Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems

Henry Ernest Dudeney: Modern Puzzles: Arithmetical and Algebraical Problems

Money Puzzles

$1$ - Concerning a Cheque

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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            ****
            ----

$72$ - 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?

Henry Ernest Dudeney/Modern Puzzles/Arithmetical and Algebraical Problems

Henry Ernest Dudeney: Modern Puzzles: Arithmetical and Algebraical Problems

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