.

The sum of all the pips on the ten cards is 55. Suppose we are trying to
get 14 pips on every side. Then 4 times 14 is 56. But each of the four
corner cards is added in twice, so that 55 deducted from 56, or 1, must
represent the sum of the four corner cards. This is clearly impossible;
therefore 14 is also impossible. But suppose we came to trying 18. Then
4 times 18 is 72, and if we deduct 55 we get 17 as the sum of the
corners. We need then only try different arrangements with the four
corners always summing to 17, and we soon discover the following
solution:--

[Illustration:

    +-------+-------+-------+
    |   2   |  10   |   6   |
    +---+---+------ +---+---+
    |   |               |   |
    | 3 |               | 7 |
    |   |               |   |
    +---+               +---+
    |   |               |   |
    | 8 |               | 1 |
    |   |               |   |
    +---+---+-------+--+----+
    |   5   |   9   |   4   |
    +-------+-------+-------+

]

The final trials are very limited in number, and must with a little
judgment either bring us to a correct solution or satisfy us that a
solution is impossible under the conditions we are attempting. The two
centre cards on the upright sides can, of course, always be
interchanged, but I do not call these different solutions. If you
reflect in a mirror you get another arrangement, which also is not
considered different. In the answer given, however, we may exchange the
5 with the 8 and the 4 with the 1. This is a different solution. There
are two solutions with 18, four with 19, two with 20, and two with
22--ten arrangements in all. Readers may like to find all these for
themselves.


382.--THE CROSS OF CARDS.

There are eighteen fundamental arrangements, as follows, where I only
give the numbers in the horizontal bar, since the remainder must
naturally fall into their places.

    5 6 1 7 4   2 4 5 6 8
    3 5 1 6 8   3 4 5 6 7
    3 4 1 7 8   1 4 7 6 8
    2 5 1 7 8   2 3 7 6 8
    2 5 3 6 8   2 4 7 5 8
    1 5 3 7 8   3 4 9 5 6
    2 4 3 7 8   2 4 9 5 7
    1 4 5 7 8   1 4 9 6 7
    2 3 5 7 8   2 3 9 6 7

It will be noticed that there must always be an odd number in the
centre, that there are four ways each of adding up 23, 25, and 27, but
only three ways each of summing to 24 and 26.


383.--THE "T" CARD PUZZLE.

If we remove the ace, the remaining cards may he divided into two groups
(each adding up alike) in four ways; if we remove 3, there are three
ways; if 5, there are four ways; if 7, there are three ways; and if we
remove 9, there are four ways of making two equal groups. There are thus
eighteen different ways of grouping, and if we take any one of these and
keep the odd card (that I have called "removed") at the head of the
column, then one set of numbers can be varied in order in twenty-four
ways in the column and the other four twenty-four ways in the
horizontal, or together they may be varied in 24 × 24 = 576 ways. And as
there are eighteen such cases, we multiply this number by 18 and get
10,368, the correct number of ways of placing the cards. As this number
includes the reflections, we must divide by 2, but we have also to
remember that every horizontal row can change places with a vertical
row, necessitating our multiplying by 2; so one operation cancels the
other.


384.--CARD TRIANGLES.

The following arrangements of the cards show (1) the smallest possible
sum, 17; and (2) the largest possible, 23.

       1        7
      9 6      4 2
     4   8    3   6
    3 7 5 2  9 5 1 8

It will be seen that the two cards in the middle of any side may always
be interchanged without affecting the conditions. Thus there are eight
ways of presenting every fundamental arrangement. The number of
fundamentals is eighteen, as follows: two summing to 17, four summing to
19, six summing to 20, four summing to 21, and two summing to 23. These
eighteen fundamentals, multiplied by eight (for the reason stated
above), give 144 as the total number of different ways of placing the
cards.


385.--"STRAND" PATIENCE.

The reader may find a solution quite easy in a little over 200 moves,
but, surprising as it may at first appear, not more than 62 moves are
required. Here is the play: By "4 C up" I mean a transfer of the 4 of
clubs with all the cards that rest on it. 1 D on space, 2 S on space, 3
D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S on
space, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5 D
exchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H) on
space (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1), 6 C
up on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7 H on
8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C up on 5
D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 62 moves in all. This is
my record; perhaps the reader can beat it.


386.--A TRICK WITH DICE.

All you have to do is to deduct 250 from the result given, and the three
figures in the answer will be the three points thrown with the dice.
Thus, in the throw we gave, the number given would be 386; and when we
deduct 250 we get 136, from which we know that the throws were 1, 3, and
6.

The process merely consists in giving 100a + 10b + c + 250, where a, b,
and c represent the three throws. The result is obvious.


387.--THE VILLAGE CRICKET MATCH.

[Illustration:

      | Mr. Dumkins >>-->
      |------------------------>    |
      |    <-------------------     |
      |     ------------------->    |
    1 |<-----------------------     |
      |                             |
      |    <------------------------|
      |     ------------------->    |
      |    <-------------------     |
      |     ----------------------->|
      |            <--<< Mr. Podder |


      | Mr. Luffey >>-->
      |------------------------>    |
      |    <-------------------     |
      |     ----------------------->|
    2 |                             |
      |<-----------------------     |
      |     ------------------->    |
      |    <------------------------|
                <--<< Mr. Struggles |

]

The diagram No. 1 will show that as neither Mr. Podder nor Mr. Dumkins
can ever have been within the crease opposite to that from which he
started, Mr. Dumkins would score nothing by his performance. Diagram No.
2 will, however, make it clear that since Mr. Luffey and Mr. Struggles
have, notwithstanding their energetic but careless movements, contrived
to change places, the manoeuvre must increase Mr. Struggles's total by
one run.


388.--SLOW CRICKET.

The captain must have been "not out" and scored 21. Thus:--

    2 men (each lbw)         19
    4 men (each caught)      17
    1 man (run out)           0
    3 men (each bowled)       9
    1 man (captain--not out) 21
    --                       --
    11                       66

The captain thus scored exactly 15 more than the average of the team.
The "others" who were bowled could only refer to three men, as the
eleventh man would be "not out." The reader can discover for himself why
the captain must have been that eleventh man. It would not necessarily
follow with any figures.


389.--THE FOOTBALL PLAYERS.

The smallest possible number of men is seven. They could be accounted
for in three different ways: 1. Two with both arms sound, one with
broken right arm, and four with both arms broken. 2. One with both arms
sound, one with broken left arm, two with broken right arm, and three
with both arms broken. 3. Two with left arm broken, three with right arm
broken, and two with both arms broken. But if every man was injured, the
last case is the only one that would apply.


390.--THE HORSE-RACE PUZZLE.

The answer is: £12 on Acorn, £15 on Bluebottle, £20 on Capsule.

391.--THE MOTOR-CAR RACE.

The first point is to appreciate the fact that, in a race round a
circular track, there are the same number of cars behind one as there
are before. All the others are both behind and before. There were
thirteen cars in the race, including Gogglesmith's car. Then one-third
of twelve added to three-quarters of twelve will give us thirteen--the
correct answer.


392.--THE PEBBLE GAME.

In the case of fifteen pebbles, the first player wins if he first takes
two. Then when he holds an odd number and leaves 1, 8, or 9 he wins, and
when he holds an even number and leaves 4, 5, or 12 he also wins. He can
always do one or other of these things until the end of the game, and so
defeat his opponent. In the case of thirteen pebbles the first player
must lose if his opponent plays correctly. In fact, the only numbers
with which the first player ought to lose are 5 and multiples of 8 added
to 5, such as 13, 21, 29, etc.


393.--THE TWO ROOKS.

The second player can always win, but to ensure his doing so he must
always place his rook, at the start and on every subsequent move, on the
same diagonal as his opponent's rook. He can then force his opponent
into a corner and win. Supposing the diagram to represent the positions
of the rooks at the start, then, if Black played first, White might have
placed his rook at A and won next move. Any square on that diagonal from
A to H will win, but the best play is always to restrict the moves of
the opposing rook as much as possible. If White played first, then Black
should have placed his rook at B (F would not be so good, as it gives
White more scope); then if White goes to C, Black moves to D; White to
E, Black to F; White to G, Black to C; White to H, Black to I; and Black
must win next move. If at any time Black had failed to move on to the
same diagonal as White, then White could take Black's diagonal and win.

    r: black rook
    R: white rook

    +-+-+-+-+-+-+-+-+
    |r| | | | | | | |
    +-+-+-+-+-+-+-+-+
    | |A| | | | | | |
    +-+-+-+-+-+-+-+-+
    | | | | | | | | |
    +-+-+-+-+-+-+-+-+
    | | | | | | | | |
    +-+-+-+-+-+-+-+-+
    | | | | |B|D|F| |
    +-+-+-+-+-+-+-+-+
    | | | | | |R|C|E|
    +-+-+-+-+-+-+-+-+
    | | | | | | |I|G|
    +-+-+-+-+-+-+-+-+
    | | | | | | | |H|
    +-+-+-+-+-+-+-+-+

THE TWO ROOKS.


394.--PUSS IN THE CORNER.

No matter whether he plays first or second, the player A, who starts the
game at 55, must win. Assuming that B adopts the very best lines of play
in order to prolong as much as possible his existence, A, if he has
first move, can always on his 12th move capture B; and if he has the
second move, A can always on his 14th move make the capture. His point
is always to get diagonally in line with his opponent, and by going to
33, if he has first move, he prevents B getting diagonally in line with
himself. Here are two good games. The number in front of the hyphen is
always A's move; that after the hyphen is B's:--

33-8, 32-15, 31-22, 30-21, 29-14, 22-7, 15-6, 14-2, 7-3, 6-4, 11-, and A
must capture on his next (12th) move, -13, 54-20, 53-27, 52-34, 51-41,
50-34, 42-27, 35-20, 28-13, 21-6, 14-2, 7-3, 6-4, 11-, and A must
capture on his next (14th) move.


395.--A WAR PUZZLE GAME.

The Britisher can always catch the enemy, no matter how clever and
elusive that astute individual may be; but curious though it may seem,
the British general can only do so after he has paid a somewhat
mysterious visit to the particular town marked "1" in the map, going in
by 3 and leaving by 2, or entering by 2 and leaving by 3. The three
towns that are shaded and have no numbers do not really come into the
question, as some may suppose, for the simple reason that the Britisher
never needs to enter any one of them, while the enemy cannot be forced
to go into them, and would be clearly ill-advised to do so voluntarily.
We may therefore leave these out of consideration altogether. No matter
what the enemy may do, the Britisher should make the following first
nine moves: He should visit towns 24, 20, 19, 15, 11, 7, 3, 1, 2. If the
enemy takes it into his head also to go to town 1, it will be found that
he will have to beat a precipitate retreat _the same way that he went
in_, or the Britisher will infallibly catch him in towns 2 or 3, as the
case may be. So the enemy will be wise to avoid that north-west corner
of the map altogether.

[Illustration]

Now, when the British general has made the nine moves that I have given,
the enemy will be, after his own ninth move, in one of the towns marked
5, 8, 11, 13, 14, 16, 19, 21, 24, or 27. Of course, if he imprudently
goes to 3 or 6 at this point he will be caught at once. Wherever he may
happen to be, the Britisher "goes for him," and has no longer any
difficulty in catching him in eight more moves at most (seventeen in
all) in one of the following ways. The Britisher will get to 8 when the
enemy is at 5, and win next move; or he will get to 19 when the enemy is
at 22, and win next move; or he will get to 24 when the enemy is at 27,
and so win next move. It will be found that he can be forced into one or
other of these fatal positions.

In short, the strategy really amounts to this: the Britisher plays the
first nine moves that I have given, and although the enemy does his very
best to escape, our general goes after his antagonist and always driving
him away from that north-west corner ultimately closes in with him, and
wins. As I have said, the Britisher never need make more than seventeen
moves in all, and may win in fewer moves if the enemy plays badly. But
after playing those first nine moves it does not matter even if the
Britisher makes a few bad ones. He may lose time, but cannot lose his
advantage so long as he now keeps the enemy from town 1, and must
eventually catch him.

This is a complete explanation of the puzzle. It may seem a little
complex in print, but in practice the winning play will now be quite
easy to the reader. Make those nine moves, and there ought to be no
difficulty whatever in finding the concluding line of play. Indeed, it
might almost be said that then it is difficult for the British general
_not_ to catch the enemy. It is a question of what in chess we call the
"opposition," and the visit by the Britisher to town 1 "gives him the
jump" on the enemy, as the man in the street would say.

Here is an illustrative example in which the enemy avoids capture as
long as it is possible for him to do so. The Britisher's moves are above
the line and the enemy's below it. Play them alternately.

    24 20 19 15 11  7  3  1  2  6 10 14 18 19 20 24
    -----------------------------------------------
    13  9 13 17 21 20 24 23 19 15 19 23 24 25 27

The enemy must now go to 25 or B, in either of which towns he is
immediately captured.


396.--A MATCH MYSTERY.

If you form the three heaps (and are therefore the second to draw), any
one of the following thirteen groupings will give you a win if you play
correctly: 15, 14, 1; 15, 13, 2; 15, 12, 3; 15, 11, 4; 15, 10, 5; 15, 9,
6; 15, 8, 7; 14, 13, 3; 14, 11, 5; 14, 9, 7; 13, 11, 6; 13, 10, 7; 12,
11, 7.

The beautiful general solution of this problem is as follows. Express
the number in every heap in powers of 2, avoiding repetitions and
remembering that 2^0 = 1. Then if you so leave the matches to your
opponent that there is an even number of every power, you can win. And
if at the start you leave the powers even, you can always continue to do
so throughout the game. Take, as example, the last grouping given
above--12, 11, 7. Expressed in powers of 2 we have--

    12 = 8 4 - -
    11 = 8 - 2 1
     7 = - 4 2 1
         -------
         2 2 2 2
         -------

As there are thus two of every power, you must win. Say your opponent
takes 7 from the 12 heap. He then leaves--

     5 = - 4 - 1
    11 = 8 - 2 1
     7 = - 4 2 1
         -------
         1 2 2 3
         -------

Here the powers are not all even in number, but by taking 9 from the 11
heap you immediately restore your winning position, thus--

    5 = - 4 - 1
    2 = - - 2 -
    7 = - 4 2 1
        -------
        - 2 2 2
        -------

And so on to the end. This solution is quite general, and applies to any
number of matches and any number of heaps. A correspondent informs me
that this puzzle game was first propounded by Mr. W.M.F. Mellor, but
when or where it was published I have not been able to ascertain.


397.--THE MONTENEGRIN DICE GAME.

The players should select the pairs 5 and 9, and 13 and 15, if the
chances of winning are to be quite equal. There are 216 different ways
in which the three dice may fall. They may add up 5 in 6 different ways
and 9 in 25 different ways, making 31 chances out of 216 for the player
who selects these numbers. Also the dice may add up 13 in 21 different
ways, and 15 in 10 different ways, thus giving the other player also 31
chances in 216.


398.--THE CIGAR PUZZLE.

Not a single member of the club mastered this puzzle, and yet I shall
show that it is so simple that the merest child can understand its
solution--when it is pointed out to him! The large majority of my
friends expressed their entire bewilderment. Many considered that "the
theoretical result, in any case, is determined by the relationship
between the table and the cigars;" others, regarding it as a problem in
the theory of Probabilities, arrived at the conclusion that the chances
are slightly in favour of the first or second player, as the case may
be. One man took a table and a cigar of particular dimensions, divided
the table into equal sections, and proceeded to make the two players
fill up these sections so that the second player should win. But why
should the first player be so accommodating? At any stage he has only to
throw down a cigar obliquely across several of these sections entirely
to upset Mr. 2's calculations! We have to assume that each player plays
the best possible; not that one accommodates the other.

The theories of some other friends would be quite sound if the shape of
the cigar were that of a torpedo--perfectly symmetrical and pointed at
both ends.

I will show that the first player should infallibly win, if he always
plays in the best possible manner. Examine carefully the following
diagram, No. 1, and all will be clear.

[Illustration: 1]

[Illustration: 2]

The first player must place his first cigar _on end_ in the exact centre
of the table, as indicated by the little circle. Now, whatever the
second player may do throughout, the first player must always repeat it
in an exactly diametrically opposite position. Thus, if the second
player places a cigar at A, I put one at AA; he places one at B, I put
one at BB; he places one at C, I put one at CC; he places one at D, I
put one at DD; he places one at E, I put one at EE; and so on until no
more cigars can be placed without touching. As the cigars are supposed
to be exactly alike in every respect, it is perfectly clear that for
every move that the second player may choose to make, it is possible
exactly to repeat it on a line drawn through the centre of the table.
The second player can always duplicate the first player's move, no
matter where he may place a cigar, or whether he places it on end or on
its side. As the cigars are all alike in every respect, one will
obviously balance over the edge of the table at precisely the same point
as another. Of course, as each player is supposed to play in the best
possible manner, it becomes a matter of theory. It is no valid objection
to say that in actual practice one would not be sufficiently exact to be
sure of winning. If as the first player you did not win, it would be in
consequence of your _not_ having played the best possible.

The second diagram will serve to show why the first cigar must be placed
on end. (And here I will say that the first cigar that I selected from a
box I was able so to stand on end, and I am allowed to assume that all
the other cigars would do the same.) If the first cigar were placed on
its side, as at F, then the second player could place a cigar as at
G--as near as possible, but not actually touching F. Now, in this
position you cannot repeat his play on the opposite side, because the
two ends of the cigar are not alike. It will be seen that GG, when
placed on the opposite side in the same relation to the centre,
intersects, or lies on top of, F, whereas the cigars are not allowed to
touch. You must therefore put the cigar farther away from the centre,
which would result in your having insufficient room between the centre
and the bottom left-hand corner to repeat everything that the other
player would do between G and the top right-hand corner. Therefore the
result would not be a certain win for the first player.


399.--THE TROUBLESOME EIGHT.

[Illustration:

    +---+---+---+
    | 4½| 8 | 2½|
    +---+---+---+
    | 3 | 5 | 7 |
    +---+---+---+
    | 7½| 2 | 5½|
    +---+---+---+

]

The conditions were to place a different number in each of the nine
cells so that the three rows, three columns, and two diagonals should
each add up 15. Probably the reader at first set himself an impossible
task through reading into these conditions something which is not
there--a common error in puzzle-solving. If I had said "a different
figure," instead of "a different number," it would have been quite
impossible with the 8 placed anywhere but in a corner. And it would have
been equally impossible if I had said "a different whole number." But a
number may, of course, be fractional, and therein lies the secret of the
puzzle. The arrangement shown in the figure will be found to comply
exactly with the conditions: all the numbers are different, and the
square adds up 15 in all the required eight ways.


400.--THE MAGIC STRIPS.

There are of course six different places between the seven figures in
which a cut may be made, and the secret lies in keeping one strip intact
and cutting each of the other six in a different place. After the cuts
have been made there are a large number of ways in which the thirteen
pieces may be placed together so as to form a magic square. Here is one
of them:--

[Illustration:

    +-------------+
    |1 2 3 4 5 6 7|
    +---------+---+
    |3 4 5 6 7|1 2|
    +-----+---+---+
    |5 6 7|1 2 3 4|
    +-+---+-------+
    |7|1 2 3 4 5 6|
    +-+---------+-+
    |2 3 4 5 6 7|1|
    +-------+---+-+
    |4 5 6 7|1 2 3|
    +---+---+-----+
    |6 7|1 2 3 4 5|
    +---+---------+

]

The arrangement has some rather interesting features. It will be seen
that the uncut strip is at the top, but it will be found that if the
bottom row of figures be placed at the top the numbers will still form a
magic square, and that every successive removal from the bottom to the
top (carrying the uncut strip stage by stage to the bottom) will produce
the same result. If we imagine the numbers to be on seven complete
_perpendicular_ strips, it will be found that these columns could also
be moved in succession from left to right or from right to left, each
time producing a magic square.


401.--EIGHT JOLLY GAOL-BIRDS.

There are eight ways of forming the magic square--all merely different
aspects of one fundamental arrangement. Thus, if you give our first
square a quarter turn you will get the second square; and as the four
sides may be in turn brought to the top, there are four aspects. These
four in turn reflected in a mirror produce the remaining four aspects.
Now, of these eight arrangements only four can possibly be reached under
the conditions, and only two of these four can be reached in the fewest
possible moves, which is nineteen. These two arrangements are shown.
Move the men in the following order: 5, 3, 2, 5, 7, 6, 4, 1, 5, 7, 6, 4,
1, 6, 4, 8, 3, 2, 7, and you get the first square. Move them thus: 4, 1,
2, 4, 1, 6, 7, 1, 5, 8, 1, 5, 6, 7, 5, 6, 4, 2, 7, and you have the
arrangement in the second square. In the first case every man has moved,
but in the second case the man numbered 3 has never left his cell.
Therefore No. 3 must be the obstinate prisoner, and the second square
must be the required arrangement.

[Illustration:

    +---+---+---+     +---+---+---+
    |   |   |   |     |   |   |   |
    | 5       7 |     | 7   4   3 |
    |   |   |   |     |   |   |   |
    +- -+- -+- -+     +- -+- -+- -+
    |   |   |   |     |   |   |   |
    | 6   4   2 |     |     4   8 |
    |   |   |   |     |   |   |   |
    +- -+- -+- -+     +- -+- -+- -+
    |   |   |   |     |   |   |   |
    | 1   8   3 |     | 5   6   1 |
    |   |   |   |     |   |   |   |
    +---+---+---+     +---+---+---+

]


402.--NINE JOLLY GAOL BIRDS.

There is a pitfall set for the unwary in this little puzzle. At the
start one man is allowed to be placed on the shoulders of another, so as
to give always one empty cell to enable the prisoners to move about
without any two ever being in a cell together. The two united prisoners
are allowed to add their numbers together, and are, of course, permitted
to remain together at the completion of the magic square. But they are
obviously not compelled so to remain together, provided that one of the
pair on his final move does not break the condition of entering a cell
already occupied. After the acute solver has noticed this point, it is
for him to determine which method is the better one--for the two to be
together at the count or to separate. As a matter of fact, the puzzle
can be solved in seventeen moves if the men are to remain together; but
if they separate at the end, they may actually save a move and perform
the feat in sixteen! The trick consists in placing the man in the centre
on the back of one of the corner men, and then working the pair into the
centre before their final separation.

[Illustration:

          A                 B
    +---+---+---+     +---+---+---+
    |   |   |   |     |   |   |   |
    | 2   9   4 |     | 6   7   2 |
    |   |   |   |     |   |   |   |
    +- -+- -+- -+     +- -+- -+- -+
    |   |   |   |     |   |   |   |
    | 7   5   3 |     | 1   5   9 |
    |   |   |   |     |   |   |   |
    +- -+- -+- -+     +- -+- -+- -+
    |   |   |   |     |   |   |   |
    | 6   1   8 |     | 8   3   4 |
    |   |   |   |     |   |   |   |
    +---+---+---+     +---+---+---+

]


Here are the moves for getting the men into one or other of the above
two positions. The numbers are those of the men in the order in which
they move into the cell that is for the time being vacant. The pair is
shown in brackets:--

Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6, 1.

Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4,
9.

Place 5 on 3. Then, 6, (8), 2, 6, 4, 7, 8, 4, 7, 1, 6, 7, (8), 9, 4, 3.

Place 5 on 7. Then, 4, (12), 8, 4, 6, 3, 2, 6, 3, 9, 4, 3, (12), 1, 6,
7.

The first and second solutions produce Diagram A; the second and third
produce Diagram B. There are only sixteen moves in every case. Having
found the fewest moves, we had to consider how we were to make the
burdened man do as little work as possible. It will at once be seen that
as the pair have to go into the centre before separating they must take
at fewest two moves. The labour of the burdened man can only be reduced
by adopting the other method of solution, which, however, forces us to
take another move.


403.--THE SPANISH DUNGEON.

[Illustration]

    +-----+-----+-----+-----+    +-----+-----+-----+-----+
    |     |     |     |     |    |     |     |     |     |
    |  1  |  2  |  3  |  4  |    | 10  |  9  |  7  |  4  |
    |_____|_____|_____|_____|    |_____|_____|_____|_____|
    |     |     |     |     |    |     |     |     |     |
    |  5  |  6  |  7  |  8  |    |  6  |  5  | 11  |  8  |
    |_____|_____|_____|_____|    |_____|_____|_____|_____|
    |     |     |     |     |    |     |     |     |     |
    |  9  | 10  | 11  | 12  |    |  1  |  2  | 12  | 15  |
    |_____|_____|_____|_____|    |_____|_____|_____|_____|
    |     |     |     |     |    |     |     |     |     |
    | 13  | 14  | 15  |     |    | 13  | 14  |     |  3  |
    |     |     |     |     |    |     |     |     |     |
    +-----+-----+-----+-----+    +-----+-----+-----+-----+

This can best be solved by working backwards--that is to say, you must
first catch your square, and then work back to the original position. We
must first construct those squares which are found to require the least
amount of readjustment of the numbers. Many of these we know cannot
possibly be reached. When we have before us the most favourable possible
arrangements, it then becomes a question of careful analysis to discover
which position can be reached in the fewest moves. I am afraid, however,
it is only after considerable study and experience that the solver is
able to get such a grasp of the various "areas of disturbance" and
methods of circulation that his judgment is of much value to him.

The second diagram is a most favourable magic square position. It will
be seen that prisoners 4, 8, 13, and 14 are left in their original
cells. This position may be reached in as few as thirty-seven moves.
Here are the moves: 15, 14, 10, 6, 7, 3, 2, 7, 6, 11, 3, 2, 7, 6, 11,
10, 14, 3, 2, 11, 10, 9, 5, 1, 6, 10, 9, 5, 1, 6, 10, 9, 5, 2, 12, 15,
3. This short solution will probably surprise many readers who may not
find a way under from sixty to a hundred moves. The clever prisoner was
No. 6, who in the original illustration will be seen with his arms
extended calling out the moves. He and No. 10 did most of the work, each
changing his cell five times. No. 12, the man with the crooked leg, was
lame, and therefore fortunately had only to pass from his cell into the
next one when his time came round.


404.--THE SIBERIAN DUNGEONS.

[Illustration]

    +-----+-----+-----+-----+
    |     |     |     |     |
    |  8  |  5  | 10  | 11  |
    |_____|_____|_____|_____|
    |     |     |     |     |
    | 16  | 13  |  2  |  3  |
    |_____|_____|_____|_____|
    |     |     |     |     |
    |  1  | 12  |  7  | 14  |
    |_____|_____|_____|_____|
    |     |     |     |     |
    |  9  |  4  | 15  |  6  |
    |     |     |     |     |
    +-----+-----+-----+-----+

In attempting to solve this puzzle it is clearly necessary to seek such
magic squares as seem the most favourable for our purpose, and then
carefully examine and try them for "fewest moves." Of course it at once
occurs to us that if we can adopt a square in which a certain number of
men need not leave their original cells, we may save moves on the one
hand, but we may obstruct our movements on the other. For example, a
magic square may be formed with the 6, 7, 13, and 16 unmoved; but in
such case it is obvious that a solution is impossible, since cells 14
and 15 can neither be left nor entered without breaking the condition of
no two men ever being in the same cell together.

The following solution in fourteen moves was found by Mr. G.
Wotherspoon: 8-17, 16-21, 6-16, 14-8, 5-18, 4-14, 3-24, 11-20, 10-19,
2-23, 13-22, 12-6, 1-5, 9-13. As this solution is in what I consider the
theoretical minimum number of moves, I am confident that it cannot be
improved upon, and on this point Mr. Wotherspoon is of the same opinion.


405.--CARD MAGIC SQUARES.

Arrange the cards as follows for the three new squares:--

    3  2  4      6  5  7    9  8 10
    4  3  2      7  6  5   10  9  8
    2  4  3      5  7  6    8 10  9

Three aces and one ten are not used. The summations of the four squares
are thus: 9, 15, 18, and 27--all different, as required.


406.--THE EIGHTEEN DOMINOES.

[Illustration]

The illustration explains itself. It will be found that the pips in
every column, row, and long diagonal add up 18, as required.


407.--TWO NEW MAGIC SQUARES.

Here are two solutions that fulfil the conditions:--

[Illustration:

    SUBTRACTING       DIVIDING
    11  4 14 13     36  8 54  27
    16  7  1  2    216 12  1   2
     6  5  3 12      6  3  4  72
     9 19  8 15      9 18 24 108

]

The first, by subtracting, has a constant 8, and the associated pairs
all have a difference of 4. The second square, by dividing, has a
constant 9, and all the associated pairs produce 3 by division. These
are two remarkable and instructive squares.


408.--MAGIC SQUARES OF TWO DEGREES.

The following is the square that I constructed. As it stands the
constant is 260. If for every number you substitute, in its allotted
place, its square, then the constant will be 11,180. Readers can write
out for themselves the second degree square.

[Illustration:

     7 53 | 41 27 |  2 52 | 48 30
    12 58 | 38 24 | 13 63 | 35 17
    ------+-------+-------+------
    51  1 | 29 47 | 54  8 | 28 42
    64 14 | 18 36 | 57 11 | 23 37
    ------+-------+-------+------
    25 43 | 55  5 | 32 46 | 50  4
    22 40 | 60 10 | 19 33 | 61 15
    ------+-------+-------+------
    45 31 |  3 49 | 44 26 |  6 56
    34 20 | 16 62 | 39 21 |  9 59

]

The main key to the solution is the pretty law that if eight numbers sum
to 260 and their squares to 11,180, then the same will happen in the
case of the eight numbers that are complementary to 65. Thus 1 + 18 + 23
+ 26 + 31 + 48 + 56 + 57 = 260, and the sum of their squares is 11,180.
Therefore 64 + 47 + 42 + 39 + 34 + 17 + 9 + 8 (obtained by subtracting
each of the above numbers from 65) will sum to 260 and their squares to
11,180. Note that in every one of the sixteen smaller squares the two
diagonals sum to 65. There are four columns and four rows with their
complementary columns and rows. Let us pick out the numbers found in the
2nd, 1st, 4th, and 3rd rows and arrange them thus :--

[Illustration:

    1    8    28    29    42    47    51    54
    2    7    27    30    41    48    52    53
    3    6    26    31    44    45    49    56
    4    5    25    32    43    46    50    55

]

Here each column contains four consecutive numbers cyclically arranged,
four running in one direction and four in the other. The numbers in the
2nd, 5th, 3rd, and 8th columns of the square may be similarly grouped.
The great difficulty lies in discovering the conditions governing these
groups of numbers, the pairing of the complementaries in the squares of
four and the formation of the diagonals. But when a correct solution is
shown, as above, it discloses all the more important keys to the
mystery. I am inclined to think this square of two degrees the most
elegant thing that exists in magics. I believe such a magic square
cannot be constructed in the case of any order lower than 8.


409.--THE BASKETS OF PLUMS.

As the merchant told his man to distribute the contents of one of the
baskets of plums "among some children," it would not be permissible to
give the complete basketful to one child; and as it was also directed
that the man was to give "plums to every child, so that each should
receive an equal number," it would also not be allowed to select just as
many children as there were plums in a basket and give each child a
single plum. Consequently, if the number of plums in every basket was a
prime number, then the man would be correct in saying that the proposed
distribution was quite impossible. Our puzzle, therefore, resolves
itself into forming a magic square with nine different prime numbers.

[Illustration]

             A                      B
    +-----+-----+-----+    +-----+-----+-----+
    |     |     |     |    |     |     |     |
    |  7  | 61  | 43  |    | 83  | 29  | 101 |
    |_____|_____|_____|    |_____|_____|_____|
    |     |     |     |    |     |     |     |
    | 73  | 37  |  1  |    | 89  | 71  | 53  |
    |_____|_____|_____|    |_____|_____|_____|
    |     |     |     |    |     |     |     |
    | 31  | 13  | 67  |    | 41  | 113 | 59  |
    |     |     |     |    |     |     |     |
    +-----+-----+-----+    +-----+-----+-----+

             C                      D
    +-----+-----+-----+    +-----+-----+-----+
    |     |     |     |    |     |     |     |
    | 103 | 79  | 37  |    |1669 | 199 |1249 |
    |_____|_____|_____|    |_____|_____|_____|
    |     |     |     |    |     |     |     |
    |  7  | 73  | 139 |    | 619 |1039 |1459 |
    |_____|_____|_____|    |_____|_____|_____|
    |     |     |     |    |     |     |     |
    | 109 | 67  | 43  |    | 829 |1879 | 409 |
    |     |     |     |    |     |     |     |
    +-----+-----+-----+    +-----+-----+-----+

In Diagram A we have a magic square in prime numbers, and it is the one
giving the smallest constant sum that is possible. As to the little trap
I mentioned, it is clear that Diagram A is barred out by the words
"every basket contained plums," for one plum is not plums. And as we
were referred to the baskets, "as shown in the illustration," it is
perfectly evident, without actually attempting to count the plums, that
there are at any rate more than 7 plums in every basket. Therefore C is
also, strictly speaking, barred. Numbers over 20 and under, say, 250
would certainly come well within the range of possibility, and a large
number of arrangements would come within these limits. Diagram B is one
of them. Of course we can allow for the false bottoms that are so
frequently used in the baskets of fruitsellers to make the basket appear
to contain more fruit than it really does.

Several correspondents assumed (on what grounds I cannot think) that in
the case of this problem the numbers cannot be in consecutive
arithmetical progression, so I give Diagram D to show that they were
mistaken. The numbers are 199, 409, 619, 829, 1,039, 1,249, 1,459,
1,669, and 1,879--all primes with a common difference of 210.


410.--THE MANDARIN'S "T" PUZZLE.

There are many different ways of arranging the numbers, and either the 2
or the 3 may be omitted from the "T" enclosure. The arrangement that I
give is a "nasik" square. Out of the total of 28,800 nasik squares of
the fifth order this is the only one (with its one reflection) that
fulfils the "T" condition. This puzzle was suggested to me by Dr. C.
Planck.


[Illustration: THE MANDARIN'S "T" PUZZLE.

    +-----+-----+-----+-----+-----+
    |     |     |     |     |     |
    |  19 |  23 |  11 |  5  |  7  |
    |_____|_____|_____|_____|_____|
    |     |     |     |     |     |
    |  1  |  10 |  17 |  24 |  13 |
    |_____|_____|_____|_____|_____|
    |     |     |     |     |     |
    |  22 |  14 |  3  |  6  |  20 |
    |_____|_____|_____|_____|_____|
    |     |     |     |     |     |
    |  8  |  16 |  25 |  12 |  4  |
    |_____|_____|_____|_____|_____|
    |     |     |     |     |     |
    |  15 |  2  |  9  |  18 |  21 |
    |     |     |     |     |     |
    +-----+-----+-----+-----+-----+


411.--A MAGIC SQUARE OF COMPOSITES.

The problem really amounts to finding the smallest prime such that the
next higher prime shall exceed it by 10 at least. If we write out a
little list of primes, we shall not need to exceed 150 to discover what
we require, for after 113 the next prime is 127. We can then form the
square in the diagram, where every number is composite. This is the
solution in the smallest numbers. We thus see that the answer is arrived
at quite easily, in a square of the third order, by trial. But I propose
to show how we may get an answer (not, it is true, the one in smallest
numbers) without any tables or trials, but in a very direct and rapid
manner.

[Illustration]

    +-----+-----+-----+
    |     |     |     |
    | 121 | 114 | 119 |
    |_____|_____|_____|
    |     |     |     |
    | 116 | 118 | 120 |
    |_____|_____|_____|
    |     |     |     |
    | 117 | 122 | 115 |
    |     |     |     |
    +-----+-----+-----+

First write down any consecutive numbers, the smallest being greater
than 1--say, 2, 3, 4, 5, 6, 7, 8, 9, 10. The only factors in these
numbers are 2, 3, 5, and 7. We therefore multiply these four numbers
together and add the product, 210, to each of the nine numbers. The
result is the nine consecutive composite numbers, 212 to 220 inclusive,
with which we can form the required square. Every number will
necessarily be divisible by its difference from 210. It will be very
obvious that by this method we may find as many consecutive composites
as ever we please. Suppose, for example, we wish to form a magic square
of sixteen such numbers; then the numbers 2 to 17 contain the factors 2,
3, 5, 7, 11, 13, and 17, which, multiplied together, make 510510 to be
added to produce the sixteen numbers 510512 to 510527 inclusive, all of
which are composite as before.

But, as I have said, these are not the answers in the smallest numbers:
for if we add 523 to the numbers 1 to 16, we get sixteen consecutive
composites; and if we add 1,327 to the numbers 1 to 25, we get
twenty-five consecutive composites, in each case the smallest numbers
possible. Yet if we required to form a magic square of a hundred such
numbers, we should find it a big task by means of tables, though by the
process I have shown it is quite a simple matter. Even to find
thirty-six such numbers you will search the tables up to 10,000 without
success, and the difficulty increases in an accelerating ratio with each
square of a larger order.


412.--THE MAGIC KNIGHT'S TOUR.

    +----+----+----+----+----+----+----+----+
    | 46 | 55 | 44 | 19 | 58 |  9 | 22 |  7 |
    +----+----+----+----+----+----+----+----+
    | 43 | 18 | 47 | 56 | 21 |  6 | 59 | 10 |
    +----+----+----+----+----+----+----+----+
    | 54 | 45 | 20 | 41 | 12 | 57 |  8 | 23 |
    +----+----+----+----+----+----+----+----+
    | 17 | 42 | 53 | 48 |  5 | 24 | 11 | 60 |
    +----+----+----+----+----+----+----+----+
    | 52 |  3 | 32 | 13 | 40 | 61 | 34 | 25 |
    +----+----+----+----+----+----+----+----+
    | 31 | 16 | 49 |  4 | 33 | 28 | 37 | 62 |
    +----+----+----+----+----+----+----+----+
    |  2 | 51 | 14 | 29 | 64 | 39 | 26 | 35 |
    +----+----+----+----+----+----+----+----+
    | 15 | 30 |  1 | 50 | 27 | 36 | 63 | 38 |
    +----+----+----+----+----+----+----+----+

Here each successive number (in numerical order) is a knight's move from
the preceding number, and as 64 is a knight's move from 1, the tour is
"re-entrant." All the columns and rows add up 260. Unfortunately, it is
not a perfect magic square, because the diagonals are incorrect, one
adding up 264 and the other 256--requiring only the transfer of 4 from
one diagonal to the other. I think this is the best result that has ever
been obtained (either re-entrant or not), and nobody can yet say whether
a perfect solution is possible or impossible.


413.--A CHESSBOARD FALLACY.

[Illustration]

The explanation of this little fallacy is as follows. The error lies in
assuming that the little triangular piece, marked C, is exactly the same
height as one of the little squares of the board. As a matter of fact,
its height (if we make the sixty-four squares each a square inch) will
be 1+1/7 in. Consequently the rectangle is really 9+1/7 in. by 7 in., so
that the area is sixty-four square inches in either case. Now, although
the pieces do fit together exactly to form the perfect rectangle, yet
the directions of the horizontal lines in the pieces will not coincide.
The new diagram above will make everything quite clear to the reader.


414.--WHO WAS FIRST?

Biggs, who saw the smoke, would be first; Carpenter, who saw the bullet
strike the water, would be second; and Anderson, who heard the report,
would be last of all.


415.--A WONDERFUL VILLAGE.

When the sun is in the horizon of any place (whether in Japan or
elsewhere), he is the length of half the earth's diameter more distant
from that place than in his meridian at noon. As the earth's
semi-diameter is nearly 4,000 miles, the sun must be considerably more
than 3,000 miles nearer at noon than at his rising, there being no
valley even the hundredth part of 1,000 miles deep.


416.--A CALENDAR PUZZLE.

The first day of a century can never fall on a Sunday; nor on a
Wednesday or a Friday.


417.--THE TIRING-IRONS.

I will give my complete working of the solution, so that readers may see
how easy it is when you know how to proceed. And first of all, as there
is an even number of rings, I will say that they may all be taken off in
one-third of (2^(n + 1) - 2) moves; and since n in our case is 14, all
the rings may be taken off in 10,922 moves. Then I say 10,922 - 9,999 =
923, and proceed to find the position when only 923 out of the 10,922
moves remain to be made. Here is the curious method of doing this. It is
based on the binary scale method used by Monsieur L. Gros, for an
account of which see W.W. Rouse Ball's _Mathematical Recreations_.

Divide 923 by 2, and we get 461 and the remainder 1; divide 461 by 2,
and we get 230 and the remainder 1; divide 230 by 2, and we get 115
and the remainder nought. Keep on dividing by 2 in this way as long as
possible, and all the remainders will be found to be 1, 1, 1, 0, 0, 1,
1, 0, 1, 1, the last remainder being to the left and the first
remainder to the right. As there are fourteen rings and only ten
figures, we place the difference, in the form of four noughts, in
brackets to the left, and bracket all those figures that repeat a
figure on their left. Then we get the following arrangement: (0 0 0 0)
1 (1 1) 0 (0) 1 (1) 0 1 (1). This is the correct answer to the puzzle,
for if we now place rings below the line to represent the figures in
brackets and rings on the line for the other figures, we get the
solution in the required form, as below:--

         O    O   O   OO
    -------------------------
    OOOO   OO   O   O    O

This is the exact position of the rings after the 9,999th move has been
made, and the reader will find that the method shown will solve any
similar question, no matter how many rings are on the tiring-irons. But
in working the inverse process, where you are required to ascertain the
number of moves necessary in order to reach a given position of the
rings, the rule will require a little modification, because it does not
necessarily follow that the position is one that is actually reached in
course of taking off all the rings on the irons, as the reader will
presently see. I will here state that where the total number of rings is
odd the number of moves required to take them all off is one-third of
(2^(n + 1) - 1).

With n rings (where n is _odd_) there are 2^n positions counting all on
and all off. In (1/3)(2^(n + 1) + 2) positions they are all removed. The
number of positions not used is (1/3)(2^n - 2).

With n rings (where n is _even_) there are 2^n positions counting all on
and all off. In (2^(n + 1) + 1) positions they are all removed. The
number of positions not used is here (1/3)(2^n - 1).

It will be convenient to tabulate a few cases.

    +--------+------------+-----------+-----------+
    | No. of | Total      | Positions | Positions |
    | Rings. | Positions. | used.     | not used. |
    +--------+------------+-----------+-----------+
    |   1    |      2     |      2    |     0     |
    |   3    |      8     |      6    |     2     |
    |   5    |     32     |     22    |    10     |
    |   7    |    128     |     86    |    42     |
    |   9    |    512     |    342    |   170     |
    |        |            |           |           |
    |   2    |      4     |      3    |     1     |
    |   4    |     16     |     11    |     5     |
    |   6    |     64     |     43    |     21    |
    |   8    |    256     |    171    |     85    |
    |  10    |   1024     |    683    |    341    |
    +--------+------------+-----------+-----------+

Note first that the number of _positions used_ is one more than the
number of _moves_ required to take all the rings off, because we are
including "all on" which is a position but not a move. Then note that
the number of _positions not used_ is the same as the number of _moves
used_ to take off a set that has one ring fewer. For example, it takes
85 moves to remove 7 rings, and the 42 positions not used are exactly
the number of moves required to take off a set of 6 rings. The fact is
that if there are 7 rings and you take off the first 6, and then wish to
remove the 7th ring, there is no course open to you but to reverse all
those 42 moves that never ought to have been made. In other words, you
must replace all the 7 rings on the loop and start afresh! You ought
first to have taken off 5 rings, to do which you should have taken off 3
rings, and previously to that 1 ring. To take off 6 you first remove 2
and then 4 rings.


418.--SUCH A GETTING UPSTAIRS.

Number the treads in regular order upwards, 1 to 8. Then proceed as
follows: 1 (step back to floor), 1, 2, 3 (2), 3, 4, 5 (4), 5, 6, 7 (6),
7, 8, landing (8), landing. The steps in brackets are taken in a
backward direction. It will thus be seen that by returning to the floor
after the first step, and then always going three steps forward for one
step backward, we perform the required feat in nineteen steps.


419.--THE FIVE PENNIES.

[Illustration]

First lay three of the pennies in the way shown in Fig. 1. Now hold the
remaining two pennies in the position shown in Fig. 2, so that they
touch one another at the top, and at the base are in contact with the
three horizontally placed coins. Then the five pennies will be
equidistant, for every penny will touch every other penny.


420.--THE INDUSTRIOUS BOOKWORM.

The hasty reader will assume that the bookworm, in boring from the first
to the last page of a book in three volumes, standing in their proper
order on the shelves, has to go through all three volumes and four
covers. This, in our case, would mean a distance of 9½ in., which is
a long way from the correct answer. You will find, on examining any
three consecutive volumes on your shelves, that the first page of Vol.
I. and the last page of Vol. III. are actually the pages that are
nearest to Vol. II., so that the worm would only have to penetrate four
covers (together, ½ in.) and the leaves in the second volume (3 in.),
or a distance of 3½ inches, in order to tunnel from the first page to
the last.


421.--A CHAIN PUZZLE.

To open and rejoin a link costs threepence. Therefore to join the nine
pieces into an endless chain would cost 2s. 3d., whereas a new chain
would cost 2s. 2d. But if we break up the piece of eight links, these
eight will join together the remaining eight pieces at a cost of 2s. But
there is a subtle way of even improving on this. Break up the two pieces
containing three and four links respectively, and these seven will join
together the remaining seven pieces at a cost of only 1s. 9d.


422.--THE SABBATH PUZZLE.

The way the author of the old poser proposed to solve the difficulty was
as follows: From the Jew's abode let the Christian and the Turk set out
on a tour round the globe, the Christian going due east and the Turk due
west. Readers of Edgar Allan Poe's story, _Three Sundays in a Week_, or
of Jules Verne's _Round the World in Eighty Days_, will know that such a
proceeding will result in the Christian's gaining a day and in the
Turk's losing a day, so that when they meet again at the house of the
Jew their reckoning will agree with his, and all three may keep their
Sabbath on the same day. The correctness of this answer, of course,
depends on the popular notion as to the definition of a day--the average
duration between successive sun-rises. It is an old quibble, and quite
sound enough for puzzle purposes. Strictly speaking, the two travellers
ought to change their reckonings on passing the 180th meridian;
otherwise we have to admit that at the North or South Pole there would
only be one Sabbath in seven years.


423.--THE RUBY BROOCH.

In this case we were shown a sketch of the brooch exactly as it appeared
after the four rubies had been stolen from it. The reader was asked to
show the positions from which the stones "may have been taken;" for it
is not possible to show precisely how the gems were originally placed,
because there are many such ways. But an important point was the
statement by Lady Littlewood's brother: "I know the brooch well. It
originally contained forty-five stones, and there are now only
forty-one. Somebody has stolen four rubies, and then reset as small a
number as possible in such a way that there shall always be eight stones
in any of the directions you have mentioned."

[Illustration]

The diagram shows the arrangement before the robbery. It will be seen
that it was only necessary to reset one ruby--the one in the centre. Any
solution involving the resetting of more than one stone is not in
accordance with the brother's statement, and must therefore be wrong.
The original arrangement was, of course, a little unsymmetrical, and for
this reason the brooch was described as "rather eccentric."


424.--THE DOVETAILED BLOCK.

[Illustration]

The mystery is made clear by the illustration. It will be seen at once
how the two pieces slide together in a diagonal direction.


425.--JACK AND THE BEANSTALK.

The serious blunder that the artist made in this drawing was in
depicting the tendrils of

[Illustration]

the bean climbing spirally as at A above, whereas the French bean, or
scarlet runner, the variety clearly selected by the artist in the
absence of any authoritative information on the point, always climbs as
shown at B. Very few seem to be aware of this curious little fact.
Though the bean always insists on a sinistrorsal growth, as B, the hop
prefers to climb in a dextrorsal manner, as A. Why, is one of the
mysteries that Nature has not yet unfolded.


426.--THE HYMN-BOARD POSER.

This puzzle is not nearly so easy as it looks at first sight. It was
required to find the smallest possible number of plates that would be
necessary to form a set for three hymn-boards, each of which would show
the five hymns sung at any particular service, and then to discover the
lowest possible cost for the same. The hymn-book contains 700 hymns, and
therefore no higher number than 700 could possibly be needed.

Now, as we are required to use every legitimate and practical method of
economy, it should at once occur to us that the plates must be painted
on both sides; indeed, this is such a common practice in cases of this
kind that it would readily occur to most solvers. We should also
remember that some of the figures may possibly be reversed to form other
figures; but as we were given a sketch of the actual shapes of these
figures when painted on the plates, it would be seen that though the 6's
may be turned upside down to make 9's, none of the other figures can be
so treated.

It will be found that in the case of the figures 1, 2, 3, 4, and 5,
thirty-three of each will be required in order to provide for every
possible emergency; in the case of 7, 8, and 0, we can only need thirty
of each; while in the case of the figure 6 (which may be reversed for
the figure 9) it is necessary to provide exactly forty-two.

It is therefore clear that the total number of figures necessary is 297;
but as the figures are painted on both sides of the plates, only 149
such plates are required. At first it would appear as if one of the
plates need only have a number on one side, the other side being left
blank. But here we come to a rather subtle point in the problem.

Readers may have remarked that in real life it is sometimes cheaper when
making a purchase to buy more articles than we require, on the principle
of a reduction on taking a quantity: we get more articles and we pay
less. Thus, if we want to buy ten apples, and the price asked is a
penny each if bought singly, or ninepence a dozen, we should both save a
penny and get two apples more than we wanted by buying the full twelve.
In the same way, since there is a regular scale of reduction for plates
painted alike, we actually save by having two figures painted on that
odd plate. Supposing, for example, that we have thirty plates painted
alike with 5 on one side and 6 on the other. The rate would be 4¾d., and
the cost 11s. 10½d. But if the odd plate with, say, only a 5 on one side
of it have a 6 painted on the other side, we get thirty-one plates at
the reduced rate of 4½d., thus saving a farthing on each of the previous
thirty, and reducing the cost of the last one from 1s. to 4½d.

But even after these points are all seen there comes in a new
difficulty: for although it will be found that all the 8's may be on the
backs of the 7's, we cannot have all the 2's on the backs of the 1's,
nor all the 4 on the backs of the 3's, etc. There is a great danger, in
our attempts to get as many as possible painted alike, of our so
adjusting the figures that some particular combination of hymns cannot
be represented.

Here is the solution of the difficulty that was sent to the vicar of
Chumpley St. Winifred. Where the sign X is placed between two figures,
it implies that one of these figures is on one side of the plate and the
other on the other side.

                               d.   £ s.  d.
    31 plates painted 5 X 9 @  4½ = 0 11  7½
    30       "        7 X 8 @  4¾ = 0 11 10½
    21       "        1 X 2 @  7  = 0 12  3
    21       "        3 X 0 @  7  = 0 12  3
    12       "        1 X 3 @  9¼ = 0  9  3
    12       "        2 X 4 @  9¼ = 0  9  3
    12       "        9 X 4 @  9¼ = 0  9  3
     8       "        4 X 0 @ 10¼ = 0  6 10
     1       "        5 X 4 @ 12  = 0  1  0
     1       "        5 X 0 @ 12  = 0  1  0
    149 plates @ 6d. each         = 3 14  6
                                  ----------
                                   £7 19  1

Of course, if we could increase the number of plates, we might get the
painting done for nothing, but such a contingency is prevented by the
condition that the fewest possible plates must be provided.

This puzzle appeared in _Tit-Bits_, and the following remarks, made by
me in the issue for 11th December 1897, may be of interest.

The "Hymn-Board Poser" seems to have created extraordinary interest. The
immense number of attempts at its solution sent to me from all parts of
the United Kingdom and from several Continental countries show a very
kind disposition amongst our readers to help the worthy vicar of
Chumpley St. Winifred over his parochial difficulty. Every conceivable
estimate, from a few shillings up to as high a sum as £1,347, 10s.,
seems to have come to hand. But the astonishing part of it is that,
after going carefully through the tremendous pile of correspondence, I
find that only one competitor has succeeded in maintaining the
reputation of the _Tit-Bits_ solvers for their capacity to solve
anything, and his solution is substantially the same as the one given
above, the cost being identical. Some of his figures are differently
combined, but his grouping of the plates, as shown in the first column,
is exactly the same. Though a large majority of competitors clearly hit
upon all the essential points of the puzzle, they completely collapsed
in the actual arrangement of the figures. According to their methods,
some possible selection of hymns, such as 111, 112, 121, 122,211, cannot
be set up. A few correspondents suggested that it might be possible so
to paint the 7's that upside down they would appear as 2's or 4's; but
this would, of course, be barred out by the fact that a representation
of the actual figures to be used was given.


427.--PHEASANT-SHOOTING.

The arithmetic of this puzzle is very easy indeed. There were clearly 24
pheasants at the start. Of these 16 were shot dead, 1 was wounded in the
wing, and 7 got away. The reader may have concluded that the answer is,
therefore, that "seven remained." But as they flew away it is clearly
absurd to say that they "remained." Had they done so they would
certainly have been killed. Must we then conclude that the 17 that were
shot remained, because the others flew away? No; because the question
was not "how many remained?" but "how many still remained?" Now the poor
bird that was wounded in the wing, though unable to fly, was very active
in its painful struggles to run away. The answer is, therefore, that the
16 birds that were shot dead "still remained," or "remained still."


428.--THE GARDENER AND THE COOK.

Nobody succeeded in solving the puzzle, so I had to let the cat out of
the bag--an operation that was dimly foreshadowed by the puss in the
original illustration. But I first reminded the reader that this puzzle
appeared on April 1, a day on which none of us ever resents being made
an "April Fool;" though, as I practically "gave the thing away" by
specially drawing attention to the fact that it was All Fools' Day, it
was quite remarkable that my correspondents, without a single exception,
fell into the trap.

One large body of correspondents held that what the cook loses in stride
is exactly made up in greater speed; consequently both advance at the
same rate, and the result must be a tie. But another considerable
section saw that, though this might be so in a race 200 ft. straight
away, it could not really be, because they each go a stated distance at
"every bound," and as 100 is not an exact multiple of 3, the gardener at
his thirty-fourth bound will go 2 ft. beyond the mark. The gardener
will, therefore, run to a point 102 ft. straight away and return (204
ft. in all), and so lose by 4 ft. This point certainly comes into the
puzzle. But the most important fact of all is this, that it so happens
that the gardener was a pupil from the Horticultural College for Lady
Gardeners at, if I remember aright, Swanley; while the cook was a very
accomplished French chef of the hemale persuasion! Therefore "she (the
gardener) made three bounds to his (the cook's) two." It will now be
found that while the gardener is running her 204 ft. in 68 bounds of 3
ft., the somewhat infirm old cook can only make 45+1/3 of his 2 ft.
bounds, which equals 90 ft. 8 in. The result is that the lady gardener
wins the race by 109 ft. 4 in. at a moment when the cook is in the air,
one-third through his 46th bound.

The moral of this puzzle is twofold: (1) Never take things for granted
in attempting to solve puzzles; (2) always remember All Fools' Day when
it comes round. I was not writing of _any_ gardener and cook, but of a
_particular_ couple, in "a race that I witnessed." The statement of the
eye-witness must therefore be accepted: as the reader was not there, he
cannot contradict it. Of course the information supplied was
insufficient, but the correct reply was: "Assuming the gardener to be
the 'he,' the cook wins by 4 ft.; but if the gardener is the 'she,' then
the gardener wins by 109 ft. 4 in." This would have won the prize.
Curiously enough, one solitary competitor got on to the right track, but
failed to follow it up. He said: "Is this a regular April 1 catch,
meaning that they only ran 6 ft. each, and consequently the race was
unfinished? If not, I think the following must be the solution,
supposing the gardener to be the 'he' and the cook the 'she.'" Though
his solution was wrong even in the case he supposed, yet he was the only
person who suspected the question of sex.


429.--PLACING HALFPENNIES.

Thirteen coins may be placed as shown on page 252.


430.--FIND THE MAN'S WIFE.

There is no guessing required in this puzzle. It is all a question of
elimination. If we can pair off any five of the ladies with their
respective husbands, other than husband No. 10, then the remaining lady
must be No. 10's wife.

[Illustration: PLACING HALFPENNIES.]

I will show how this may be done. No. 8 is seen carrying a lady's
parasol in the same hand with his walking-stick. But every lady is
provided with a parasol, except No. 3; therefore No. 3 may be safely
said to be the wife of No. 8. Then No. 12 is holding a bicycle, and the
dress-guard and make disclose the fact that it is a lady's bicycle. The
only lady in a cycling skirt is No. 5; therefore we conclude that No. 5
is No. 12's wife. Next, the man No. 6 has a dog, and lady No. 11 is seen
carrying a dog chain. So we may safely pair No. 6 with No. 11. Then we
see that man No. 2 is paying a newsboy for a paper. But we do not pay
for newspapers in this way before receiving them, and the gentleman has
apparently not taken one from the boy. But lady No. 9 is seen reading a
paper. The inference is obvious--that she has sent the boy to her
husband for a penny. We therefore pair No. 2 with No. 9. We have now
disposed of all the ladies except Nos. 1 and 7, and of all the men
except Nos. 4 and 10. On looking at No. 4 we find that he is carrying a
coat over his arm, and that the buttons are on the left side;--not on
the right, as a man wears them. So it is a lady's coat. But the coat
clearly does not belong to No. 1, as she is seen to be wearing a coat
already, while No. 7 lady is very lightly clad. We therefore pair No. 7
lady with man No. 4. Now the only lady left is No. 1, and we are
consequently forced to the conclusion that she is the wife of No. 10.
This is therefore the correct answer.



INDEX.


        Abbot's Puzzle, The, 20, 161.
          ---- Window, The, 87, 213.

        Academic Courtesies, 18, 160.

        Acrostic Puzzle, An, 84, 210.

        Adam and Eve and the Apples, 18.

        Aeroplanes, The Two, 2, 148.

        Age and Kinship Puzzles, 6.
          ---- Concerning Tommy's, 7, 153.
          ---- Mamma's, 7, 152.
          ---- Mrs. Timpkins's, 7, 152.
          ---- Rover's, 7, 152.

        Ages, The Family, 7, 152.
          ---- Their, 7, 152.

        Alcuin, Abbot, 20, 112.

        Almonds, The Nine, 64, 195.

        Amazons, The, 94, 221.

        Andrews, W.S., 125.

        Apples, A Deal in, 3, 149.
          ---- Buying, 6, 151.
          ---- The Ten, 64, 195.

        Approximations in Dissection, 28.

        Arithmetical and Algebraical Problems, 1.
          ---- Various, 17.

        Arthur's Knights, King, 77, 203.

        Artillerymen's Dilemma, 26, 167.

        Asparagus, Bundles of, 140.

        Aspects all due South, 137.

        Associated Magic Squares, 120.

        Axiom, A Puzzling, 138.


        Bachet de Méziriac, 90, 109, 112.

        Bachet's Square, 90, 216.

        Ball Problem, The, 51, 183.

        Ball, W.W. Rouse, 109, 204, 248.

        Balls, The Glass, 78, 204.

        Banker's Puzzle, The, 25, 165.

        Bank Holiday Puzzle, A, 73, 201.

        Banner Puzzle, The, 46, 179.
          ---- St. George's, 50, 182.

        Barrel Puzzle, The, 109, 235.

        Barrels of Balsam, The, 82, 208.

        Beanfeast Puzzle, A, 2, 148.

        Beef and Sausages, 3, 149.

        Beer, The Barrel of, 13, 155.

        Bell-ropes, Stealing the, 49, 181.

        Bells, The Peal of, 78, 204.

        Bergholt, E., 116, 119, 125.

        Betsy Ross Puzzle, The, 40, 176.

        Bicycle Thief, The, 6, 152.

        Bishops--Guarded, 88, 214.
          ---- in Convocation, 89, 215.
          ---- Puzzle, A New, 98, 225.
          ---- Unguarded, 88, 214.

        Board, The Chess-, 85.
          ---- in Compartments, The, 102, 228.
          ---- Setting the, 105, 231.

        Boards with Odd Number of Squares, 86, 212.

        Boat, Three Men in a, 78, 204.

        Bookworm, The Industrious, 143, 248.

        Boothby, Guy, 154.

        Box, The Cardboard, 49, 181.
          ---- The Paper, 40.

        Boys and Girls, 67, 197.

        Bridges, The Monk and the, 75, 202.

        Brigands, The Five, 25, 164.

        Brocade, The Squares of, 47, 180.

        Bun Puzzle, The, 35, 170.

        Busschop, Paul, 172.

        Buttons and String Method, 230.


        Cab Numbers, The, 15, 157.

        Calendar Puzzle, A, 142, 247.

        _Canterbury Puzzles, The_, 14, 28, 58, 117, 121, 195, 202, 205, 206, 212, 213, 217, 233.

        Card Frame Puzzle, The, 114, 238.
          ---- Magic Squares, 123, 244.
          ---- Players, A Puzzle for, 78, 203.
          ---- Puzzle, The "T," 115, 239.
          ---- Triangles, 115, 239.

        Cards, The Cross of, 115, 238.

        Cardan, 142.

        Carroll, Lewis, 43.

        Castle Treasure, Stealing the, 113, 237.

        Cats, the Wizard's, 42, 178.

        Cattle, Judkins's, 6, 151.
          ---- Market, At a, 1, 148.

        Census Puzzle, A, 7, 152.

        Century Puzzle, The, 16, 158.
          ---- The Digital, 16, 159.

        Chain Puzzle, A, 144, 249.
          ---- The Antiquary's, 83, 209.
          ---- The Cardboard, 40, 176.

        Change, Giving, 4, 150.
          ---- Ways of giving, 151.

        Changing Places, 10, 154.

        Channel Island, 138.

        Charitable Bequest, A, 2, 148

        Charity, Indiscriminate, 2, 148.

        Checkmate, 107, 233.

        Cheesemonger, The Eccentric, 66, 196.

        Chequered Board Divisions, 85, 210.

        Cherries and Plums, 56, 189.

        Chess Puzzles, Dynamical, 96.
          ---- Statical, 88.
          ---- Various, 105.
          ---- Queer, 107, 233.

        Chessboard, The, 85.
          ---- Fallacy, A, 141, 247.
          ---- Guarded, 95.
          ---- Non-attacking Arrangements, 96.
          ---- Problems, 84.
          ---- Sentence, The, 87, 214.
          ---- Solitaire, 108, 234.
          ---- The Chinese, 87, 213.
          ---- The Crowded, 91, 217.

        Chestnuts, Buying, 6, 152.

        Chinese Money, 4, 150.
          ---- Puzzle, Ancient, 107, 233.
          ---- ---- _The Fashionable_, 43.

        Christmas Boxes, The, 4, 150.
          ---- Present, Mrs. Smiley's, 46, 179.
          ---- Pudding, The, 43, 178.

        Cigar Puzzle, The, 119, 242.

        Circle, The Dissected, 69, 197.

        Cisterns, How to Make, 54, 188.

        Civil Service "Howler," 154.

        Clare, John, 58.

        Clock Formulæ, 154.
          ---- Puzzles, 9.
          ---- The Club, 10, 154.
          ---- The Railway Station, 11, 155.

        Clocks, The Three, 11, 154.

        Clothes Line Puzzle, The, 50, 182.

        Coast, Round the, 63, 195.

        Coincidence, A Queer, 2, 148.

        Coins, The Broken, 5, 150.
          ---- The Ten, 57, 190.
          ---- Two Ancient, 140.

        Combination and Group Problems, 76.

        Compasses Puzzle, The, 53, 186.

        Composite Magic Squares, 127, 246.

        Cone Puzzle, The, 55, 188.

        Corn, Reaping the, 20, 161.

        Cornfields, Farmer Lawrence's, 101, 227.

        Costermonger's Puzzle, The, 6, 152.

        Counter Problems, Moving, 58.
          ---- Puzzle, A New, 98, 225.
          ---- Solitaire, 107, 234.

        Counters, The Coloured, 91, 217.
          ---- The Forty-nine, 92, 217.
          ---- The Nine, 14, 156.
          ---- The Ten, 15, 156.

        Crescent Puzzle, The, 52, 184.

        Crescents of Byzantium, The Five, 92, 219.

        Cricket Match, The Village, 116, 239.
          ---- Slow, 116, 239.

        Cross and Triangle, 35, 169.
          ---- of Cards, 115, 238.
          ---- The Folded, 35, 169.
          ---- The Southern, 93, 220.

        Crosses, Counter,  81, 207.
          ---- from One, Two, 35, 168.
          ---- ---- Three, 169.

        Crossing River Problems, 112.

        Crusader, The, 106, 232.

        Cubes, Sums of, 165.

        Cushion Covers, The, 46, 179.

        Cutting-out Puzzle, A, 37, 172.

        Cyclists' Feast, The, 2, 148.


        Dairyman, The Honest, 110, 235.

        Definition, A Question of, 23, 163.

        De Fonteney, 112.

        Deified Puzzle, The, 74, 202.

        Delannoy, 112.

        De Morgan, A., 27.

        De Tudor, Sir Edwyn, 12, 155.

        Diabolique Magic Squares, 120.

        Diamond Puzzle, The, 74, 202.

        Dice, A Trick with, 116, 239.
          ---- Game, The Montenegrin, 119, 242.
          ---- Numbers, The, 17, 160.

        Die, Painting the, 84, 210,

        Digital Analysis, 157, 158.
          ---- Division, 16, 158.
          ---- Multiplication, 15, 156.
          ---- Puzzles, 13.

        Digits, Adding the, 16, 158.
          ---- and Squares, 14, 155.
          ---- Odd and Even, 14, 156.

        Dilemma, An Amazing, 106, 233.

        Diophantine Problem, 164.

        Dissection Puzzle, An Easy, 35, 170.
          ---- Puzzles, 27.
          ---- ---- Various, 35.

        Dividing Magic Squares, 124.

        Division, Digital, 16, 158.
          ---- Simple, 23, 163.

        Doctor's Query, The, 109, 235.

        Dogs Puzzle, The Five, 92, 218.

        Domestic Economy, 5, 151.

        Domino Frame Puzzle, The, 114, 238.

        Dominoes in Progression, 114, 237.
          ---- The Eighteen, 123, 245.
          ---- The Fifteen, 83, 209.
          ---- The Five, 114, 238.

        Donkey Riding, 13, 155.

        Dormitory Puzzle, A, 81, 208.

        Dovetailed Block, The, 145, 249.

        Drayton's _Polyolbion_, 58.

        Dungeon Puzzle, A, 97, 224.

        Dungeons, The Siberian, 123, 244.
          ---- The Spanish, 122, 244.

        Dutchmen's Wives, The, 26, 167.

        Dynamical Chess Puzzles, 96.


        Earth's Girdle, The, 139.

        _Educational Times Reprints_, 204.

        Eggs, A Deal in, 3, 149.
          ---- Obtaining the, 140.

        Election, The Muddletown, 19, 161.
          ---- The Parish Council, 19, 161.

        Eleven, The Mystic, 16, 159.

        Elopements, The Four, 113, 237.

        Elrick, E., 231.

        Engines, The Eight, 61, 194.

        Episcopal Visitation, An, 98, 225.

        Estate, Farmer Wurzel's, 51, 184.

        Estates, The Yorkshire, 51, 183.

        Euclid, 31, 138.

        Euler, L., 165.

        Exchange Puzzle, The, 66, 196.


        Fallacy, A Chessboard, 141, 247.

        Family Party, A, 8, 153.

        Fare, The Passenger's, 13, 155.

        Farmer and his Sheep, The, 22, 163.

        Fence Problem, A, 21, 162.

        Fences, The Landowner's, 42, 178.

        Fermat, 164, 168.

        Find the Man's Wife, 147, 251.

        Fly on the Octahedron, The, 70, 198.

        Fog, Mr. Gubbins in a, 18, 161.

        Football Players, The, 116, 240.

        Fraction, A Puzzling, 138.

        Fractions, More Mixed, 16, 159.

        Frame Puzzle, The Card, 114, 238.
          ---- ---- The Domino, 114, 238.

        Frankenstein, E.N., 232.

        Frénicle, B., 119, 168.

        Frogs, The Educated, 59, 194-
          ---- The Four, 103, 229.
          ---- The Six, 59, 193.

        Frost, A.H., 120.


        Games, Puzzle, 117.
          ---- Problems concerning, 114.

        Garden, Lady Belinda's, 52, 186.
          ---- Puzzle, The, 49, 182.

        Gardener and the Cook, The, 146, 251.

        Geometrical Problems, 27.
          ---- Puzzles, Various, 49.

        George and the Dragon, St., 101, 227.

        Getting Upstairs, Such a, 143, 248.

        Girdle, the Earth's, 139.

        Goat, The Tethered, 53, 186.

        Grand Lama's Problem, The, 86, 212.

        Grasshopper Puzzle, The, 59, 193.

        Greek Cross Puzzles, 28.
          ---- ---- Three from One, 169.

        Greyhound Puzzle, The, 101, 227.

        Grocer and Draper, The, 5, 151.

        Gros, L., 248.

        Group Problems, Combination and, 76.

        Groups, The Three, 14, 156.

        Guarini, 229.


        Hairdresser's Puzzle, The, 137.

        Halfpennies, Placing, 147, 251.

        Hampton Court Maze solved, 133.

        Hannah's Puzzle, 75, 202.

        Hastings, The Battle of, 23, 164.

        Hatfield Maze solved, 136.

        Hat Puzzle, The, 67, 196.

        Hat-peg Puzzle, The, 93, 221.

        Hats, The Wrong, 78, 203.

        Hay, The Trusses of, 18, 161.

        Heads or Tails, 22, 163.

        Hearthrug, Mrs. Hobson's, 37, 172.

        Helmholtz, Von, 41.

        Honey, The Barrels of, 111, 236.

        Honeycomb Puzzle, The, 75, 202.

        Horse Race Puzzle, The, 117, 240.

        Horseshoes, The Two, 40, 175.

        Houdin, 68.

        Hydroplane Question, The, 12, 155.

        Hymn-board Poser, The, 145, 250.


        Icosahedron Puzzle, The, 70, 198.


        Jack and the Beanstalk, 145, 249.

        Jackson, John, 56.

        Jaenisch, C.F. de, 92.

        Jampots, Arranging the, 68, 197.

        Jealous Husbands, Five, 113, 236.

        Joiner's Problem, The, 36, 171.
          ---- ---- Another, 37, 171.

        Jolly Gaol-Birds, Eight, 122, 243.
          ---- ---- Nine, 122, 243.

        Journey, The Queen's, 100, 227.
          ---- The Rook's, 96, 224.

        Junior Clerks' Puzzle, The, 4, 150.

        Juvenile Puzzle, A, 68, 197.


        Kangaroos, The Four, 102, 228.

        Kelvin, Lord, 41.

        Kennel Puzzle, The, 105, 231.

        King and the Castles, The, 56, 189.
          ---- The Forsaken, 106, 232.

        Kite-flying Puzzle, A, 54, 187.

        Knight-guards, The, 95, 222.

        Knights, King Arthur's, 77, 203.
          ---- Tour, Magic, 127, 247.
          ---- ---- The Cubic, 103, 229.
          ---- ---- The Four, 103, 229.


        Labosne, A., 25, 90, 216.

        Labourer's Puzzle, The, 18, 160.

        _Ladies' Diary_, 26.

        Lagrange, J.L., 9.

        Laisant, C.A., 76.

        Lamp-posts, Painting the, 19, 161.

        Leap Year, 155.
          ---- ---- Ladies, The, 19, 161.

        Legacy, A Puzzling, 20, 161.

        Legal Difficulty, A, 23, 163.

        Le Plongeon, Dr., 29.

        Letter Block Puzzle, The, 60, 194.
          ---- Blocks, The Thirty-six, 91, 216.
          ---- Puzzle, The Fifteen, 79, 205.

        Level Puzzle, The, 74, 202.

        Linoleum Cutting, 48, 181.
          ---- Puzzle, Another, 49, 181.

        Lion and the Man, The, 97, 224.
          ---- Hunting, 94, 222.

        Lions and Crowns, 85, 212.
          ---- The Four, 88, 214.

        Lockers Puzzle, The, 14, 156.

        Locomotion and Speed Puzzles, 11.

        Lodging-house Difficulty, A, 61, 194.

        London and Wise, 131.

        Loyd, Sam, 8, 43, 44, 98, 144, 232, 233.

        Lucas, Edouard, 16, 76, 112, 121.

        Luncheons, The City, 77, 203.


        MacMahon, Major, 109.

        Magic Knight's Tour, 127, 247.
          ---- Square Problems, 119.
          ---- ---- Card, 123, 244.
          ---- ---- of Composites, 127, 246.
          ---- ---- of Primes, 125.
          ---- ---- of Two Degrees, 125, 245.
          ---- ---- Two New, 125, 245.
          ---- Strips, 121, 243.

        Magics, Subtracting, Multiplying, and Dividing, 124.

        Maiden, The Languishing, 97, 224.

        Mandarin's Puzzle, The, 103, 230.
          ---- "T" Puzzle, The, 126, 246.

        Marketing, Saturday, 27, 168.

        Market Women, The, 3, 149.

        Mary and Marmaduke, 7, 152.

        Mary, How Old was, 8, 153.

        Massacre of Innocents, 139.

        Match Mystery, A, 118, 241.
          ---- Puzzle, A New, 55, 188.

        Mates, Thirty-six, 106, 233.

        Mazes and how to thread Them, 127.

        Measuring, Weighing, and Packing Puzzles, 109.
          ---- Puzzle, New, 110, 235.

        Meeting, The Suffragists', 19, 161.

        Mellor, W.M.F., 242.

        Ménages, Problême de, 76.

        Mersenne, M., 168.

        Mice, Catching the, 65, 196.

        Milkmaid Puzzle, The, 50, 183.

        Millionaire's Perplexity, The, 3, 149.

        Mince Pies, The Twelve, 57, 191.

        Mine, Inspecting a, 71, 199.

        Miners' Holiday, The, 23, 163.

        Miser, The Converted, 21, 162.

        Mitre, Dissecting a, 35, 170.

        Monad, The Great, 39, 174.

        Money, A Queer Thing in, 2, 148.
          ---- Boxes, The Puzzling, 3, 149.
          ----, Pocket, 3, 149.
          ---- Puzzles, 1.
          ---- Puzzle, A New, 2, 148.
          ----, Square, 3, 149.

        _Monist, The_, 125.

        Monk and the Bridges, The, 75, 202.

        Monstrosity, The, 108, 234.

        Montenegrin Dice Game, The, 119, 242.

        Moreau, 76.

        Morris, Nine Men's, 58.

        Mosaics, A Problem in, 90, 215.

        Mother and Daughter, 7, 152.

        Motor-car Race, The, 117, 240.
          ---- Tour, The, 74, 201.
          ---- Garage Puzzle, The, 62, 195.

        Motorists, A Puzzle for, 73, 201.

        Mouse-trap Puzzle, The, 80, 206.

        Moving Counter Problems, 58.

        Multiplication, Digital, 15, 156.
          ---- Queer, 15, 157.
          ---- Simple, 23, 163.

        Multiplying Magic Squares, 124.

        Muncey, J.N., 125.

        Murray, Sir James, 44.


        Napoleon, 43, 44.

        Nasik Magic Squares, 120.

        Neighbours, Next-Door, 8, 153.

        Newton, Sir Isaac, 56.

        Nine Men's Morris, 58.

        Notation, Scales of, 149.

        Noughts and Crosses, 58, 117.

        _Nouvelles Annales de Mathématiques_, 14.

        Number Checks Puzzle, The, 16, 158.

        Numbers, Curious, 20, 162.

        Nuts, The Bag of, 8, 153.


        Observation, Defective, 4, 150.

        Octahedron, The Fly on the, 70, 198.

        Oval, How to draw an, 50, 182.

        Ovid's Game, 58.


        Packing in Russia, Gold, 111, 236.
          ---- Puzzles, Measuring, Weighing, and, 109.
          ---- Puzzle, A, 111, 236.

        Pandiagonal Magic Squares, 120.

        Papa's Puzzle, 53, 187.

        Pappus, 53.

        Paradox Party, The, 137.

        Party, A Family, 8, 153.

        Patchwork Puzzles, 46.
          ---- Puzzle, Another, 48, 180.
          ---- The Silk, 34, 168.

        Patience, _Strand_, 116, 239.

        Pawns, A Puzzle with, 94, 222.
          ---- Immovable, 106, 233.
          ---- The Six, 107, 233.
          ---- The Two, 105, 231.

        Pearls, The Thirty-three, 18, 160.

        Pebble Game, The, 117, 240.

        Pedigree, A Mixed, 8, 153.

        Pellian Equation, 164, 167.

        Pennies, The Five, 143, 248.
          ---- The Twelve, 65, 195.

        Pension, Drawing her, 12, 155.

        Pentagon and Square, The, 37, 172.
          ---- Drawing a, 37.

        Pfeffermann, M., 125.

        Pheasant-Shooting, 146, 251.

        Philadelphia Maze solved, 137.

        Pierrot's Puzzle, The, 15, 156.

        Pigs, The Seven, 41, 177.

        Planck, C., 220, 246.

        Plane Paradox, 138.

        Plantation Puzzle, A, 57, 189.
          ---- The Burmese, 58, 191.

        Plates and Coins, 65, 195.

        Plums, The Baskets of, 126, 245.

        Poe, E.A., 249.

        Points and Lines Problems, 56.

        Postage Stamps, The Four, 84, 210.

        Post-Office Perplexity, A, 1, 148.

        Potato Puzzle, The, 41, 177.

        Potatoes, The Basket of, 13, 155.

        Precocious Baby, The, 139.

        Presents, Buying, 2, 148.

        Prime Magic Squares, 125.

        Printer's Error, A, 20, 162.

        Prisoners, Exercise for, 104, 230.
          ---- The Ten, 62, 195.

        Probabilities, Two Questions in, 5, 150.

        Problems concerning Games, 114.

        Puss in the Corner, 118, 240.

        Puzzle Games, 117.

        Pyramid, Painting a, 83, 208.

        Pyramids, Square and Triangular, 167.

        Pythagoras, 31.


        "Queen, The," 120.

        Queens and Bishop Puzzle, 93, 219.
          ---- The Eight, 89, 215.

        Queen's Journey, The, 100, 227.
          ---- Tour, The, 98, 225.

        Quilt, Mrs. Perkins's, 47, 180.


        Race Puzzle, The Horse-, 117, 240.
          ---- The Motor-car, 117, 240.

        Rackbrane's Little Loss, 21, 163.

        Railway Muddle, A, 62, 194.
          ---- Puzzle, A, 61, 194.
          ---- Stations, The Three, 49, 182.

        _Rational Amusement for Winter Evenings_, 56.

        Rectangles, Counting the, 105, 232.

        Reiss, M., 58.

        Relationships, Queer, 8, 153.

        Reversals, A Puzzle in, 5, 151.

        River Axe, Crossing the, 112, 236.

        River Problems, Crossing, 112.

        Rookery, The, 105, 232.

        Rook's Journey, The, 96, 224.
          ---- Tour, The, 96, 223.

        Rooks, The Eight, 88, 214.
          ---- The Two, 117, 240.

        Round Table, The, 80, 205.

        Route Problems, Unicursal and, 68.

        Ruby Brooch, The, 144, 249.


        Sabbath Puzzle, The, 144, 249.

        Sailor's Puzzle, The, 71, 199.

        Sayles, H.A., 125.

        Schoolboys, The Nine, 80, 205.

        Schoolgirls, The Fifteen, 80, 204.

        Scramble, The Great, 19, 161.

        Sculptor's Problem, The, 23, 164.

        Second Day of Week, 139.

        See-Saw Puzzle, The, 22, 163.

        Semi-Nasik Magic Squares, 120.

        Senior and Junior, 140.

        Sevens, The Four, 17, 160.

        Sharp's Puzzle, 230.

        Sheepfold, The, 52, 184.

        Sheep Pens, The Six, 55, 189.
          ---- The Sixteen, 80, 206.
          ---- The Three, 92, 217.
          ---- Those Fifteen, 77, 203.

        Shopping Perplexity, A, 4, 150.

        Shuldham, C.D., 125, 126.

        Siberian Dungeons, The, 123, 244.

        Simpleton, The Village, 11, 155.

        Skater, The Scientific, 100, 226.

        Skeat, Professor, 127.

        Solitaire, Central, 63, 195.
          ---- Chessboard, 108, 234.
          ---- Counter, 107, 234.

        Sons, The Four, 49, 181.

        Spanish Dungeons, The, 122, 244.
          ---- Miser, The, 24, 164.

        Speed and Locomotion Puzzles, 11.
          ---- Average, 11, 155.

        Spiral, Drawing a, 50, 182.

        Spot on the Table, The, 17, 160.

        Square Numbers, Check for, 13.
          ---- ---- Digital, 16, 159.
          ---- of Veneer, The, 39, 175.
          ---- Puzzle, An Easy, 35, 170.

        Squares, A Problem in, 23, 163.
          ---- Circling the, 21, 162.
          ---- Difference of Two, 167.
          ---- Magic, 119.
          ---- Sum of Two, 165, 175.
          ---- The Chocolate, 35, 170.

        Stalemate, 106, 232.

        Stamp-licking, The Gentle Art of, 91, 217.

        Star Puzzle, The, 99, 226.

        Stars, The Eight, 89, 215.
          ---- The Forty-nine, 100, 226.

        Statical Chess Puzzles, 88.

        Sticks, The Eight, 53, 186.

        Stonemason's Problem, The, 25, 165.

        Stop-watch, The, 11, 154.

        _Strand Magazine, The_, 44, 116, 220.

        _Strand_ Patience, 116, 239.

        Stream, Crossing the, 112, 236.

        Strutt, Joseph, 59.

        Subtracting Magic Squares, 124.

        Sultan's Army, The, 25, 165.

        Suppers, The New Year's Eve, 3, 149.

        Surname, Find Ada's, 27, 168.

        Swastika, The, 29, 31, 169.


        "T" Card Puzzle, The, 115, 239.

        Table, The Round, 80, 205.

        Table-top and Stools, The, 38, 173.

        Tangram Paradox, A, 43, 178.

        Target, The Cross, 84, 210.

        Tarry, 112.

        Tartaglia, 25, 109, 112.

        Tea, Mixing the, 111, 235.

        Telegraph Posts, The, 139.

        Tennis Tournament, A, 78, 203.

        Tetrahedron, Building the, 82, 208.

        Thief, Catching the, 19, 161.

        Thrift, A Study in, 25, 166.

        Thompson, W.H., 232.

        Ticket Puzzle, The Excursion, 5, 151.

        Time Puzzle, A, 10, 153.
          ---- What was the, 10, 153.

        Tiring Irons, The, 142, 247.

        _Tit-Bits_, 58, 79, 124, 251.

        Torn Number, The, 20, 162.

        Torpedo Practice, 67, 196.

        Tour, The Cyclists', 71, 199.
          ---- The Grand, 72, 200.
          ---- The Queen's, 98, 225.
          ---- The Rook's, 96, 223.

        Towns, Visiting the, 70, 198.

        Trains, The Two, 11, 155.

        Treasure Boxes, The Nine, 24, 164.

        Trees, The Twenty-one, 57, 190.

        Trémaux, M., 133, 135.

        Triangle, The Dissected, 38, 173.

        Triangular Numbers, 13, 25, 166.
          ---- ---- Check for, 13.

        Troublesome Eight, The, 121, 242.

        Tube Inspector's Puzzle, The, 69, 198.
          ---- Railway, Heard on the, 8, 153.

        Turks and Russians, 58, 191.

        Turnings, The Fifteen, 70, 198.

        Twickenham Puzzle, The, 60, 194.

        Two Pieces Problem, The, 96.


        Unclassified Puzzles, 142.

        Unicursal and Route Problems, 68.

        Union Jack, The, 50, 69, 197.


        Vandermonde, A., 58, 103.

        Veil, Under the, 90, 216.

        Verne, Jules, 249.

        Victoria Cross Puzzle, The, 60, 194.

        Village, A Wonderful, 142, 247.

        Villages, The Three, 12, 155.

        Villas, The Eight, 80, 206.

        Vortex Rings, 40.

        Voter's Puzzle, The, 75, 202.


        Wall, The Puzzle, 52, 184.

        Wallis, J., 142.
          ---- (Another), 220.

        Walls, The Garden, 52, 185.

        Wapshaw's Wharf Mystery, The, 10, 153.

        War Puzzle Game, The, 118, 240.

        Wassail Bowl, The, 109, 235.

        Watch, A Puzzling, 10, 153.

        Water, Gas, and Electricity, 73, 200.

        _Weekly Dispatch, The_, 28, 124, 125, 146, 148.

        Weighing Puzzles, Measuring, Packing, and, 109.

        Wheels, Concerning, 55, 188.

        Who was First? 142, 247.

        Whyte, W.T., 147.

        Widow's Legacy, The, 2, 148.

        Wife, Find the Man's, 147, 251.

        Wilkinson, Rev. Mr., 193.

        Wilson, Professor, 29.

        Wilson's Poser, 9, 153.

        Wine and Water, 110, 235.
          ---- The Keg of, 110, 235.

        Wotherspoon, G., 244.


        Yacht race, The, 99, 226.

        Youthful Precocity, 1, 148.


        Zeno, 139.



THE END.





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