nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/342 - A Wily Puzzle

== by : $342$ ==
;A Wily Puzzle
:An unscrupulous advertiser offered a hundred dollars for a correct solution to this puzzle:
:A life prisoner appealed to the king for pardon.
:Not being ready to favour the appeal, the king proposed a pardon on condition that the prisoner should start at cell $A$
::and go in and out of each cell of the prison, coming back to the cell $A$ without going into any cell twice.
:
:Either the advertiser had no answer, and knew he had none,
::or he was prepared to fall back on some trick or quibble.
:What is the best answer the reader can devise that may be held to comply with the advertiser's conditions as given?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $342$. -- A Wily Puzzle
* : Combinatorial & Topological Problems: Route & Network Puzzles: $434$. A Wily Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/342 - A Wily Puzzle/Solution

== by : $342$ ==
;A Wily Puzzle
== Solution ==
:
:In the diagram it will be seen that the prisoner's course is undoubtedly all right until we get to $b$.
:If we had been the prisoner, when we got to that point we should have placed one foot at $c$, in the neighbouring cell,
::and have said, "As one foot has been in cell $c$ we have undoubtedly entered it,
::and yet when we withdraw that foot into $b$ we do not enter $b$ a second time,
::for the simple reason that we never left it since we first went in."
== Sources ==
* : Solutions: $342$. -- A Wily Puzzle
* : Answers: $434$. A Wily Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/343 - Tom Tiddler's Ground

== by : $343$ ==
;Tom Tiddler's Ground
:I am on Tom Tiddler's ground picking up gold and silver.
:Here we have a piece of land marked off with $36$ circular plots,
::on each of which is deposited a bag containing as many sovereigns as the figures indicate in the diagram.
:
:I am allowed to pick up as many bags of gold as I like,
::provided I do not take two lying on the same line.
:What is the greatest amount of money I can secure?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $343$. -- Tom Tiddler's Ground
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $469$. Tom Tiddler's Ground
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/343 - Tom Tiddler's Ground/Solution

== by : $343$ ==
;Tom Tiddler's Ground
== Solution ==
:$47$ sovereigns.
== Proof ==
These are contained in $10$ bags, all deposited on the outside plots, thus:
:$4$, $5$ and $6$ in the first row
:$5$ in the second row
:$4$ in the third row
:$3$ in the fourth row
:$5$ in the fifth row
:$5$, $6$ and $4$ in the bottom row.
Total: $47$.
If you include $5$ bags containing $6$ sovereigns each, you can get only $9$ bags and so a value of $46$ sovereigns.
== Sources ==
* : Solutions: $343$. -- Tom Tiddler's Ground
* : Answers: $469$. Tom Tiddler's Ground
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/344 - Coin and Hole

== by : $344$ ==
;Coin and Hole
:We have before us a specimen of every coin which was current in Britain in $1930$.
:And we have a sheet of paper with a circular hole cut in it $\tfrac 3 4$ of an inch in diameter.
:What is the largest coin I can pass through that hole without tearing the paper?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $344$. -- Coin and Hole
* : Unclassified Puzzles: $527$. Coin and Hole
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/344 - Coin and Hole/Solution

== by : $344$ ==
;Coin and Hole
== Solution ==
:Strange thought it may appear, the half-crown may be passed through the small hole (about the size of a sixpence) without tearing the paper.
:This is the largest coin that can be used.
:First fold the paper across the centre of the hole and drop the coin into the fold.
:
:Then, holding the paper at $A$ and $B$, bring the hands together upwards, and the coin may be shaken through the hole.
== Sources ==
* : Solutions: $344$. -- Coin and Hole
* : Answers: $527$. Coin and Hole
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/345 - The Egg Cabinet

== by : $345$ ==
;The Egg Cabinet
:A man has a cabinet for holding birds' eggs.
:There are $12$ drawers, and all -- except the first drawer, which holds the catalogue -- are divided into cells by intersecting wooden strips,
::running the entire length or width of a drawer.
:The number of cells in any drawer is greater than that of the drawer above.
:The bottom drawer, No. $12$, has $12$ times as many cells as strips,
::No. $11$ has $11$ times as many cells as strips, and so on.
:Can you show how the drawers were divided -- how many cells and strips in each drawer?
:Give the smallest possible number in each case.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $345$. -- The Egg Cabinet
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $252$. The Egg Cabinet
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/345 - The Egg Cabinet/Solution

== by : $345$ ==
;The Egg Cabinet
== Solution ==
Let the drawer number be $n$.
Then there will be $2 n - 1$ strips one way and $2 n - 3$ strips the other way.
This gives $4 n^2 - 4 n$ cells and $4 n - 4$ strips.
So in the $12$th drawer we get $23$ and $21$ strips, or $44$ altogether, and hence $528$ cells.
This applies to all drawers except the second, where we may have any number of strips one way, and one strip the other way.
So $1$ and $1$ will serve, but because all cells have intersecting strips, it cannot have just one.
Hence there are $262$ strips in all, and $2284$ cells.
== Sources ==
* : Solutions: $345$. -- The Egg Cabinet
* : Answers: $252$. The Egg Cabinet
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/346 - A Leap Year Puzzle

== by : $346$ ==
;A Leap Year Puzzle
:The month of February in $1928$ contained five Wednesdays.
:There is, of course, nothing remarkable in this fact, but it will be found interesting to discover
::when was the last year and when will be the next year that had, and that will have, $5$ Wednesdays in February.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $346$. -- A Leap Year Puzzle
* : Unclassified Puzzles: $528$. A Leap Year Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/346 - A Leap Year Puzzle/Solution

== by : $346$ ==
;A Leap Year Puzzle
:The month of February in $1928$ contained five Wednesdays.
:There is, of course, nothing remarkable in this fact, but it will be found interesting to discover
::when was the last year and when will be the next year that had, and that will have, $5$ Wednesdays in February.
== Solution ==
The next year after $1928$ that contains $5$ February Wednesdays was $1956$.
The one immediately before $1928$ was $1888$.
== Proof ==
$5$ February Wednesdays can occur only in a leap year.
A leap year February contains $29$ days.
Hence such a February always has exactly one day of the week of which there are $5$.
This happens on $1$st, $8$th, $15$th, $22$nd and $29$th February.
So the puzzle is equivalent to determining on which years $29$th February falls on a Wednesday.
Except when the year is divisible by $100$ but not $400$, leap years occur every $4$ years.
Each consecutive set of $4$ years in the Gregorian calendar contains $3 \times 365 + 366 = 1461$ days.
This is congruent to $5$ modulo $7$.
So, on consecutive leap years, $29$th February happens on:
:$(1): \quad$ Wednesday, Monday, Saturday, Thursday, Tuesday, Sunday, Friday.
Hence (apart from at the turn of the century), there are $4 \times 7 = 28$ years between consecutive years when the days coincide.
Hence after $1928$, the next one is $1956$.
$28$ years before $1928$ is $1900$.
But $1900$, as we have determined, is not a leap year.
If it were a leap year, then $28$ years before that, that is $1872$, would have had $5$ Wednesdays.
But there is a day less between $1872$ and $1900$ than there would be.
So $29$th February in $1872$ fell on a Thursday.
Hence, from the sequence of days in $(1)$ above, $29$th February falls on a Wednesday $4$ leap years later.
That is, $16$ years, which is $1888$.
== Sources ==
* : Solutions: $346$. -- A Leap Year Puzzle
* : Answers: $528$. A Leap Year Puzzle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/347 - The Iron Chain

== by : $347$ ==
;The Iron Chain
:Two pieces of iron chain were picked up on the battlefield.
:What purpose they had originally served is not certain, and does not immediately concern us.
:They were formed of circular iron links (all of the same size) out of metal half an inch thick.
:One piece of chain was exactly $3$ feet long, and the other $22$ inches in length.
:Now, as one piece contained exactly six links more than the other, how many links were there in each piece of chain?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $347$. -- The Iron Chain
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $253$. The Iron Chain
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/347 - The Iron Chain/Solution

== by : $347$ ==
;The Iron Chain
== Solution ==
The two pieces of chain contained $9$ and $15$ links respectively.
== Proof ==
The difficult bit here is getting the picture.
In order to visualise the situation, we present a diagram of a $3$-link chain:
:
We see that the length of a chain is equal to:
:the total lengths of the inner diameters of each link
plus:
:the thickness of the link for both of the links at either end.
Let $d$ inches be the inner diameter of one link.
Let $n$ be the number of links in the shorter chain.
We are given that the links are made of metal half an inch thick.
So twice the thickness of the links is $1$ inch.
Recall there are $12$ inches to the foot, so a $3$ foot length is $36$ inches
So, we have that:
So the shorter chain has $9$ links, while the longer chain has $9 + 6 = 15$ links.
== Sources ==
* : Solutions: $347$. -- The Iron Chain
* : Answers: $253$. The Iron Chain
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/348 - Blowing Out the Candle

== by : $348$ ==
;Blowing Out the Candle
:Candles were lighted on Colonel Crackham's breakfast-table one foggy morning.
:When the fog lifted, the Colonel rolled a sheet of paper into the form of a hollow cone, like a megaphone.
:He then challenged his young friends to use it in blowing out the candles.
:They failed, until he showed them the trick.
:Of course, you blow through the small end.
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $348$. -- Blowing Out the Candle
* : Unclassified Puzzles: $529$. Blowing Out the Candle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/348 - Blowing Out the Candle/Solution

== by : $348$ ==
;Blowing Out the Candle
== Solution ==
:The trick is to lower the funnel until the dotted line, $CA$, on the top in the diagram, is in line with the flame of the candle.
:Any attempt to blow the candle out with the flame opposite the centre of the opening of the funnel is hopeless.
== Proof ==
== Sources ==
* : Solutions: $348$. -- Blowing Out the Candle
* : Answers: $529$. Blowing Out the Candle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/349 - Releasing the Stick

== by : $349$ ==
;Releasing the Stick
:Here is a puzzle that will often cause a good deal of bewilderment amongst your friends,
::though it is not so generally known as it deserves to be.
:I think it was invented by the late , the American chess and puzzle genius.
:It is simply a loop of string passed through one end of a stick as here shown, but not long enough to pass round the other end.
:The puzzle is to suspend it in the manner shown from the top hole of a man's coat, and then get it free again.
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $349$. -- Releasing the Stick
* : Unclassified Puzzles: $530$. Releasing the Stick
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/349 - Releasing the Stick/Solution

== by : $349$ ==
;Releasing the Stick
== Solution ==
:Take the loop and pull it through as much of the coat above and below the buttonhole as necessary.
:It will then be possible to pass the lower end of the stick through the buttonhole and get it into the required position.
:To disentangle it, you reverse the process.
== Sources ==
* : Solutions: $349$. -- Releasing the Stick
* : Answers: $530$. Releasing the Stick
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/34 - The Banker and the Note

== by : $34$ ==
;The Banker and the Note
:A banker in a country town was walking down the street when he saw a $\pounds 5$ note on the kerb-stone.
:He picked it up, noted the number, and went to his private house for luncheon.
:His wife said that the butcher had sent in his bill for $\pounds 5$,
::and, as the only money he had was the note he had found, he gave it to her and she paid the butcher.
:The butcher paid it to a farmer in buying a calf,
::the farmer paid it to a merchant
::who in turn paid it to a laundry-woman,
::and she, remembering that she owed the bank $\pounds 5$, went there and paid the note.
:The banker recognised the note as the one he had found,
::and by that time it had paid $\pounds 25$ worth of debts.
:On careful examination he discovered that the note was counterfeit.
:Now, what was lost in the whole transaction, and by whom?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $34$. -- The Banker and the Note
* : Arithmetical and Algebraical Problems: Money Puzzles: $31$. The Banker and the Counterfeit Bill
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/34 - The Banker and the Note/Solution

== by : $34$ ==
;The Banker and the Note
== Solution ==
Because the identical forged banknote can be traced through all the transactions, all of them are invalid.
Therefore everybody stands in relation to their debtor exactly as they were before the banker picked up the note.
All, that is, except the butcher owes $\pounds 5$ to the farmer for the calf received.
== Sources ==
* : Solutions: $34$. -- The Banker and the Note
* : Answers: $31$. The Banker and the Counterfeit Bill
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/350 - The Keys and Ring

== by : $350$ ==
;The Keys and Ring
:Colonel Crackham the other day produced a ring and two keys, as here shown,
::cut out of a solid piece of cardboard, without a break or join anywhere.
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $350$. -- The Keys and Ring
* : Unclassified Puzzles: $531$. The Keys and Ring
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/350 - The Keys and Ring/Solution

== by : $350$ ==
;The Keys and Ring
== Solution ==
:
:First cut out the keys and ring in one piece, as here shown.
:Now cut only half through the cardboard along the eight little dark lines,
::and similarly cut half through on the eight little dotted lines on the other side of the cardboard.
:Then insert a penknife and split the card below the four little squares formed by these lines,
::and the keys will come apart with the keys loose on the ring and no join.
== Sources ==
* : Solutions: $350$. -- The Keys and Ring
* : Answers: $531$. The Keys and Ring
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/351 - The Entangled Scissors

== by : $351$ ==
;The Entangled Scissors
:Here is an old puzzle that many readers, who have forgotten to put on the string, will be glad to see again.
:If you start on the loop at the bottom, the string can readily be got into position.
:The puzzle is, of course, to let someone hold the two ends of the string until you disengage the scissors.
:A good length of string should be used to give you free play.
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $351$. -- The Entangled Scissors
* : Unclassified Puzzles: $532$. The Entangled Scissors
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/351 - The Entangled Scissors/Solution

== by : $351$ ==
;The Entangled Scissors
:Here is an old puzzle that many readers, who have forgotten to put on the string, will be glad to see again.
== Solution ==
:Slacken the string throughout so as to bring the scissors near the hand of the person holding the end.
:Then work the loop shown at the bottom of the diagram backwards along the double cord.
:Make the loop follow the double cord throughout, in and out through the fingerholes, until it is completely free of the scissors.
:Then pass the loop right round the points of the scissors and follow the double cord backwards.
:The string will then, if you have done this right, detach itself from the scissors.
== Sources ==
* : Solutions: $351$. -- The Entangled Scissors
* : Answers: $532$. The Entangled Scissors
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/352 - Locating the Coins

== by : $352$ ==
;Locating the Coins
:Said Dora to her brother:
::"Put a shilling in one of your pockets and a penny in the pocket in the opposite side.
::Now the shilling represents $12$ and the penny $1$.
::Triple the coin in your right pocket, and double that in your left pocket.
::Add these products together and tell me whether the result is odd or even."
:He said the result was even, and she immediately told him that the shilling was in the right pocket and the penny in the left one.
:Every time he tried it she told him correctly how the coins were located.
:How did she do it?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $352$. -- Locating the Coins
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $254$. Locating the Coins
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/352 - Locating the Coins/Solution

== by : $352$ ==
;Locating the Coins
== Solution ==
If the answer is "even", the shilling is in the right pocket and the penny in the left.
If the answer is "odd", the shilling is in the left pocket and the penny in the right.
== Proof ==
See 's from $1926$, no. $100$ - Odds and Evens, of which this a rehash.
== Also see ==
* Henry Ernest Dudeney: Modern Puzzles: $100$ - Odds and Evens
== Sources ==
* : Solutions: $352$. -- Locating the Coins
* : Answers: $254$. Locating the Coins
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/353 - The Three Sugar Basins

== by : $353$ ==
;The Three Sugar Basins
:Three basins each contain the same number of lumps of sugar,
::and nine cups are empty.
:If we transfer to each cup one-eighteenth of the number of lumps that each basin contains,
::we then find that each basin holds $12$ more lumps than each of the cups.
:How many lumps are there in each basin before they are removed?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $353$. -- The Three Sugar Basins
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $255$. The Three Sugar Basins
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/353 - The Three Sugar Basins/Solution

== by : $353$ ==
;The Three Sugar Basins
== Solution ==
:$36$
== Proof ==
Let $n$ be the number of lumps each basin contains before starting.
Each sugar bowl transfers a total of $9 \times \dfrac n {18} = \dfrac n 2$ lumps to the teacups.
So after this transaction, each sugar bowl contains $n - \dfrac n 2 = \dfrac n 2$ lumps.
Each teacup then contains a total of $3 \times \dfrac n {18} = \dfrac n 6$ lumps.
Hence:
:$\dfrac n 2 = \dfrac n 6 + 12$
which after algebra leads us to:
:$n = 36$
== Sources ==
* : Solutions: $353$. -- The Three Sugar Basins
* : Answers: $255$. The Three Sugar Basins
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/354 - The Wheels of the Car

== by : $354$ ==
;The Wheels of the Car
:"You see, sir," said the motor-car salesman, "at present the fore-wheel of the car I am selling you makes four revolutions more than the hind-wheel in going $120$ yards;
::but if you have the circumference of each wheel reduced by $3$ feet, it would make as many as six revolutions more than the hind-wheel in the same distance."
:Why the buyer wished that the difference in the number of revolutions between the two wheels should not be increased does not concern us.
:The puzzle is to discover the circumference of each wheel in the first case.
:It is quite easy.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $354$. -- The Wheels of the Car
* : Geometrical Problems: Circle Puzzles: $292$. The Wheels of the Car
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/354 - The Wheels of the Car/Solution

== by : $354$ ==
;The Wheels of the Car
== Solution ==
The fore-wheel was $15$ feet in circumference, while the hind-wheel was $18$ feet in circumference.
== Proof ==
We have that $120$ yards is $360$ feet.
We need to find $a$ and $b$ which are divisors of $360$ such that $a n = b \paren {n + 4}$.
We have that:
Thus we see that:
:$15 \times \paren {20 + 4} = 18 \times 20$
Reducing $15$ and $18$ by $3$ each gives $12$ and $15$, where:
:$12 \times \paren {24 + 6} = 15 \times 24$
confirming and corroborating our inkling that the circumferences were $15$ and $18$ feet respectively.
== Sources ==
* : Solutions: $354$. -- The Wheels of the Car
* : Answers: $292$. The Wheels of the Car
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/355 - The Seven Children

== by : $355$ ==
;The Seven Children
:Four boys and three girls are seated in a row at random.
::What are the chances that the two children at the end of each row will be girls?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $355$. -- The Seven Children
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $470$. The Seven Children
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/355 - The Seven Children/Solution

== by : $355$ ==
;The Seven Children
== Solution ==
$1$ in $7$.
== Proof ==
From Number of Permutations, there are $7! = 5040$ ways in total of seating the $7$ children.
Now, we are given that there are $3$ girls.
The number of ways of seating the two girls at either end is the number of permutations of $2$ out of $3$.
From Number of Permutations again, this is:
:${}^2 P_3 = \dfrac {3!} {\paren {3 - 2}!} = 3! = 6$
Having seated these two girls, there are $5$ children remaining to be seated in the remaining $5$ seats.
From Number of Permutations, there are $5! = 120$ ways of doing that.
Hence there are $6 \times 120 = 720$ ways of seating the children with a girl at either end.
Therefore the chances are $720$ in $5040$, or $1$ in $7$.
== Sources ==
* : Solutions: $355$. -- The Seven Children
* : Answers: $470$. The Seven Children
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions