Henry Ernest Dudeney/Modern Puzzles

A man went into a bank to cash a cheque.

In handing over the money the cashier, by mistake, gave him pounds for shillings and shillings for pounds.

He pocketed the money without examining it, and spent half a crown on his way home, when he found that he possessed exactly twice the amount of the cheque.

He had no money in his pocket before going to the bank, and it is an interesting puzzle to find out what was the exact amount of that cheque.

$2$ - Pocket-money

I went down the street with a certain amount of money in my pocket,

and when I returned home I discovered that I had spent just half of it,

and that I now had just as many shillings as I previously had pounds,

and half as many pounds as I then had shillings.

How much money had I spent?

$3$ - Dollars and Cents

An American correspondent tells me that a man went into a store and spent one-half of the money that was in his pocket.

When he came out he found that he had just as many cents as he had dollars when he went in

and half as many dollars as he had cents when he went in.

How much money did he have on him when he entered?

$4$ - Loose Cash

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 half a sovereign?

$5$ - Doubling the Value

It is a curious fact that if you double $\pounds 6 \ 13 \shillings$, you get $\pounds 13 \ 6 \shillings$, which is merely changing the shillings and the pounds.

Can you find another sum of money that has the same peculiarity that, when multiplied by any number you may choose to select, will merely exchange the shillings and the pounds?

There is only one other multiplier and sum of money, besides the case shown, that will work.

What is it?

$6$ - Generous Gifts

A generous man set aside a certain sum of money for equal distribution weekly to the needy of his acquaintance.

One day he remarked:

"If there are five fewer applicants next week, you will each receive $2$ shillings more."

Unfortunately, instead of there being fewer there were actually four more persons applying for the gift.

"This means," he pointed out, "that you will each receive one shilling less."

Now, how much did each person receive at that last distribution?

$7$ - Selling Eggs

A woman took a certain number of eggs to market and sold some of them.

The next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.

On the third day the new remainder was trebled, and she sold the same number as before.

On the fourth day the remainder was quadrupled, and her sales the same as before.

On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions, and so disposed of her entire stock.

What is the smallest number of eggs she could have taken to market the first day, and how many did she sell daily?

$8$ - Buying Buns

Buns were being sold at three prices:

one a penny,

two a penny,

and three a penny.

Some children (there were as many boys as girls) were given sevenpence to spend on these buns, each receiving exactly alike.

How many buns did each receive?

Of course no buns were divided.

$9$ - Fractional Value

What part of threepence is one-third of twopence?

$10$ - Unrewarded Labour

A man persuaded Weary Willie, with some difficulty, to try to work on a job for $30$ days at $8$ shillings a day,

on the condition that he would forfeit $10$ shillings a day for every day that he idled.

At the end of the month neither owed the other anything, which entirely convinced Willie of the folly of labour.

Now, can you tell me just how many days' work he put in, and on how many days he idled?

$11$ - The Perplexed Banker

A man went into a bank with $1000$ sovereigns and $10$ bags.

He said,

"Place this money, please, in the bags in such a way that if I call and ask for a certain number of sovereigns

you can hand me over one or more bags, giving me the exact amount called for without opening any of the bags."

How was it to be done?

We are, of course, only concerned with a single application,

but he may ask for any exact number of pounds from $\pounds 1$ to $\pounds 1000$.

$12$ - A Weird Game

Seven men engaged in play.

Whenever a player won a game he doubled the money of each of the other players.

That is, he gave each player just as much money as each had in his pocket.

They played $7$ games and, strange to say, each won a game in turn in the order of their names,

which began with the letters $\text A$, $\text B$, $\text C$, $\text D$, $\text E$, $\text F$, and $\text G$.

When they had finished it was found that each man had exactly $2$ shillings and $8$ pence in his pocket.

How much had each man had in his pocket before play?

$13$ - Find the Coins

Three men, Abel, Best and Crewe, possessed money, all in silver coins.

Abel had one coin fewer than Best and one more than Crewe.

Abel gave Best and Crewe as much money as they already had,

then Best gave Abel and Crewe the same amount of money as they they held,

and finally Crewe gave Abel and Best as much money as they then had.

Each man then held exactly $10$ shillings.

To find what amount each man started with is not difficult.

But the sting of the puzzle is in the tail.

Each man held exactly the same coins (the fewest possible) amounting to $10$ shillings.

What were the coins and how were they originally distributed?

$14$ - An Easy Settlement

Three men, Andrews, Baker and Carey, sat down to play at some game.

When they put their money on the table it was found that they each possessed $2$ coins only, making altogether $\pounds 1 \ 4 \shillings 6 \oldpence$

At the end of play Andrews had lost $5$ shillings and Carey had lost sixpence, and they all squared up by simply exchanging the coins.

What were the exact coins that each held on rising from the table?

$15$ - Sawing Logs

"Your charge," said Mr. Grigsby, "was $30$ shillings for sawing up $3$ cords of wood made up of logs $3$ feet long,

each log to be cut into pieces $1$ foot in length."

"That is so," the man replied.

"Well, here are $4$ cords of logs, all of the same thickness as before,

only they are in $6$-feet lengths, instead of $3$ feet.

What will your charge be for cutting them all up into similar $1$-foot lengths?"

It is curious that they could not at once agree as to the fair price for the job.

What does the reader think the charge ought to be?

$16$ - Digging a Ditch

Here is a curious question that is more perplexing than it looks at first sight.

Abraham, an infirm old man, undertook to dig a ditch for $2$ pounds.

He engaged Benjamin, an able-bodied fellow, to assist him and share the money fairly according to their capacities.

Abraham could dig as fast as Benjamin could shovel out the dirt,

and Benjamin could dig four times as fast as Abraham could do the shovelling.

How should they divide the money?

Of course, we must assume their relative abilities for work to be the same in digging or shovelling.

$17$ - Name their Wives

A man left a legacy of $\pounds 1000$ to $3$ relatives and their wives.

The wives received together $\pounds 396$.

Jane received $\pounds 10$ more than Catherine,

and Mary received $\pounds 10$ more than Jane.

John Smith was given just as much as his wife,

Henry Snooks got half as much again as his wife,

and Tom Crowe received twice as much as his wife.

What was the Christian name of each man's wife?

$18$ - A Curious Paradox

A man went into a shop to pay a little bill that he owed.

On placing the money on the counter he found that he had not quite sufficient,

owing to a small purchase that he had thoughtlessly made on the way.

"I am so sorry," he said, "but you see I am a little short.

"Oh, that is all right," replied the tradesman, after looking at the money, "it won't make any difference to me."

"My good man!" exclaimed the customer ...

... etc. etc. ...

"... But really it will not affect my pocket in the slightest."

Can you explain the mystery?

It may come to you in a flash.

The tradesman was certainly correct.

$19$ - Market Transactions

A farmer goes to market and buys $100$ animals at a total cost of $\pounds 100$.

The price of cows being $\pounds 5$ each,

sheep $\pounds 1$ each,

and rabbits $1 \shillings$ each,

how many of each kind does he buy?

$20$ - The Seven Applewomen

Seven applewomen,

possessing respectively $20$, $40$, $60$, $80$, $100$, $120$, and $140$ apples,

went to market and sold all their apples at the same price,

and each received the same sum of money.

What was the price?

Age and Kinship Puzzles

$21$ - Their Ages

If you add the square of Tom's age to the age of Mary,

the sum is $62$;

but if you add the square of Mary's age to the age of Tom,

the result is $176$.

Can you say what are the ages of Tom and Mary?

$22$ - Mrs. Wilson's Family

Mrs. Wilson had three children, Edgar, James and John.

Their combined ages were half of hers.

Five years later, during which time Ethel was born, Mrs. Wilson's age equalled the total of all her children's ages.

Ten years more have passed, Daisy appearing during that interval.

At the latter event Edgar was as old as John and Ethel together.

The combined ages of all the children are now double Mrs. Wilson's age, which is, in fact, only equal to that of Edgar and James together.

Edgar's age also equals that of the two daughters.

Can you find all their ages?

$23$ - De Morgan and Another

Augustus de Morgan, the mathematician, who died in $1871$, used to boast that he was $x$ years old in the year $x^2$.

My living friend, Jasper Jenkins, wishing to improve on this, tells me he was $a^2 + b^2$ in $a^4 + b^4$;

that he was $2 m$ in the year $2 m^2$;

and that he was $3 n$ years old in the year $3 n^4$.

Can you give the years in which De Morgan and Jenkins were respectively born?

$24$ - "Simple" Arithmetic

Two gentlemen with an eccentric approach to philosophy were pinned down by your investigative reporter.

They wished to riddle my mathematical understanding.

"Our two ages combined," said the first, "is $44$."

"Don't be silly," said the other, "it's $1280$."

They looked at me and said, "You see, we didn't tell you how we were combining them."

It was clear to me that the first number was their difference and the second was their product.

Now, how old were these two gentlemen?

Clock Puzzles

$25$ - A Dreamland Clock

In a dream, I was travelling in a country where they had strange ways of doing things.

One little incident was fresh in my memory when I awakened.

I saw a clock and announced the time as it appeared to be indicated.

but my guide corrected me.

He said, "You are apparently not aware that the minute hand always moves in the opposite direction to the hour hand.

Except for this improvement, our clocks are precisely the same as those you have been accustomed to."

Now, as the hands were exactly together between the hours of $4$ and $5$ o'clock,

and they started together at noon,

what was the real time?

$26$ - What is the Time?

At what time are the two hands of a clock so situated that,

reckoning as minute points past $\textit {XII}$,

one is exactly the square of the distance of the other?

$27$ - The First-Born's Legacy

Mrs. Goodheart gave birth to twins.

The clock showed clearly that Tommy was born about an hour later than Freddy.

Mr. Goodheart, who died a few months earlier, had made a will leaving $\pounds 8400$,

and had taken the precaution to provide for the possibility of there being twins.

In such a case the money was to be divided in the following proportions:

two-thirds to the widow,

one-fifth to the first-born,

one-tenth to the other twin,

and one-twelfth to his brother.

Now, what is the exact amount that should be settled on Freddy?

Locomotion and Speed Puzzles

$28$ - Hill Climbing

Weary Willie went up a certain hill at the rate of $1 \tfrac 1 2$ miles per hour

and came down at the rate of $4 \tfrac 1 2$ miles per hour,

so that it took him just $6$ hours to make the double journey.

Now, how far was it to the top of the hill?

$29$ - Timing the Motor-car

"I was walking along the road at $3 \tfrac 1 2$ miles an hour," said Mr. Pipkins,

"when the motor-car dashed past me and only missed me by a few inches."

"Do you know what speed it was going?" asked his friend.

"Well, from the moment it passed me to its disappearance round a corner I took $27$ steps, and walking on reached that corner with $135$ steps more."

"Then, assuming you walked, and the car ran, each at a uniform rate, we can easily work out the speed."

$30$ - The Staircase Race

This is a rough sketch of a race up a staircase in which $3$ men took part.

Ackworth, who is leading, went up $3$ risers at a time, as arranged;

Barnden, the second man, went $4$ risers at a time,

and Croft, who is last, went $5$ at a time.

Undoubtedly Ackworth wins.

But the point is,

How many risers are there in the stairs, counting the top landing as a riser?

$31$ - A Walking Puzzle

A man set out at noon to walk from Appleminster to Boneyham,

and a friend of his started at $2$ p.m. on the same day to walk from Boneyham to Appleminster.

They met on the road at $5$ minutes past $4$ o'clock

and each man reached his destination at exactly the same time.

Can you say what time they both arrived?

$32$ - Riding in the Wind

A man on a bicycle rode a mile in $3$ minutes with the wind at his back,

but it took him $4$ minutes to return against the wind.

How long would it take him to ride a mile if there was no wind?

$33$ - A Rowing Puzzle

A crew can row a certain course upstream in $8 \tfrac 4 7$ minutes,

and, if there were no stream, they could row it in $7$ minutes less than it takes them to drift down the stream.

How long would it take to row down with the stream?

$34$ - The Moving Stairway

On one of the moving stairways on the London Tube Railway I find that if I walk down $26$ steps I require $30$ seconds to get to the bottom,

but if I take $34$ steps I require only $18$ seconds to reach the bottom.

What is the height of the stairway in steps?

$35$ - Sharing a Bicycle

Anderson and Brown have to go $20$ miles and arrive at exactly the same time.

They have only one bicycle.

Anderson can only walk $4$ miles an hour,

while Brown can walk $5$ miles an hour,

but Anderson can ride $10$ miles an hour to Brown's $8$ miles an hour.

How are they to arrange the journey?

$36$ - More Bicycling

Referring to the last puzzle, let us now consider the case where a third rider has to share the same bicycle.

As a matter of fact, I understand that Anderson and Brown have taken a man named Carter into partnership, and the position today is this:

Anderson, Brown and Carter walk respectively $4$, $5$ and $3$ miles per hour,

and ride respectively $10$, $8$ and $12$ miles per hour.

How are they to use that single bicycle so that all shall complete the $20$ miles journey at the same time?

$37$ - A Side-car Problem

Atkins, Baldwin and Clarke had to go a journey of $52$ miles across country.

Atkins had a motor-bicycle with sidecar for one passenger.

How was he to take one of his companions a certain distance,

drop him on the road to walk the remainder of the way,

and return to pick up the second friend,

so that they should all arrive at their destination at exactly the same time?

The motor-bicycle could do $20$ miles per hour,

Baldwin could walk $5$ miles per hour,

and Clarke could walk $4$ miles per hour.

$38$ - The Despatch-Rider

If an army $40$ miles long advances $40$ miles

while a despatch-rider gallops from the rear to the front,

delivers a despatch to the commanding general,

and returns to the rear,

how far has he to travel?

$39$ - The Two Trains

Two railway trains, one $400$ feet long and the other $200$ feet long, ran on parallel rails.

It was found that when they went in opposite directions they passed each other in $5$ seconds,

but when they ran in the same direction the faster train would pass the other in $15$ seconds.

Now, a curious passenger worked out from these facts the rate per hour at which each train ran.

Can the reader discover the correct answer?

$40$ - Pickleminster to Quickville

Two trains, $A$ and $B$, leave Pickleminster for Quickville at the same time as two trains, $C$ and $D$, leave Quickville for Pickleminster.

$A$ passes $C$ $120$ miles from Pickleminster and $D$ $140$ miles from Pickleminster.

$B$ passes $C$ $126$ miles from Quickville and $D$ half-way between Pickleminster and Quickville.

Now, what is the distance from Pickleminster to Quickville?

$41$ - The Damaged Engine

We were going by train from Anglechester to Clinkerton, and an hour after starting some accident happened to the engine.

We had to continue the journey at $\tfrac 3 5$ of the former speed, and it made us $2$ hours late at Clinkerton,

and the driver said that if only the accident had happened $50$ miles farther on the train would have arrived $40$ minutes sooner.

Can you tell from that statement just how far it is from Anglechester to Clinkerton?

$42$ - The Puzzle of the Runners

Two men ran a race round a circular course, going in opposite directions.

Brown was the best runner and gave Tompkins a start of $\tfrac 1 8$ of the distance.

But Brown, with a contempt for his opponent, took things too easily at the beginning,

and when he had run $\tfrac 1 6$ of his distance he met Tompkins,

and saw that his chance of winning the race was very small.

How much faster than he went before must Brown now run in order to tie with his competitor?

$43$ - The Two Ships

A correspondent asks the following question.

Two ships sail from one port to another -- $200$ nautical miles -- and return.

The Mary Jane travels outwards at $12$ miles an hour and returns at $8$ miles an hour,

thus taking $41 \tfrac 2 3$ hours on the double journey.

The Elizabeth Ann travels both ways at $10$ miles an hour, taking $40$ hours on the double journey.

Now, seeing that both ships travel at the average speed of $10$ miles per hour, why does the Mary Jane take longer than the Elizabeth Ann?

$44$ - Find the Distance

A man named Jones set out to walk from $A$ to $B$,

and on the road he met his friend Kenward, $10$ miles from $A$, who had left $B$ at exactly the same time.

Jones executed his commission at $B$ and, without delay, set out on his return journey,

while Kenward as promptly returned from $A$ to $B$.

They met $12$ miles from $B$.

Of course, each walked at a uniform rate throughout.

Now, how far is $A$ from $B$?

$45$ - The Man and the Dog

"Yes, when I take my dog for a walk," said a mathematical friend, "he frequently supplies me with some interesting problem to solve.

One day, for example, he waited, as I left the door, to see which way I should go,

and when I started he raced to the end of the road, immediately returning to me;

again racing to the end of the road and again returning.

He did this four times in all, at a uniform speed,

then ran at my side the remaining distance, which according to my paces measured $27$ yards.

I afterwards measured the distance from my door to the end of the road and found it to be $625$ feet.

Now, if I walk $4$ miles per hour, what is the speed of my dog when racing to and fro?"

$46$ - Baxter's Dog

Anderson set off from an hotel at San Remo at nine o'clock and had been walking an hour when Baxter went after him along the same road.

Baxter's dog started at the same time as his master and ran uniformly forwards and backwards between him and Anderson until the two men were together.

Anderson's speed is $2$, Baxter's $4$, and the dog's $10$ miles an hour.

How far had the dog run when Baxter overtook Anderson?

$47$ - The Runner's Refreshment

A man runs $n$ times round a circular track whose radius is $t$ miles.

He drinks $s$ quarts of beer for every mile that he runs.

Prove that he will only need one quart!

$48$ - Railway Shunting

How are the trains in our illustration to pass one another, and proceed with their engines in front?

The small side track is large enough to hold one engine or one carriage at a time, and no tricks, such as ropes and flying-switches, are allowed.

Every reversal -- that is, change of direction -- of an engine is counted as a move in the solution.

What is the smallest number of moves necessary?

$49$ - Exploring the Desert

Nine travellers, each possessing a motor-car, meet on the eastern edge of a desert.

They wish to explore the interior, always going due west.

Each car can travel $40$ miles on the contents of the engine tank,

which holds a gallon of petrol,

and each can carry $9$ extra gallon tins of petrol and no more.

Unopened tins can alone be transferred from car to car.

What is the greatest distance at which they can enter the desert without making any depots of petrol for the return journey?

$50$ - Exploring Mount Neverest

Professor Walkingholme, one of the exploring party, was allotted the special task of making a complete circuit of the base of the mountain at a certain level.

The circuit was exactly $100$ miles in length and he had to do it all alone on foot.

He could walk $20$ miles a day, but he could only carry rations for $2$ days at a time,

the rations for each day being packed in sealed boxes for convenience in dumping.

He walked his full $20$ miles every day and consumed $1$ day's ration as he walked.

What is the shortest time in which he could complete the circuit?

Digital Puzzles

$51$ - An Exceptional Number

A number is formed of $5$ successive digits (not necessarily in regular order)

so that the number formed by the first $2$ multiplied by the central digit will produce the number expressed by the last $2$.

$52$ - The Five Cards

I have $5$ cards bearing the figures $1$, $3$, $5$, $7$ and $9$.

How can I arrange them in a row so that the number formed by the $1$st pair multipied by the number formed with the last pair,

with the central number subtracted,

will produce a number composed of repetitions of one figure?

$53$ - Squares and Digits

What is the smallest square number that terminates with the greatest possible number of similar digits?

Thus the greatest possible number might be $5$ and the smallest square number with $5$ similar digits at the end might be $24677777$.

But this is certainly not a square number.

Of course, $0$ is not to be regarded as a digit.

$54$ - The Two Additions

Can you arrange the following figures into two groups of $4$ figures each so that each group shall add to the same sum?

$1 \ 2 \ 3 \ 4 \ 5 \ 7 \ 8 \ 9$

$55$ - The Repeated Quartette

If we multiply $64253$ by $365$ we get the product $23452345$, where the first $4$ figures are repeated.

What is the largest number that we can multiply by $365$ in order to produce a similar product of $8$ figures repeated in the same order?

There is no objection to a repetition of figures -- that is, the $4$ that are repeated need not be all different, as in the case shown.

$56$ - Easy Division

To divide the number $8 \, 101 \, 265 \, 822 \, 784$ by $8$, all we need to do is transfer the $8$ from the beginning to the end!

Can you find a number beginning with $7$ that can be divided by $7$ in the same simple manner?

$57$ - A Misunderstanding

An American correspondent asks me to find a number composed of any number of digits that may be correctly divided by $2$

by simply transferring the last figure to the beginning.

He has apparently come across our last puzzle with the conditions wrongly stated.

If you are to transfer the first figure to the end it is solved by $315 \, 789 \, 473 \, 684 \, 210 \, 526$,

and a solution may easily be found from this with any given figure at the beginning.

But if the figure is to be moved from the end to the beginning, there is no possible solution for the divisor $2$.

But there is a solution for the divisor $3$.

Can you find it?

$58$ - The Two Fours

The point [of the Four Fours puzzle] is to express all possible whole numbers with four fours (no more and no fewer), using the various arithmetical signs.

Thus:

$17 = 4 \times 4 + \dfrac 4 4$

and:

$50 = 44 + 4 + \sqrt 4$

All numbers up to $112$ inclusive may be solved, using only the signs for addition, subtraction, multiplication, division, square root, decimal points, and the factorial sign $4!$ which means $1 \times 2 \times 3 \times 4$, or $24$, but $113$ is impossible.

It is necessary to discover which numbers can be formed with one four, with two fours, and with three fours, and to record these for combination as required.

It is the failure to find some of these that leads to so much difficulty.

For example, I think very few discover that $64$ can be expressed with only two fours.

Can the reader do it?

$59$ - The Two Digits

Write down any $2$-figure number (different figures and no $0$)

and then express that number by writing the same figures in reverse order,

with or without arithmetical signs.

$60$ - Digital Coincidences

If I multiply, and also add, $9$ and $9$, I get $81$ and $18$, which contain the same figures.

If I multiply and add $2$ and $47$ I get $94$ and $49$ -- the same figures.

If I multiply and add $3$ and $24$ I get the same figures -- $72$ and $27$.

Can you find two numbers that, when multiplied and added will, in this simple manner, produce the same three figures?

$61$ - Palindromic Square Numbers

This is a curious subject for investigation -- the search for square numbers the figures of which read backwards and forwards alike.

Some of them are very easily found.

For example, the squares of $1$, $11$, $111$ and $1111$ are respectively $1$, $121$, $12321$, and $1234321$, all palindromes,

and the rule applies for any number of $1$'s provided the number does not contain more than nine.

But there are other cases that we may call irregular, such as the square of $264 = 69696$ and the square of $2285 = 5221225$.

Now, all the examples I have given contain an odd number of digits.

Can the reader find a case where the square palindrome contains an even number of figures?

$62$ - Factorizing

What are the factors (the numbers that will divide it without any remainder) of this number -- $1000000000001$?

This is easily done if you happen to know something about numbers of this peculiar form.

In fact, it is just as easy for me to give two factors if you insert, say $101$ noughts, instead of $11$, between the two ones.

There is a curious, easy, and beautiful rule for these cases.

Can you find it?

$63$ - Find the Factors

Find $2$ whole numbers with the smallest possible difference between them

which, when multiplied together, will produce $1234567890$.

$64$ - Dividing by Eleven

If the $9$ digits are written at haphazard in any order,

for example $4 \ 1 \ 2 \ 5 \ 3 \ 9 \ 7 \ 6 \ 8$, what are the chances that the number that happens to be produced will be divisible by $11$ without remainder?

The number I have written at random is not, I see, so divisible, but if I had happened to make the $1$ and the $8$ change places it would be.

$65$ - Dividing by $37$

I want to know whether the number $49,129,308,213$ is exactly divisible by $37$,

or if not, what is the remainder when so divided.

How may I do this quite easily without any process of actual division whatever?

$66$ - Another $37$ Division

If the $9$ digits are written at haphazard in any order, for example $412539768$,

what are the chances that the number that happens to be produced will be divisible by $37$ without remainder?

$67$ - A Digital Difficulty

Arrange the $10$ digits, $1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8 \ 9 \ 0$, in such order that they shall form a number

that may be divided by every number from $2$ to $18$ without in any case a remainder.

$68$ - Threes and Sevens

What is the smallest number composed only of the digits $3$ and $7$ that may be divided by $3$ and $7$,

and also the sum of its digits by $3$ and $7$, without any remainder.

$69$ - Root Extraction

In a conversation I had with Professor Simon Greathead, the eminent mathematician, ...

the extraction of the cube root.

"Ah," said the professor, "it is astounding what ignorance prevails ...

... the simple fact that, to extract the cube root of a number, all you have to do is to add together the digits.

Thus, ignoring the obvious case of the number $1$, if we want the cube root of $512$, add the digits -- $8$, and there you are!"

I suggested that that was a special case.

"Not at all," he replied. "Take another number at random -- $4913$ -- and the digits add to $17$, the cube of which is $4913$."

I did not presume to argue the point with the learned man,

but I will just ask the reader to discover all the other numbers whose cube root is the same as the sum of their digits.

$70$ - The Solitary Seven

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     ----
       ***
       ***
       ---
       ****
        ***
       ----
         ****
         ****
         ----

$71$ - A Complete Skeleton

       ****.****
    ---------
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     ***
     ---
       ***
       ***
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        ***
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        ---
         ** *
         ** *
         ----
            ****
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            ----

$72$ - Alphabetical Sums

      RSR
   ------
 PR)MTVVR
    MVR
    ---
     KKV
     KMD
     ---
      MVR
      MVR
      ---

$73$ - Alphabetical Arithmetic

                            F G
 Less A B multiplied by C = D E
                            ---
                   Leaving  H I
                            ---

$74$ - Queer Division

Find the smallest number which when divided successively by $45$, $454$, $4545$, and $45454$

leaves the remainders $4$, $45$, $454$, and $4545$ respectively.

$75$ - A Teasing Legacy

Professor Rackbrain left his typist what he called a trifle of a legacy if she was able to claim it.

The legacy was the largest amount that she could find in an addition sum,

where pounds, shillings and pence were all represented and no digit used more than once.

Every digit must be used once, a single $0$ may or may not appear, as in the examples below, and the dash may be employed in the manner shown.

 £  s. d.      £  s. d.
 -  3  7       4  2  5
 -  4  8       6  7  3
 -  5  9      --------
 1  6  -     £10  9  8
--------
£2  -  -

The young lady was cleverer than he thought.

What was the largest amount that she could claim?

$76$ - The Nine Volumes

In a small bookcase were arranged $9$ volumes of some big work,

numbered from $1$ to $9$ inclusive,

on $3$ shelves exactly as shown:

  26  5  9
         7
 184     3

The $9$ digits express money value.

You will see that they are so arranged that $\pounds 26 \ 5 \shillings \ 9 \oldpence$ multiplied by $7$ will produce $\pounds 184 \ 0 \shillings \ 3 \oldpence$

Every digit represented once, and yet they form a correct sum in the multiplication of money.

But in the blank shillings space in the bottom row is a slight defect, and I want to correct it.

The puzzle is to use the multiplier $3$, instead of $7$, and get a correct result with the $9$ volumes, without any blank space;

with pounds, shillings and pence all represented in both the top and bottom line.

$77$ - The Ten Volumes

As an extension of the last puzzle, let us introduce a $10$th volume marked $0$.

If we arrange the $10$ volumes as follows, we get a sum of money correctly multiplied by $2$.

  54  3  9
         2
 108  7  6

Can you do the same thing with the multiplier $4$ so that the $9$ digits and $0$ are all represented,

once and once only?

Various Arithmetical and Algebraical Problems

$78$ - The Miller's Toll

A miller was accustomed to take as toll one-tenth of the flour that he ground for his customers.

How much did he grind for a man who had just one bushel after the toll had been taken?

$79$ - Egg Laying

If a hen and a half lays an egg and a half in a day and a half,

how many and a half who lay better by half will lay half a score and a half in a week and a half?

$80$ - The Flocks of Sheep

Four brothers were comparing the number of sheep that they owned.

It was found that Claude had ten more sheep than Dan.

If Claude gave a quarter of his sheep to Ben,

then Claude and Adam would together have the same number as Ben and Dan together.

If, then, Adam gave one-third to Ben,

and Ben gave a quarter of what he then held to Claude,

who then passed on a fifth of his holding to Dan,

and then Ben divided one-quarter of the number he then possessed equally among Adam, Claude and Dan,

they would all have an equal number of sheep.

How many sheep did each possess?

$81$ - Pussy and the Mouse

"There's a mouse in one of these barrels," said the dog.

"Which barrel?" asked the cat.

"Why, the five hundredth barrel."

"What do you mean, the five hundredth? There are only five barrels in all."

"It's the five hundredth if you count backwards and forwards this way."

And the dog explained that if you count like this:

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

so that the seventh barrel would be the one marked $3$ and the twelfth barrel the one numbered $4$.

The story goes on laboriously to its inevitable conclusion that the mouse escapes before the cat has finished counting, until:

Now, which was the five hundredth barrel?

Can you find a quick way of arriving at the answer without making the actual count?

$82$ - Army Figures

A certain division in an army was composed of a little over twenty thousand men, made up of five brigades.

It was know that one third of the first brigade,

two-sevenths of the second brigade,

seven-twelfths of the third,

nine-thirteenths of the fourth,

and fifteen-twenty-seconds of the fifth brigades happened in every case to be the same number of men.

Can you discover how many men there were in every brigade?

$83$ - A Critical Vote

A meeting of a charitable society was held to decide whether the members should expand their operations.

It was arranged that during the count those in favour of the motion should remain standing,

and those who voted against should sit down.

"Ladies and gentlemen," said the chairman in due course, "I have the pleasure to announce that the motion is carried by a majority exactly equal to exactly a quarter of the opposition."

"Excuse me, sir," called somebody from the back, "but some of us over here could not sit down, because there are not enough chairs."

"Then those who wanted to sit down but couldn't are to hold up their hands ... I find there are a dozen of you, so the motion is lost by a majority of one."

Now, how many people voted at that meeting?

$84$ - The Three Brothers

The discussion arose before one of the tribunals as to which of a tradesman's three sons could best be spared for service in the Army.

"All I know as to their capabilities," said the father, "is this:

Arthur and Benjamin can do a certain quantity of work in eight days,

which Arthur and Charles will do in nine days,

and which Benjamin and Charles will take ten days over."

Of course, it was at once seen that as longer time was taken over the job whenever Charles was one of the pair,

he must be the slowest worker.

This was all they wanted to know, but it is an interesting puzzle to ascertain just how long each son would be required to do that job alone.

Can you discover?

$85$ - The House Number

A man said the house of his friend was in a long street,

numbered on his side one, two, three, and so on,

and that all the numbers on one side of him added up exactly the same as all the numbers on the other side of him.

He said he knew there were more than fifty houses on that side of the street,

but not as many as five hundred.

Can you discover the number of that house?

$86$ - A New Street Puzzle

Brown lived in a street which contained more than twenty houses, but fewer than five hundred,

all numbered one, two, three, four, etc., throughout.

Brown discovered that all the numbers from one upwards to his own number inclusive summed to exactly half the sum of all the numbers from one up to, and including, the last house.

Now what was the number of his house?

$87$ - Another Street Puzzle

A long street in Brussels has all the odd numbers of the houses on one side and all the even numbers on the other.

$(1)$ If a man lives in an odd-numbered house and all the numbers on one side of him, added together, equal the numbers on the other side,

how many houses are there, and what is the number of his house?

$(2)$ If a man lives on the even side and all the numbers on one side of him equal those on the other side,

how many houses are there, and what is his number?

We will assume that there are more than fifty houses on each side of the street and fewer than five hundred.

$88$ - Correcting an Error

Hermione was given a certain number to multiply by $409$,

but she made a blunder that is very common with mudbloods when learning the elements of simple arithmetic:

she placed the first figure of her product by $4$ below the second figure from the right instead of below the third.

We have all done that as youngsters (speak for yourself, Harry, old boy) when there has happened to be a $0$ in the multiplier.

The result of Hermione's mistake was that her answer was wrong by $328,320$, entirely in consequence of that little slip.

Now, what was the multiplicand?

$89$ - The Seventeen Horses

"I suppose you all know this old puzzle," said Jeffries.

"A farmer left seventeen horses to be divided among his three sons in the following proportions:

To the eldest, one-half;

to the second, one-third;

and to the youngest, one-ninth.

How should they be divided?

"Yes; I think we all know that," said Robinson, "but it can't be done.

The answer always given is a fallacy."

(Considerable pointless argument ensues.)

... The terms of the will can be exactly carried out, without any mutilation of a horse.

... How should the horses be divided in strict accordance with the directions?

$90$ - Equal Perimeters

Rational right-angled triangles have been a fascinating subject for study since the time of Pythagoras, before the Christian era.

Every schoolboy knows that the sides of these, generally expressed in whole numbers,

are such that the square of the hypotenuse is exactly equal to the sum of the squares of the other two sides.

Now, can you find $6$ rational right-angled triangles each with a common perimeter, and the smallest possible?

$91$ - Counting the Wounded

When visiting with a friend one of our hospitals for wounded soldiers, I was informed that

exactly two-thirds of the men had lost an eye,

three-fourths had lost an arm,

and four-fifths had lost a leg.

"Then," I remarked to my friend, "it follows that at least twenty-six of the men must have lost all three -- an eye, an arm, and a leg."

That being so, can you say exactly how many men were in the hospital?

$92$ - A Cow's Progeny

"Supposing," said my friend Farmer Hodge, "that cow of mine to have a she-calf at the age of two years,

and supposing she goes on having the like every year,

and supposing every one of her young to have a she-calf at the age of two years,

and afterwards every year likewise, and so on.

Now, how many do you suppose would spring from that cow and all her descendants in the space of twenty-five years?"

$93$ - Sum Equals Product

There are two numbers whose sum equals their product, that is, $2$ and $2$.

What other numbers have that property?

$94$ - Adding their Cubes

The numbers $407$ and $370$ have this peculiarity, that they exactly equal the sum of the cubes of their digits.

Thus the cube of $4$ is $64$, the cube of $0$ is $0$, and the cube of $7$ is $343$.

Add together $64$, $0$ and $343$, and you get $407$.

Again, the cube of $3$ ($27$), added to the cube of $7$ ($343$), is $370$.

Can you find a number not containing a nought that will work in the same way?

Of course, we bar the absurd case of $1$.

$95$ - Squares and Cubes

Can you find two whole numbers, such that the difference of their squares is a cube and the difference of their cubes is a square?

What is the answer in the smallest possible numbers?

$96$ - Concerning a Cube

What is the length in feet of the side of a cube when

$(1)$ the superficial area equals the cubical contents;

$(2)$ the superficial area equals the square of the cubical contents;

$(3)$ the square of the superficial area equals the cubical contents?

$97$ - A Common Divisor

Find a common divisor for the three numbers $480 \, 608$, $508 \, 811$, and $723 \, 217$, so that the remainder shall be the same in every case.

$98$ - Curious Multiplication

If a person can add correctly but is incapable of multiplying or dividing by a number higher than $2$,

it is possible to obtain the product of any two numbers in this curious way.

Multiply $97$ by $23$.

In the first column we divide by $2$, rejecting the remainders, until $1$ is reached.

In the second column we multiply $23$ by $2$ the same number of times.

If we now strike out those products that are opposite ton the even numbers in the first column

(we have enclosed these in brackets for convenience in printing)

and add up the remaining numbers we get $2231$, which is the correct answer.

Why is this?

$99$ - The Rejected Gun

An inventor offered a new large gun to the committee appointed by our Government for the consideration of such things.

He declared that when once loaded it would fire sixty shots at the rate of a shot a minute.

The War Office put it to the test and found that it fired sixty shots an hour,

but declined it, "as it did not fulfil the promised condition."

"Absurd, said the inventor, "for you have shown that it clearly does all that we undertook it should do."

"Nothing of the sort," said the experts. "It has failed."

Now, can you explain this extraordinary mystery?

Was the inventor, or were the experts, right?

$100$ - Odds and Evens

Ask a friend to take an even number of coins in one hand and an odd number in the other.

You then undertake to tell him which hand holds the odd and which the even.

Tell him to multiply the number in the right hand by $7$ and the number in the left by $6$,

add the two products together, and tell you the result.

You can then immediately give him the required answer.

How are you to do it?

$101$ - Twenty Questions

I think of a number containing six figures.

Can you discover what it is by putting to me twenty questions,

each of which can only be answered by "yes" or "no"?

After the twentieth question you must give the number.

$102$ - The Nine Barrels

In how many different ways may these nine barrels be arranged in three tiers of three

so that no barrel shall have a smaller number than its own below it or to the right of it?

The first correct arrangement that will occur to you is $1 \ 2 \ 3$ at the top then $4 \ 5 \ 6$ in the second row, and $7 \ 8 \ 9$ at the bottom,

and my sketch gives a second arrangement.

How many are there altogether?

Geometrical Problems

Dissection Puzzles

$103$ - A New Cutting-out Puzzle

Cut the figure into four pieces that will fit together and form a square.

$104$ - The Square Table-Top

A man had three pieces of beautiful wood, measuring $12$ units, $15$ units and $16$ units square respectively.

He wanted to cut those into the fewest pieces possible that would fit together and form a small square table-tip $25$ units by $25$ units.

How was he to do it?

$105$ - The Squares of Veneer

A man has two squares of valuable veneer, each measuring $25$ units by $25$ units.

One piece he cut, in the manner shown in our illustration, in four parts that will form two squares,

one $20$ units by $20$ units, and the other $15$ units by $15$ units.

Simply join $C$ to $A$ and $D$ to $B$.

How is he to cut the other square into four pieces that will form again two other squares, with sides in exact units,

but not $20$ and $15$ as before?

$106$ - Dissecting the Moon

In how large a number of pieces can this crescent moon be cut with five straight cuts of the knife?

The pieces may not be piled or shifted after a cut.

$107$ - Dissecting the Letter E

Can you cut this letter $\text E$ into only five pieces so that they will fit together to form a perfect square?

All the measurements have been given so that there should be no doubt as to the correct proportions of the letter.

In this case you are not allowed to turn over any piece.

$108$ - Hexagon to Square

Can you cut a regular hexagon into $5$ pieces that will fit together to form a square?

$109$ - Squaring a Star

This six-pointed star can be cut into as few as five pieces that will fit together to form a perfect square.

To perform the feat in $7$ pieces is quite easy,

but to do it in $5$ is more difficult.

The dotted lines are there to show the true shape of the star, which is made of $12$ equilateral triangles.

$110$ - The Mutilated Cross

Here is a regular Greek cross from which has been cut a square piece exactly equal to one of the arms of the cross.

The puzzle is to cut what remains into four pieces that will fit together and form a square.

$111$ - The Victoria Cross

Cut the cross shown into seven pieces that will fit together and form a perfect square.

Of course, there must be no trickery or waste of material.

$112$ - Squaring the Swastika

Cut out the swastika and then cut it up into four pieces that will fit together and form a square.

$113$ - The Maltese Cross

Can you cut the star into four pieces and place them inside the frame so as to show a perfect Maltese cross?

$114$ - The Pirates' Flag

Here is a flag taken from a band of pirates on the high seas.

The twelve stripes represented the number of men in the band,

and when a new man was admitted or dropped out a new stripe was added or one removed, as the case might be.

Can you discover how the flag should be cut into as few pieces as possible so that they may be put together again and show only ten stripes?

No part of the material may be wasted, and the flag must retain its oblong shape.

$115$ - The Carpenter's Puzzle

A ship's carpenter had to stop a hole $12$ inches square,

and the only piece of wood that was available measured $9 \ \mathrm{in.}$ in breadth by $16 \ \mathrm{in.}$ length.

How did he cut it into only two pieces that would exactly fit the hole?

The answer is based on the "step principle", as shown in the diagram.

If you move the piece marked $B$ up one step to the left,

it will exactly fit on $A$ and form a perfect square measuring $12$ inches on every side.

This is very simple and obvious.

But nobody has ever attempted to explain the general law of the thing.

As a consequence, the notion seems to have got abroad that the method will apply to any rectangle where the proportion of length to breadth is within reasonable limits.

This is not so, and I have had to expose some bad blunders in the case of published puzzles that were supposed to be solved by an application of this step principle,

but were really impossible of solution.$^*$

Let the reader take different measurements, instead of $9 \ \mathrm{in.}$ by $16 \ \mathrm{in.}$,

and see if he [or she] can find other cases in which this trick will work in two pieces and form a perfect square.

$116$ - The Crescent and the Star

Here is a little puzzle on the Crescent and the Star.

Look at the illustration, and see if you can determine which is the larger, the Crescent or the Star.

If both were cut out of a sheet of solid gold, which would be more valuable?

As it is very difficult to guess by the eye,

I will state that the outer arc is a semicircle;

the radius of the inner arc is equal to the straight line $BC$;

the distance in a straight line from $A$ to $B$ is $12$ units,

and the point of the star, $D$, contains $3$ square units.

Patchwork Puzzles

$117$ - The Patchwork Quilt

Here is a patchwork quilt that was produced by two young ladies for some charitable purpose.

When they came to join their work it was found that each lady had contributed a portion that was exactly the same size and shape.

It is an amusing puzzle to discover just where these two portions are joined together.

Can you divide the quilt into two parts, simply by cutting the stitches, so that the portions shall be of the same size and shape?

$118$ - The Improvised Draughts-Board

Some Englishmen at the front during the Great War wished to pass a restful hour at a game of draughts.

They had coins and small stones for the men, but no board.

However, one of them found a piece of linoleum as shown n the illustration,

and, as it contained the right number of squares, it was decided to cut it and fit the pieces together to form a board,

blacking some of the squares afterwards for convenience in playing.

An ingenious Scotsman showed how this could be done by cutting the stuff in two pieces only,

and it is a really good puzzle to discover how he did it.

Cut the linoleum along the lines into two pieces that will fit together and form the board, eight by eight.

$119$ - Tessellated Pavements

The reader must often have noticed, in looking at tessellated pavements and elsewhere,

that a square space had sometimes to be covered with square tiles under such conditions that a certain number of the tiles have to be cut in two parts.

A familiar example is shown in the illustration, where a square has been formed with ten square tiles.

As ten is not a square number a certain number of tiles must be cut.

In this case it is six.

It will be seen that the pieces $1$ and $1$ are cut from one tile, $2$ and $2$ from another, and so on.

Now, if you had to cover a square space with exactly twenty-nine square tiles of equal size, how would you do it?

What is the smallest number of tiles that you need cut in two parts?

Paper-Folding Puzzles

$120$ - The Ribbon Pentagon

I want to form a regular pentagon, but the only thing at hand happens to be a rectangular strip of paper.

How am I to do it without pencil, compasses, scissors, or anything else whatever but my fingers?

$121$ - Paper Folding

Suppose you are given a perfectly square piece of paper,

how are you going to fold it so as to indicate by creases a regular hexagon,

as shown in the illustration, all ready to be cut out?

$122$ - Folding a Pentagon

If you are given a perfectly square piece of paper,

how are you going to fold it so as to indicate by creases a regular pentagon,

all ready to be cut out?

$123$ - Making an Octagon

Can you cut the regular octagon from a square piece of paper without using compasses or ruler,

or anything but scissors?

You can fold the paper to make creases.

Various Geometrical Puzzles

$124$ - Drawing a Straight Line

If we want to describe a circle we use an instrument that we call a pair of compasses,

but if we need a straight line we use no such instrument --

we employ a ruler or other straight edge.

In other words, we first seek a straight line to produce our required straight line,

which is equivalent to using a coin, saucer of other circular object to draw a circle.

Now, imagine yourself in such a position that you cannot obtain a straight edge --

not even a piece of thread.

Could you devise a simple instrument that would draw your straight line,

just as the compasses describe a circle?

$125$ - Making a Pentagon

How do you construct a regular pentagon on a given unit straight line?

$126$ - Drawing an Oval

It is well-known that you can draw an ellipse by sticking two pins into the paper, enclosing them with a loop of thread,

and keeping the loop taut, running a pencil all the way round till you get back to the starting point.

Suppose you want an ellipse with a given major axis and minor axis.

How do you arrange the position of the pins, and what would be the length of the thread?

$127$ - With Compasses Only

Can you show how to mark off the four corners of a square, using the compasses only?

$128$ - Lines and Squares

With how few straight lines can you make exactly one hundred squares?

Thus, in the first diagram it will be found that with nine straight lines I have made twenty squares

(twelve with sides of the length $AB$, six with sides $AC$, and two with sides of the length $AD$).

In the second diagram, although I use one more line, I only get seventeen squares.

$129$ - The Circle and Discs

During a recent visit to a fair we saw a man with a table,

on the oilcloth covering of which was painted a large red circle,

and he invited the public to cover this circle entirely with five tin discs which he provided,

and offered a substantial prize to anyone who was successful.

The circular discs were all of the same size, and each, of course, smaller than the red circle.

he showed that it was "quite easy when you know how," by covering up the circle himself without any apparent difficulty,

but many tried over and over again and failed every time.

It was a condition that when once you had placed any disc you were not allowed to shift it,

otherwise, by sliding them about after they had been placed, it might be tolerably easy to do.

Let us assume that the red circle is six units in diameter.

Now, what is the smallest possible diameter for the five discs in order to make a solution possible?

$130$ - Mr. Grindle's Garden

"My neighbour," said Mr. Grindle, "generously offered me, for a garden,

as much land as I could enclose with four straight walls measuring $7$, $8$, $9$ and $10$ rods in length respectively."

"And what was the largest area you were able to enclose?" asked his friend.

Perhaps the reader can discover Mr. Grindle's correct answer.

$131$ - The Garden Path

A man has a rectangular garden, $55$ yards by $40$ yards,

and he makes a diagonal path, one yard wide, exactly in the manner indicated in the diagram.

What is the area of the path?

$132$ - The Garden Bed

A man has a triangular lawn of the proportions shown,

and he wants to make the largest possible rectangular flower-bed without enclosing the tree.

$133$ - A Problem for Surveyors

A man bought a little field, and here is a scale map that was given to me.

I asked my surveyor to tell me the area of the field,

but he said it was impossible without some further measurements;

the mere length of one side, $7$ rods, was insufficient.

What was his surprise when I showed him in about two minutes what was the area!

Can you tell how it is to be done?

$134$ - A Fence Problem

A man has a square field, $60 \ \mathrm {ft.}$ by $60 \ \mathrm {ft.}$, with other property, adjoining the highway.

For some reason he put up a straight fence in the line of three trees, as shown,

and the length of fence from the middle tree to the tree on the road was just $91$ feet.

What is the distance in exact feet from the middle tree to the gate on the road?

$135$ - The Domino Swastika

Form a square frame with twelve dominoes, as shown in the illustration.

Now, with only four extra dominoes, form within the frame a swastika.

$136$ - A New Match Puzzle

I have a box of matches.

I find that I can form with them any given pair of these four regular figures, using all the matches every time.

This, if there were eleven matches, I could form with them, as shown, the triangle and pentagon

or the pentagon and hexagon, or the square and triangle (by using only three matches in the triangle);

but could not with eleven matches form the triangle and hexagon,

or the square and pentagon, or the square and hexagon.

Of course there must be the same number of matches in every side of a figure.

Now, what is the smallest number of matches I can have in the box?

$137$ - Hurdles and Sheep

A farmer says that four of his hurdles will form a square enclosure just sufficient for one sheep.

That being so, what is the smallest number of hurdles that he will require for enclosing ten sheep?

$138$ - The Four Draughtsmen

The four draughtsmen are shown exactly as they stood on a square chequered board --

not necessarily eight squares by eight --

but the ink with which the board was drawn was evanescent,

so that all the diagram except the men has disappeared.

How many squares were there in the board and how am I to reconstruct it?

I know that each man stood in the middle of a square,

one on the edge of each side of the board and no man in a corner.

$139$ - A Crease Problem

Fold a page, so that the bottom outside corner touches the inside edge and the crease is the shortest possible.

$140$ - The Four-Colour Map Theorem

In colouring any map under the condition that no contiguous countries shall be coloured alike,

not more than four colours can ever be necessary.

Countries only touching at a point ... are not contiguous.

I will give, in condensed form, a suggested proof of my own

which several good mathematicians to whom I have shown it accept it as quite valid.

Two others, for whose opinion I have great respect, think it fails for a reason that the former maintain will not "hold water".

The proof is in a form that anybody can understand.

It should be remembered that it is one thing to be convinced, as everybody is, that the thing is true,

but quite another to give a rigid proof of it.

$141$ - The Six Submarines

If five submarines, sunk on the same day, all went down at the same spot where another had previously been sunk,

how might they all lie at rest so that every one of the six U-boats should touch every other one?

To simplify we will say, place six ordinary wooden matches so that every match shall touch every other match.

No bending or breaking allowed.

$142$ - Economy in String

Owing to the scarcity of string a lady found herself in this dilemma.

In making up a parcel for her son, she was limited to using $12$ feet of string, exclusive of knots,

which passed round the parcel once lengthways and twice round its girth, as shown in the illustration.

What was the largest rectangular parcel that she could make up, subject to these conditions?

$143$ - The Stone Pedestal

In laying the base and cubic pedestal for a certain public memorial,

the stonemason used cubic blocks of stone all measuring one foot on each side.

There was exactly the same number of these blocks (all uncut) in the pedestal as in the square base on the centre of which it stood.

Look at the sketch and try to determine the total number of blocks actually used.

The base is only a single block in depth.

$144$ - The Bricklayer's Task

When a man walled in his estate, one of the walls was partly level and partly over a small rise or hill,

precisely as shown in the drawing herewith, wherein it will be observed that the distance from $A$ to $B$ is the same as from $B$ to $C$.

Now, the master-builder desired and claimed that he should be paid more for the part that was on the hill than for the part that was level,

since (at least, so he held) it demanded the use of more material.

But the employer insisted that he should pay less for that part.

It was a nice point, over which they nearly had recourse to the law.

Which of them was in the right?

$145$ - A Cube Paradox

I had two solid cubes of lead, one very slightly larger than the other.

Through one of them I cut a hole (without destroying the continuity of the four sides)

so that the other cube could be passed right through it.

On weighing them afterwards it was found that the larger cube was still the heavier of the two.

How was this possible?

$146$ - The Cardboard Box

If I have a closed cubical cardboard box, by running the penknife along seven of the twelve edges (it must always be seven)

I can lay it out in one flat piece in various shapes.

Thus, in the diagram, if I pass the knife along the darkened edges and down the invisible edge indicated by the dotted line, I get the shape $A$.

Another way of cutting produces $B$ or $C$.

It will be seen that $D$ is simply $C$ turned over, so we will not call that a different shape.

Now, how many shapes can be produced?

$147$ - The Austrian Pretzel

Here is a twisted Vienna bread roll, known as a Pretzel.

The twist, like the curl in a pig's tail, is entirely for ornament.

The Wiener Pretzel, like some other things, is doomed to be cut up or broken, and the interest lies in the number of resultant pieces.

Suppose you had the Pretzel depicted in the illustration lying on the table before you,

what is the greatest number of pieces into which you could cut it with a single straight cut of a knife?

In what direction would you make the cut?

$148$ - Cutting the Cheese

I have a piece of cheese in the shape of a cube.

How am I to cut it in two pieces with one straight cut of the knife

so that the two new surfaces produced by the cut shall each be a perfect hexagon?

$149$ - A Tree-Planting Puzzle

How do you plant $13$ trees so as to form $9$ straight rows of $4$ trees each?

Moving Counter Problems

$150$ - Counter Solitaire

The puzzle is to remove all but one counter by a succession of leaps.

A counter can leap over another adjoining it to the next square beyond, if vacant,

and in making the leap you remove the one jumped over.

But no leap may be made in a diagonal direction.

The following is a solution in eight moves:

$5 - 13$, $(6 - 14, 6 - 5)$, $16 - 15$, $(3 - 13, 3 - 6)$, $2 - 10$, $(8 - 7, 8 - 16, 8 - 3)$, $(1 - 9, 1 - 2, 1 - 8)$, $(4 - 12, 4 - 1)$

This means that $5$ leaps over $13$ and $13$ is removed, then $6$ leaps over $14$ and $14$ is removed, and so on.

The leaps within a bracket count as one move, because the leaps are made with the same counter in succession.

It will be seen that No. $4$ makes the last leap.

Now try to find a solution, in seven moves, in which No. $1$ makes the last leap.

$151$ - Sinking the Fishing-Boats

There are forty-nine fishing-boats in the North Sea.

How could an enemy ram and sink the lot in twelve straight courses,

starting at $A$ and finishing up at the same place?

$152$ - A New Leap-Frog Puzzle

Make a rough board, as shown, and place seventeen counters on the squares indicated.

The puzzle is to remove all but one by a series of leaping moves, as in draughts or solitaire.

A counter can be made to leap over another to the next square beyond, if vacant, and you then remove the one jumped over.

It will be seen that the first leap must be made by the central counter, No. $9$, and one has the choice of $8$ directions.

A continuous series of leaps with the same counter will count as a single move.

It is required to take off $26$ counters in $4$ moves, leaving the No. $9$ on its original centre square.

Every play must be a leap.

$153$ - Transferring the Counters

Divide a sheet of paper into six compartments, as shown in the diagram,

and place a pile of $15$ counters, numbered consecutively $1$, $2$, $3$, $\ldots$, $15$ downwards, in compartment $A$.

The puzzle is to transfer the complete pile, in the fewest possible moves, to compartment $F$.

You can move the counters one at a time to any compartment,

but may never place a counter on one that bears a smaller number than itself.

Thus, if you place $1$ on $B$ and $2$ on $C$, you can then place $1$ on $2$, but not $2$ on $1$.

Unicursal and Route Problems

$154$ - The Way to Tipperary

The popular bard assures us that "it's a long way to Tipperary."

Look at the accompanying chart and see if you can discover the best way from London to "the sweetest girl I know."

The lines represent stages from town to town, and it is necessary to get from London to Tipperary in an even number of stages.

You will find no difficulty getting there in $3$, $5$, $7$, $9$ or $11$ stages,

but these are odd numbers and will not do.

The reason that they are odd is that they all omit the sea passage, a very necessary stage.

If you get to your destination in an even number of stages, it will be because you have crossed the Irish Sea.

Which stage is the Irish Sea?

$155$ - Marking a Tennis Court

The lines of our tennis court are faint and want re-marking.

My marker is of such a kind that, though I can start anywhere and finish anywhere,

it cannot be lifted off the line when working without making a mess.

I therefore have to go over some of the lines twice.

Where should I start and what route should I take, without lifting the marker,

to mark the court completely and yet go over the minimum distance twice?

I give the correct proportions of a tennis court in feet.

What is the best route?

$156$ - Water, Gas and Electricity

It is required 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.

$157$ - Crossing the Lines

You are asked to draw the diagram of Figure $1$ (exclusive of the little crosses) with three continuous strokes of the pencil,

without removing the pencil from the paper during a stroke, or going over a line twice.

As generally understood, it is quite impossible.

Wherever I have placed a cross there is an "odd node", and the law for all such cases is that half as many lines will be necessary as there are odd nodes --

that is, points from which you can depart in an odd number of ways.

Here we have, as indicated, $8$ odd nodes, from each of which you can proceed in three directions (an odd number),

and therefore, four lines will be required.

But, as I have shown in my book of Amusements in Mathematics, it may be solved by a trick, overriding the conditions as understood.

You first fold the paper, and with a thick lead-pencil draw $CD$ and $EF$, in Figure $2$, with a single stroke.

Then draw the line from $A$ to $B$ as the second stroke, and $GH$ as the third!

During the last few years this puzzle has taken a new form.

You are given the same diagram and asked to start where you like and try to pass through every short line comprising the figure,

once and once only, without crossing your own path.

Figure $3$ will make quite clear what is meant.

It is an attempted solution, but it fails because the line from $K$ to $L$ has not been crossed.

We might have crossed it instead of $KM$, but that would be no better.

Is it possible?

Many who write to me about the puzzle say that though they have satisfied themselves as a "pious opinion", that it cannot be done,

yet they see no way whatever of proving the impossibility, which is quite another matter.

I will show my way of settling the question.

$158$ - The Nine Bridges

The diagram represents the map of a district with a peculiar system of irrigation.

The lines are waterways enclosing the four islands $A$, $B$, $C$, and $D$, each with its house,

and it will be seen that there are nine bridges available.

Whenever Tompkins leaves his house to visit his friend Johnson, who lives in one of the others,

he always carries out the eccentric rule of crossing every one of the bridges once, and once only,

before arriving at his destination.

How many different routes has he to select from?

You may choose any house you like as the residence of Tompkins.

$159$ - The Five Regiments

The diagram represents a map of a certain district.

The dots and circles are towns and the lines are roads.

During a war five regiments marched to new positions in the night.

The body stationed at the upper $A$ marched to the lower $A$,

that at the upper $B$ to the lower $B$,

that at the upper $C$ to the lower $C$,

that at the upper $D$ to the lower $D$,

and the regiment at the left-hand $E$ marched to the right-hand $E$.

Yet no regiment saw anything of any other regiment.

Can you mark out the route taken by each so that no two regiments ever go along the same road anywhere?

$160$ - Going to Church

A man living in the house shown as $H$ in the diagram wants to know what is the greatest number of different routes by which he can go to the church at $C$.

The possible roads are indicated by the lines, and he always walks either due $N$, due $E$, or $N.E.$;

that is, he goes so that every step brings him nearer to the church.

Can you count the total number of different routes from which he may select?

$161$ - A Motor-Car Puzzle

A traveller starts in his car from the point $A$ and wishes to go as far as possible while making only $15$ turnings, and never going along the same road twice.

The dots represent towns and are one mile apart.

Supposing, for example, that he went straight to $B$, then straight to $C$, then to $D$, $E$, $F$, and $G$,

then you will find that he has gone $37$ miles in five turnings.

How far can he go in $15$ turnings?

$162$ - The Fly and the Honey

I have a cylindrical cup four inches high and six inches in circumference.

On the inside of the vessel, one inch from the top, is a drop of honey,

and on the opposite side of the vessel, one inch from the bottom on the outside, is a fly.

Can you tell exactly how far the fly must walk to reach the honey?

$163$ - The Russian Motor-Cyclists

Two Russian Army motor-cyclists, on the road $A$, wish to go to $B$.

Now Pyotr said: "I shall go to $D$, which is $6$ miles, and then take the straight road to $B$, another $15$ miles.

But Sergei thought he would try the upper road by way of $C$.

Curiously enough, they found on reference to their odometers that the distance either way was exactly the same.

This being so, they ought to have been able easily to answer the General's simple question,

"How far is it from $A$ to $C$?"

it can be done in the head in a few moments, if you only know how.

Can the reader state correctly the distance?

$164$ - Those Russian Cyclists Again

In the section from a map given in the diagram we are shown three long straight roads, forming a right-angled triangle.

The General asked the two men how far it was from $A$ to $B$.

Pyotr replied that all he knew was that riding round the triangle, from $A$ to $B$,

from there to $C$ and home to $A$, his odometer showed exactly $60$ miles,

while Sergei could only say that he happened to know that $C$ was exactly $12$ miles from the road $A$ to $B$ --

that is, to the point $D$, as shown by the dotted line.

Whereupon the General made a very simple calculation in his head and declared that the distance from $A$ to $B$ must be ...

Can the reader discover so easily how far it was?

$165$ - The Despatch-Rider in Flanders

A despatch-rider on horseback, somewhere in Flanders, had to ride with all possible speed from $A$ to $B$.

The distances are marked on the map.

Now, he can ride just twice as as fast over the soft turf (the shaded bit) as he can ride over the loose sand.

Can you show what is the quickest possible route for him to take?

Combination and Group Problems

$166$ - Picture Presentation

A wealthy collector had ten valuable pictures.

He proposed to make a presentation to a public gallery, but could not make up his mind as to how many he would give.

So it amused him to work out the exact number of different ways.

You see, he could give any one picture, any two, any three, and so on, or give the whole ten.

$167$ - A General Election

In how many different ways may a Parliament of $615$ members be elected if there are only $4$ parties:

Conservatives, Liberals, Socialists, and Independents?

You see you might have $\text C. 310$, $\text L. 152$, $\text S. 150$, $\text I. 3$;

or $\text C. 0$, $\text L. 0$, $\text S. 0$, $\text I. 615$;

or $\text C. 205$, $\text L. 205$, $\text S. 205$, $\text I. 0$; and so on.

The candidates are indistinguishable, as we are only concerned with the party numbers.

$168$ - The Magisterial Bench

A bench of magistrates consists of two Englishmen, two Scotsmen, two Welshmen, one Frenchman, one Italian, one Spaniard, and one American.

The Englishmen will not sit beside one another, the Scotsmen will not sit beside one another, and the Welshmen also object to sitting together.

Now, in how many different ways may the ten men sit in a straight line so that no two men of the same nationality shall ever be next to one another?

$169$ - The Card Pentagon

Make a rough pentagon on a large sheet of paper.

Then throw down the ten non-court cards of a suit at the places indicated in the diagram,

so that the pips on every row of three cards on the sides of the pentagon shall add up alike.

The example will be found faulty.

$170$ - A Heptagon Puzzle

Using the fourteen numbers, $1$, $2$, $3$, up to $14$, place a different number in every circle

so that the three numbers in every one of the seven sides add up to $19$.

Magic Square Problems

$171$ - An Irregular Magic Square

Here we have a perfect magic square composed of the numbers $1$ to $16$ inclusive.

$\qquad \begin{array}{|c|c|c|c|} \hline 1 & 14 & 7 & 12 \\ \hline 15 & 4 & 9 & 6 \\ \hline 10 & 5 & 16 & 3 \\ \hline 8 & 11 & 2 & 13 \\ \hline \end{array}$

The rows, columns, and two long diagonals all add up to $34$.

Now, supposing you were forbidden to use the two numbers $2$ and $15$, but allowed, in their place, to repeat any two numbers already used,

how would you construct your square so that rows, columns, and diagonals should still add up to $34$?

Your success will depend on which two numbers you select as substitutes for $2$ and $15$.

$172$ - A Magic Square Delusion

Here is a magic square of the fifth order.

$\qquad \begin{array}{|c|c|c|c|c|} \hline 17 & 24 & 1 & 8 & 15 \\ \hline 23 & 5 & 7 & 14 & 16 \\ \hline 4 & 6 & 13 & 20 & 22 \\ \hline 10 & 12 & 19 & 21 & 3 \\ \hline 11 & 18 & 25 & 2 & 9 \\ \hline \end{array}$

I have found that a great many people who have not gone very profoundly into these things believe that the central number in all squares of this order must be $13$.

One correspondent who had devoted years to amusing himself with this particular square was astounded when I told him that any number from $1$ to $25$ might be in the centre.

I will show that this is so.

Try to form such a magic square with $1$ in the central cell.

$173$ - Difference Squares

Can you rearrange the nine digits in the square so that in all the eight directions the difference between one of the digits and the sum of the remaining two shall always be the same?

$\qquad \begin{array}{|c|c|c|} \hline 4 & 3 & 2 \\ \hline 7 & 1 & 9 \\ \hline 6 & 5 & 8 \\ \hline \end{array}$

In the example shown it will be found that all the rows and columns give the difference $3$:

(thus $4 + 2 - 3$, and $1 + 9 - 7$, and $6 + 5 - 8$, etc.),

but the two diagonals are wrong, because $8 - \paren {4 + 1}$ and $6 - \paren {1 + 2}$ is not allowed:

the sum of the two must not be taken from the single digit, but the single digit from the sum.

How many solutions are there?

$174$ - Swastika Magic Square

It is a magic square, the rows, columns, and two diagonals all adding up to $65$,

and all prime numbers that occur between $1$ and $25$

(viz. $1$, $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$)

are to be found in the swastika except $11$.

"This number," he says, "in occult lore is ominous and is associated with the eleven Curses of Ebal,

so it is just as well it does not come into this potent charm of good fortune."

He is clearly under the impression that $11$ cannot be got into the swastika with the other primes.

But in this he is wrong, and the reader may like to try to reconstruct the square

so that the swastika contains all the prime numbers

and yet forms a correct magic square, for it is quite possible.

$175$ - Is it Very Easy?

Here is a simple magic square, the three columns, three rows, and two diagonals adding up to $72$.

$\qquad \begin{array}{|c|c|c|} \hline 27 & 20 & 25\\ \hline 22 & 24 & 26 \\ \hline 23 & 28 & 21 \\ \hline \end{array}$

The puzzle is to convert it into a multiplying magic square,

in which the numbers in all the eight lines if multiplied together give the same product in every case.

You are not allowed to change, or add to, any of the figures in a cell

or use any arithmetical sign whatever!

But you may shift the two figures within a cell.

Thus, you may write $27$ as $72$, if you like.

Magic Star Problems

$176$ - The Five-Pointed Star

It is required to place a different number in every circle so that the four circles in a line shall add up to $24$ in all the five directions.

No solution is possible with $10$ consecutive numbers, but you can use any whole numbers you like.

$177$ - The Six-Pointed Star

In this case we can always use the twelve consecutive numbers $1$ to $12$ and the sum of the four numbers in every line will always be $26$.

The numbers at the six points of the star may add up to any even number from $24$ to $54$ inclusive, except $28$ and $50$, which are impossible.

It will be seen in the example that the six points add up to $24$.

If for every number in its present position you substitute its difference from $13$ you will get another solution, its complementary,

with the points adding up to $54$, which is $78$ less $24$.

The two complementary totals will always sum to $78$.

I will give the total number of different solutions and point out some of the pretty laws which govern the problem,

but I will leave the reader this puzzle to solve.

There are six arrangements, and six only, in which all the lines of four and the six points also add up to $26$.

Can you find one or all of them?

$178$ - The Seven-Pointed Star

All you have to do is place the numbers $1$, $2$, $3$, up to $14$ in the fourteen discs so that every line of four disks shall add up to $30$.

$179$ - Two Eight-Pointed Stars

The star may be formed in two different ways, as shown in our diagram, and the first example is a solution.

The numbers $1$ to $16$ are so placed that every straight line of four adds up to $34$.

If you substitute for every number its difference from $17$ you will get the complementary solution.

$\qquad$

Let the reader try to discover some of the other solutions, and he will find it a very hard nut, even with this one to help him.

But I will present the puzzle in an easy and entertaining form.

When you know how, every arrangement in the first star can be transferred to the second one automatically.

Every line of four numbers in the one case will appear in the other, only the order of the numbers will have to be changed.

Now, with this information given, it is not a difficult puzzle to find a solution for the second star.

Measuring, Weighing, and Packing Problems

$180$ - The Damaged Measure

A young man has a yardstick from which $3$ inches have been broken off,

so that it is only $33$ inches in length.

Some of the graduation marks are also obliterated, so that only eight of these marks are legible;

yet he is able to measure any given number of inches from $1$ inch up to $33$ inches.

Where are these marks placed?

$181$ - The Six Cottages

A circular road, $27$ miles long, surrounds a tract of wild and desolate country,

and on this road are $6$ cottages so placed that one cottage or another is at a distance of one, two, three up to $26$ miles inclusive from some other cottage.

Thus, Brown may be a mile from Stiggins, Jones two miles from Rogers, Wilson three miles from Jones, and so on.

Of course, they can walk in either direction if required.

Can you place the cottages at distances that will fulfil the conditions?

The illustration is intended to give no clue as to the relative distances.

$182$ - A New Domino Puzzle

Two dominoes have been placed together so that by taking the pips in unbroken conjunction

I can get all the numbers from $1$ to $9$ inclusive.

Thus, $1$, $2$ and $3$ can be taken alone;

then $1$ and $3$ make $4$; $3$ and $2$ make $5$; $3$ and $3$ make $6$;

$1$, $3$ and $3$ make $7$; $3$, $3$ and $2$ make $8$, and $1$, $3$, $3$ and $2$ make $9$.

It would not have been allowed to take the $1$ and the $2$ to make $3$, nor to take the first $3$ and the $2$ to make $5$.

The numbers would not have been in conjunction.

Now try to arrange four dominoes so that you can make the pips in this way sum to any number from $1$ to $24$ inclusive.

The dominoes need not be placed $1$ against $1$, $2$ against $2$, and so on, as in play.

$183$ - At the Brook

A man goes to the brook with two measures of $15$ pints and $16$ pints.

How is he to measure exactly $8$ pints of water, in the fewest possible transactions?

Filling or emptying a vessel or pouring any quantity from one vessel to another counts as a transaction.

$184$ - A Prohibition Poser

The American Prohibition authorities discovered a full barrel of beer,

and were about to destroy the liquor by letting it run down a drain

when the owner pointed to two vessels standing by and begged to be allowed to retain in them a small quantity for the immediate consumption of his household.

One vessel was a $7$-quart and the other a $5$-quart measure.

The officer was a wag, and, believing it to be impossible, said that if the man could measure an exact quart into each vessel

(without any pouring back into the barrel) he might do so.

How was it to be done in the fewest possible transactions without any marking or other tricks?

Pouring down the drain counts as one transaction.

Perhaps I should state that an American barrel of beer contains exactly $120$ quarts.

$185$ - Prohibition Again

Let us now try to discover the fewest possible manipulations under the same conditions as in the last puzzle,

except that we may now pour back into the barrel.

$186$ - The False Scales

A pudding, when placed into one of the pans of a balance, appeared to weigh $4$ ounces more than $\tfrac 9 {11}$ of its true weight,

but when placed into the other pan it appeared to weigh $3$ pounds more than in the first pan.

What was its true weight?

$187$ - Weighing the Goods

A tradesman whose morals had become corrupted during the war by a course of profiteering went to the length of introducing a pair of false scales.

It will be seen from the diagram that one arm is longer than the other,

though they are purposely drawn so as to give no clue as to the answer.

As a consequence, it happened that in one of the cases exhibited eight of the little packets

(it does not matter what they contain)

exactly balanced three of the canisters,

while in the other case one packet appeared to be of the same weight as $6$ canisters.

Now, as the true weight of one canister was known to be exactly one ounce, what was the true weight of the eight packets?

$188$ - Monkey and Pulley

A rope is passed over a pulley.

It has a weight at one end and a monkey at the other.

There is the same length of rope on either side and equilibrium is maintained.

The rope weighs four ounces per foot.

The age of the monkey and the age of the monkey's mother total four years.

The weight of the monkey is as many pounds as the monkey's mother is years old.

The monkey's mother is twice as old as the monkey was

when the monkey's mother was half as old as the monkey will be

when the monkey is three times as old as the monkey's mother was

when the monkey's mother was three times as old as the monkey.

The weight of the rope and the weight at the end was half as much again as the difference in weight

between the weight of the weight and the weight and the weight of the monkey.

Now, what was the length of the rope?

$189$ - Weighing the Baby

There was a family group at the automatic weighing machine, trying to weigh the baby.

Whenever they put the baby on the machine, she always yelled and rolled off,

while the father was holding off the dog, who always insisted on being included in the operations.

At last the man, with the baby and Fido, were on the machine together.

The dial read $180$ pounds.

The man turned to his wife and said,

"Baby and I weigh $162$ pounds more than the dog,

while the dog weighs $70$ per cent less than the baby.

We must try to work it out at home."

What was the actual weight of the baby?

$190$ - Packing Cigarettes

A manufacturer sends out his cigarettes in boxes of $160$;

they are packed in $8$ rows of $20$ each, and exactly fill the box.

Could he, by packing differently, get more cigarettes than $160$ into the box?

If so, what is the greatest number that he could add?

Crossing River Problems

$191$ - Crossing the Ferry

Six persons, all related, have to cross a river in a small boat that will only hold two.

Mr. Webster, who had to plan the little affair, had quarrelled with his father-in-law and his son,

and, I am sorry to say, Mrs. Webster was not on speaking terms with her mother or her daugther-in-law.

In fact, the relations were so strained that it was not safe to permit any of the belligerents to pass over together

or to remain together on the same side of the river.

And to prevent further discord, no man was to be left with two women or two men with three women.

How are they to perform the feat in the fewest possible crossings?

No tricks, such as making use of a rope or current, or swimming across, are allowed.

$192$ - Missionaries and Cannibals

There is a strange story of three missionaries and three cannibals,

who had to cross a river in a small boat that would only carry two men at a time.

Being acquainted with the peculiar appetites of the cannibals,

the missionaries could never allow their companions to be in a majority on either side of the river.

Only one of the missionaries and one of the cannibals could row the boat.

How did they manage to get across?

Problems Concerning Games

$193$ - A Domino Square

Select any eighteen dominoes you please from an ordinary box,

and arrange them any way you like in a square so that no number shall be repeated in any row or any column.

The example given is imperfect, for it will be seen that though no number is repeated in any one of the columns yet three of the rows break the condition.

There are two $4$'s and two blanks in the first row, two $5$s and two $6$'s in the third row, and two $3$'s in the fourth row.

Can you form an arrangement without such errors?

Blank counts as a number.

$194$ - A Domino Star

Place the $28$ dominoes, as shown in the diagram, so as to form a star with alternate rays of $4$ and $3$ dominoes.

Every ray must contain $21$ pips (in the example only one ray contains this number)

and the central numbers must be $1$, $2$, $3$, $4$, $5$, $6$, and two blanks, as at present, and these may be in any order.

In every ray the dominoes must be placed according to the ordinary rule, $6$ against $6$, blank against blank, and so on.

$195$ - Domino Groups

If the dominoes are laid out in the manner shown in the diagram and I then break the line into $4$ lengths of $7$ dominoes each,

it will be found that the sum of the pips in the first group is $49$, in the second $34$, in the third $46$, and in the fourth $39$.

Now I want to play them out so that all the four groups of seven when the line is broken shall contain the same number of pips.

Can you find a way of doing it?

$196$ - Les Quadrilles

It is required to arrange a complete set of $28$ dominoes so as to form the figure shown in the diagram,

with all the numbers forming a series of squares.

Thus, in the upper two rows we have a square of blanks, and a square of four $3$'s, and a square of $4$'s, and a square of $1$'s and so on.

This is, in fact, a perfect solution under the conditions usually imposed,

but what I now ask for is an arrangement with no blanks anywhere on the outer edge.

At present every number from blank to $6$ inclusive will be found somewhere on the margin.

Can you construct an arrangement with all the blanks inside?

$197$ - A Puzzle with Cards

Take from the pack the $13$ cards forming the suit of diamonds and arrange them in this order face downwards with the $3$ at the top and $5$ at the bottom:

$3$, $8$, $7$, ace, queen, $6$, $4$, $2$, jack, king, $10$, $9$, $5$.

Now play them out in a row on the table in this way.

As you spell "ace" transfer for each letter a card from the top to the bottom of the pack -- A-C-E -- and play the fourth card on to the table.

Then spell T-W-O, while transferring three more cards to the bottom, and place the next card on the table.

Then spell T-H-R-E-E, while transferring five to the bottom, and so on until all are laid out in a row,

and you will find that they will be all in regular order.

Of course, you will spell out the knave as J-A-C-K.

Can you arrange the whole pack so that they will play out correctly in order,

first all the diamonds, then the hearts, then the spades, and lastly the clubs?

$198$ - A Card Trick

Take an ordinary pack of playing-cards and regard all the court cards as tens.

Now, look at the top card -- say it is a seven -- place it on the table face downwards and play more cards on top of it, counting up to twelve.

Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile.

Then look again at the top card of pack -- say it is a queen -- then count $10$, $11$, $12$ (three cards in all) and complete the second pile.

Continue this, always counting up to twelve, and if at last you have not put sufficient cards to complete a pile, put these apart.

Now, if I am told how many piles have been made and how many unused cards remain over,

I can at once tell you the sum of all the bottom cards in the piles.

I simply multiply by $13$ the number of piles less $4$, and add the number of cards left over.

Thus, if there were $6$ piles and $4$ cards over, then $13$ times $2$ (i.e. $6$ less $4$) added to $5$ equals $31$, the sum of the bottom cards.

Why is this?

This is the question.

$199$ - Golf Competition Puzzle

I was asked to construct some schedules for players in American golf competitions.

The conditions are:

$(1)$ Every player plays every player once, and once only.

$(2)$ There are half as many links as players, and every player plays twice on every links except one, on which he plays but once.

$(3)$ All the players play simultaneously in every round, and the last round is the one in which every player is playing on a links for the first time.

I have written out schedules for a long series of even numbers of players up to $26$,

but the problem is too difficult for this page except for in its most simple form -- for six players.

Can the reader, calling the players $A$, $B$, $C$, $D$, $E$, and $F$,

and pairing these in all possible ways, such as $AB$, $CD$, $EF$, $AF$, $BD$, $CE$, etc.,

complete this table for six players?

$\qquad \begin {array} {r | c |} \text {Rounds} & 1 & 2 & 3 & 4 & 5 \\ \hline \text{$1$st Links} & & & & & \\ \hline \text{$2$nd Links} & & & & & \\ \hline \text{$3$rd Links} & & & & & \\ \hline \end{array}$

$200$ - Cricket Scores

In a country match Great Muddleton, who went in first, made a score of which they were proud.

Then Little Wurzleford had their innings and scored a quarter less.

The Muddletonians in their next attempt made a quarter less than their opponents,

who, curiously enough, were only rewarded on their second attempt by a quarter less than their last score.

Thus, every innings was a quarter less fruitful in runs than the one that preceded it.

Yet the Muddletonians won the match by $50$ runs.

Can you give the exact score for every one of the four innings?

$201$ - Football Results

The following table appeared in a newspaper:

$\qquad \begin {array} {r | cccc |cc|c} & \text {Played} & \text {Won} & \text {Lost} & \text {Drawn} & \text {Goals For} & \text {Against} & \text {Points} \\ \hline \text{Scotland} & 3 & 3 & 0 & 0 & 7 & 1 & 6 \\ \hline \text{England} & 3 & 1 & 1 & 1 & 2 & 3 & 3 \\ \hline \text{Wales} & 3 & 1 & 1 & 1 & 3 & 3 & 3 \\ \hline \text{Ireland} & 3 & 0 & 3 & 0 & 1 & 6 & 0 \\ \hline \end{array}$

It is known that Scotland had beaten England $3 - 0$.

It is now possible to find the scores in the other five matches from the table.

How many goals were won, drawn, or lost by each side in every match?

Puzzle Games

$202$ - Noughts and Crosses

Every child knows how to play this ancient game.

You make a square of nine cells, and each of the two players, playing alternately, puts his mark

(a nought or a cross, as the case may be) in a cell with the object of getting three in a line.

Whichever player gets three in a line wins.

In this game, cross has won:

$\qquad \begin {array} {|c|c|c|} \hline \text X & \text O & \text O \\ \hline \text X & \text X & \text O \\ \hline \text O & & \text X \\ \hline \end{array}$

I have said in my book, The Canterbury Puzzles,

that between two players who thoroughly understand the play every game should be drawn,

for neither party could ever win except through the blundering of his opponent.

Can you prove this?

Can you be sure of not losing a game against an expert opponent?

$203$ - The Horse-Shoe Game

This little game is an interesting companion to our "Noughts and Crosses".

There are two players.

One has two white counters, the other two black.

Playing alternately, each places a counter on a vacant point, where he leaves it.

When all are played, you slide only, and the player is beaten who is so blocked that he cannot move.

In the example, Black has just placed his lower counter.

White now slides his lower one to the centre, and wins.

Black should have played to the centre himself, and won.

Now, which player ought to win at this game?

$204$ - Turning the Die

This is played with a single die.

The first player calls any number he chooses, from $1$ to $6$, and the second player throws the die at hazard.

Then they take it in turns to roll over the die in any direction they choose, but never giving it more than a quarter turn.

The score increases as they proceed, and the player wins who manages to score $25$ or forces his opponent to score beyond $25$.

I will give an example game.

Player $A$ calls $6$, and $B$ happens to throw $3$, making the score $9$.

Now $A$ decides to turn up $1$, scoring $10$;

$B$ turns up $3$, scoring $13$;

$A$ turns up $6$, scoring $19$;

$B$ turns up $3$, scoring $22$;

$A$ turns up $1$, scoring $23$;

and $B$ turns up $2$, scoring $25$ and winning.

What call should $A$ make in order to have the best chance at winning?

Remember that the numbers on opposite sides of a correct die always sum to $7$, that is, $1 - 6$, $2 - 5$, $3 - 4$.

$205$ - The Three Dice

Mason and Jackson were playing with three dice.

The player won whenever the numbers thrown added up to one of the two numbers he selected at the beginning of the game.

As a matter of fact, Mason selected $7$ and $13$, and one of his winning throws was $6$, $4$, $3$.

What were his chances of winning a throw?

And what two other numbers should Jackson have selected for his own throws to make his chances of winning exactly equal?

$206$ - The $37$ Puzzle Game

Here is a beautiful new puzzle game, absurdly simple to play but quite fascinating.

To most people it will seem to be practically a game of chance -- equal for both players --

but there are pretty subtleties in it, and I will show how to win with certainty.

Place the five dominoes $1$, $2$, $3$, $4$, $5$, on the table.

There are two players, who play alternately.

The first player places a coin on any domino, say the $5$, which scores $5$;

then the second player removes the coin to another domino, say to the $3$,

and adds that domino, scoring $8$;

then the first player removes the coin again, say to the $1$, scoring $9$; and so on.

The player who scores $37$, or forces his opponent to score more than $37$, wins.

Remember, the coin must be removed to a different domino at each play.

$207$ - The Twenty-Two Game

Here is a variation of our little "Thirty-one Game" (The Canterbury Puzzles: No. $79$).

Lay out the $16$ cards as shown.

$\qquad \begin{matrix} \boxed {A \heartsuit} & \boxed {A \spadesuit} & \boxed {A \diamondsuit} & \boxed {A \clubsuit} \\ \boxed {2 \heartsuit} & \boxed {2 \spadesuit} & \boxed {2 \diamondsuit} & \boxed {2 \clubsuit} \\ \boxed {3 \heartsuit} & \boxed {3 \spadesuit} & \boxed {3 \diamondsuit} & \boxed {3 \clubsuit} \\ \boxed {4 \heartsuit} & \boxed {4 \spadesuit} & \boxed {4 \diamondsuit} & \boxed {4 \clubsuit} \\ \end{matrix}$

Two players alternately turn down a card and add it to the common score,

and the player who makes the score of $22$, or forces his opponent to go beyond that number, wins.

For example, $A$ turns down a $4$, $B$ turns down $3$ (counting $7$), $A$ turns down a $4$ (counting $11$),

$B$ plays a $2$ (counting $13$), $A$ plays $1$ (counting $14$), $B$ plays $3$ ($17$), and whatever $A$ does, $B$ scores the winning $22$ next play.

Now, which player should always win, and how?

$208$ - The Nine Squares Game

Make the simple square diagram show and provide a box of matches.

The side of the large square is three matches in length.

The game is, playing one match at a time alternately, to enclose more of those small squares than your opponent.

For every small square that you enclose, you not only score one point, but you play again.

This illustration shows an illustrative game in progress.

Twelve matches are placed, my opponent and myself having made six plays each, and, as I had first play, it is now my turn to play a match.

What is my best line of play in order to win most squares?

If I play $FG$ my opponent will play $BF$ and score one point.

Then, as he has the right to play again, he will score another with $EF$ and again with $IJ$, and still again with $IJ$, and still again with $GK$.

If he now plays $CD$, I have nothing better than $DH$ (scoring one), but, as I have to play again, I am compelled, whatever I do, to give him all the rest.

So he will win by $8$ to $1$ -- a bad defeat for me.

Now, what should I have played instead of that disastrous $FG$?

There is room for a lot of skilful play in the game, and it can never end in a draw.

Wheel Paradox Problems

$209$ - A Wheel Fallacy

The wheel shown in the diagram makes one complete revolution in passing from $A$ to $B$.

It is therefore obvious that the line $AB$ is exactly equal in length to the circumference of the wheel.

Now the inner circle (the large hub in the diagram) also makes one complete revolution along the dotted line $CD$ and,

since the line CD is equal to the line $AB$, the circumference of the larger and smaller circles are the same.

This is clearly not true.

Wherein lies the fallacy?

$210$ - A Famous Paradox

When a bicycle is in motion, does the upper part of each wheel move faster than the bottom part near the ground?

$211$ - Another Wheel Paradox

When a railway train is in motion, it is always the case that some parts of the train are travelling backwards.

Explain why.

Unclassified Problems

$212$ - A Chain Puzzle

A man has $80$ links of old chain in $13$ fragments, as shown in the diagram.

It will cost him $1 \oldpence$ to open a link and $2 \oldpence$ to weld one together again.

What is the lowest price it must cost him to join all the pieces together so as to form an endless chain?

A new chain will cost him $3 \shillings$ (that is, $36 \oldpence$)

What is the cheapest method of procedure?

Remember that the small and large links must run alternately.

$213$ - The Six Pennies

Lay six pennies on the table, and arrange them as shown by the $6$ white circles in the diagram,

so that a seventh penny can be dropped into the centre and touch each of the other $6$.

It is required to get it exact, without any dependence on the eye.

In this case it is not allowed to lift any penny off the table, nor can any measuring or marking be employed.

However, you are allowed to slide the pennies after they have been placed on the table.

You require only the six pennies.

$214$ - Folding Postage Stamps

Take a $4 \times 2$ sheet of $8$ postage stamps, labelled $1$ to $8$, as shown in the diagram.

It is an interesting exercise to count how many ways they may be folded up so they will all lie under the one stamp, as shown.

There are in fact $40$ ways to do this so that No. $1$ is always on the top.

Numbers $5$, $2$, $7$ and $4$ will always be face down.

You can always arrange for any stamp except No. $6$ to lie next to $1$,

although there are only two ways each in which $7$ and $8$ can be made to lie in that position.

They can be folded in the order $1$, $5$, $6$, $4$, $8$, $7$, $3$, $2$ and also $1$, $3$, $7$, $5$, $6$, $8$, $4$, $2$, with $1$ at the top face upwards,

but it is a puzzle to work out how.

Can you fold them like that without tearing any of the perforations?

$215$ - An Ingenious Match Puzzle

Place $6$ matches as shown, and then shift just one match without touching the others so that the new arrangement shall represent a fraction equal to $1$.

The match forming the horizontal fraction bar must not be the one moved.

$216$ - Fifty-Seven to Nothing

Place $6$ matches as shown, so as to represent the number $57$ in Roman numerals.

Remove and replace any $2$ of them (without disturbing the others) to make an expression representing zero ($0$).

There are two completely different solutions.

$217$ - The Five Squares

$12$ matches are arranged to form $4$ squares.

Rearrange the same matches (so they lie flat on the table) to make $5$ squares.

All the squares must be empty.

$218$ - A Square with Four Pennies

Can you place four (old) pennies together so as to show a square?

They must all lie flat on the table.

$219$ - A Calendar Puzzle

I have stated in my book, Amusements in Mathematics, that, under our present calendar rules,

the first day of a century cannot fall on a Sunday or a Wednesday or a Friday.

I am frequently asked the reason why.

Try to explain the mystery in as simple a way as possible.

Note that $1$st January $1901$ was the first day of the $20$th century, not $1900$.

$220$ - The Fly's Tour

I had a ribbon of paper, divided into squares on each side.

I joined the ends together to make a ring, and tossed it down onto the table.

Then I watched a fly land on the ring and walk in a line over every one of those squares on both sides,

returning to the point where it started, without ever passing over the edge of the paper.

Its course passed through the centre of the squares all the time.

How was this possible?

$221$ - A Musical Enigma

Here is an old musical enigma that has been pretty well known in Germany for some years.

$222$ - A Mechanical Paradox

A remarkable mechanical paradox, invented by James Ferguson about the year $1751$, ought to be known by everyone, but, unfortunately, it is not.

It was contrived by him as a challenge to a sceptical watchmaker during a metaphysical controversy.

"Suppose," Ferguson said, "I make one wheel as thick as three others and cut teeth in them all,

and then put the three wheels all loose upon one axis and set the thick wheel to turn them,

so that its teeth may take into those of the three thin ones.

Now, if I turn the thick wheel round, how must it turn the others?"

The watchmaker replied that it was obvious that all three must be turned the contrary way.

Then Ferguson produced his simple machine, which anyone can make in a few hours,

showing that, turning the thick wheel which way you would,

one of the thin wheels revolved in the same way, the second the contrary way, and the third remained stationary.

Although the watchmaker took the machine away for careful examination, he failed to detect the cause of the strange paradox.

Henry Ernest Dudeney/Modern Puzzles

Henry Ernest Dudeney: Modern Puzzles