nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Puzzles and Curious Problems/308 - Domino Hollow Squares

== by : $308$ ==
;Domino Hollow Squares
:It is required with the $28$ dominoes to form $7$ hollow squares, like the example given,
::so that the pips in the four sides of every square shall add up alike.
:
:All these seven squares need not have the same sum, and, of course, the example given need not be one of your set.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $308$. -- Domino Hollow Squares
* : Domino Puzzles: $487$. Domino Hollow Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/308 - Domino Hollow Squares/Solution

== by : $308$ ==
;Domino Hollow Squares
== Solution ==
: $\quad$ 
: $\quad$ 
: $\quad$ 
:
:Remember this rule:
:If the pips on your dominoes sum, say, to $7$ (as in the first example),
::and you wish the sides to add up to $3$, then $4 \times 3 - 7$ gives us $5$ as the sum of the four corners.
== Sources ==
* : Solutions: $308$. -- Domino Hollow Squares
* : Answers: $487$. Domino Hollow Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/309 - Domino Sequences

== by : $309$ ==
;Domino Sequences
:A boy who had a complete set of dominoes, up to double $9$, was trying to arrange them all in sequence, in the usual way --
::$6$ against $6$, $3$ against $3$, blank against blank, and so on.
:His father said to him, "You are attempting an impossibility, but if you let me pick out $4$ dominoes it can them be done.
::And those I take shall contain the smallest total number of pips possible in the circumstances.
:Now, which dominoes might the father have selected?
:Remember that the dominoes in common use stop at double $6$, but we are here using a set up to double $9$.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $309$. -- Domino Sequences
* : Domino Puzzles: $488$. Domino Sequences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/309 - Domino Sequences/Solution

== by : $309$ ==
;Domino Sequences
== Solution ==
Remove from the set the four dominoes:
:$7 - 6$, $5 - 4$, $3 - 2$, $1 - 0$.
The remaining dominoes can be put together in proper sequence.
Any other combinations of these particular numbers will also do as well, for example:
:$7 - 0$, $6 - 1$, $5 - 2$, $4 - 3$
Generally, for any set of dominoes ending in a double odd number, those removed must contain together every number once from blank to two less than the highest number in the set.
== Sources ==
* : Solutions: $309$. -- Domino Sequences
* : Answers: $488$. Domino Sequences
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/30 - Subscriptions

== by : $30$ ==
;Subscriptions
:Seven men agreed to subscribe towards a certain fund,
::and the first six gave $\pounds 10$ each.
:The other man gave $\pounds 3$ more than the average of the seven.
:What amount did the seventh man subscribe?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $30$. -- Subscriptions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/30 - Subscriptions/Solution

== by : $30$ ==
;Subscriptions
== Solution ==
:$\pounds 13, 10 \shillings$
== Proof ==
Let $T$ be the total amount subscribed.
Let $S$ be the amount contributed by the seventh man.
Then:
== Sources ==
* : Solutions: $30$. -- Subscriptions
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/310 - Two Domino Squares

== by : $310$ ==
;Two Domino Squares
:Arrange the $28$ dominoes as shown in the diagram to form two squares
::so that the pips in every one of the eight sides shall add up alike.
:
:The constant addition must of course be within limits to make the puzzle possible,
::and it will be interesting to find those limits.
:Of course, the dominoes need not be laid according to the rule, $6$ against $6$, blank against blank, and so on.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $310$. -- Two Domino Squares
* : Domino Puzzles: $489$. Two Domino Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/310 - Two Domino Squares/Solution

== by : $310$ ==
;Two Domino Squares
== Solution ==
This solution has $22$ as the constant sum.''
:
:With $22$ as this constant sum, the corners must add to $8$,
::with $23$ to $16$, with $24$ to $24$, with $25$ to $32$, with $26$ to $40$.
:The constant sum cannot be less than $22$ or more than $26$.
== Sources ==
* : Solutions: $310$. -- Two Domino Squares
* : Answers: $489$. Two Domino Squares
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/311 - Domino Multiplication

== by : $311$ ==
;Domino Multiplication
:We have received the following entertaining puzzle from W.D.W., of Philadelphia, Pa.:
:Four dominoes may be so placed as to form a simple multiplication sum if we regard the pips as figures.
:The example here shown will make everything perfectly clear.
:
:Now, the puzzle is, using all the $28$ dominoes to arrange them so as to form $7$ such little sums in multiplication.
:No blank may be placed at the left end of the multiplicand or product.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $311$. -- Domino Multiplication
* : Domino Puzzles: $490$. Domino Multiplication
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/312 - Domino Rectangle

== by : $312$ ==
;Domino Rectangle
:Arrange the $28$ dominoes exactly as shown in the diagram, where the pips are omitted,
::so that the pips in every one of the seven columns shall sum to $24$, and the pips in every one of the eight rows to $21$.
:
:The dominoes need not be $6$ against $6$, $4$ against $4$, and so on.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $312$. -- Domino Rectangle
* : Domino Puzzles: $491$. Domino Rectangle
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/313 - The Domino Column

== by : $313$ ==
;The Domino Column
:Arrange the $28$ dominoes in a column so that the three sets of pips, taken anywhere,
::shall add up alike on the left side and on the right.
:Such a column has been started in the diagram.
:
:This is merely an example, so you can start afresh if you like.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $313$. -- The Domino Column
* : Domino Puzzles: $492$. The Domino Column
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/313 - The Domino Column/Solution

== by : $313$ ==
;The Domino Column
== Solution ==
:
:The column has been broken into three for convenience in presentation.
:Place the second column under the first, and the third under the second.
== Sources ==
* : Solutions: $313$. -- The Domino Column
* : Answers: $492$. The Domino Column
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/314 - Card Shuffling

== by : $314$ ==
;Card Shuffling
:The rudimentary method of shuffling a pack of cards is to take the pack face downwards in the left hand and then transfer them one by one to the right hand,
::putting the second on top of the third, the third under, the fourth above, and so on until all are transferred.
:If you do this with any even number of cards and keep on repeating the shuffle in the same way,
:the cards will in due time return to their original order.
:Try with $4$ cards, and you will find the order is restored in $3$ shuffles.
:In fact, where the number of cards is $2$, $4$, $8$, $16$, $32$, $64$,
:the number of shuffles to get them back to the original arrangement is $2$, $3$, $4$, $5$, $6$, $7$ respectively.
:Now, how many shuffles are necessary in the case of $14$ cards?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $314$. -- Card Shuffling
* : Unclassified Puzzles: $512$. Card Shuffling
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/314 - Card Shuffling/Historical Note

== Historical Note on by : $314$ - Card Shuffling ==
 informs us that a shuffle of this kind is known as Monge's shuffle, for , who was apparently the first to investigate it.
Its mathematical theory is investigated in 's , $1960$, edited by .
== Sources ==
* : Answers: $512$. Card Shuffling
Category:Historical Notes
Henry Ernest Dudeney/Puzzles and Curious Problems/314 - Card Shuffling/Solution

== by : $314$ ==
;Card Shuffling
== Solution ==
:$14$ shuffles are needed for $14$ cards.
== Proof ==
Each shuffle can be represented as a permutation in $S_{14}$, the symmetric group on $14$ letters.
In two-row notation, this can be written as:
:$\begin {pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\
14 & 12 & 10 & 8 & 6 & 4 & 2 & 1 & 3 & 5 & 7 & 9 & 11 & 13
\end {pmatrix}$
which, when expressed in cycle notation, would be:
:$\begin {pmatrix} 1 & 14 & 13 & 11 & 7 & 2 & 12 & 9 & 3 & 10 & 5 & 6 & 4 & 8 \end {pmatrix}$
which is a single (disjoint) $14$-cycle.
Thus this permutation has order $14$.
This means that $14$ shuffles are required to get the cards back to their original arrangement.
== Historical Note ==
== Sources ==
* : Solutions: $314$. -- Card Shuffling
* : Answers: $512$. Card Shuffling
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/315 - Arranging the Dominoes

== by : $315$ ==
;Arranging the Dominoes
:The number of ways the set of $28$ dominoes may be arranged in a straight line, in accordance with the original rule of the game,
::left to right and right to left, in any arrangement counting as different ways,
::is $7 \, 959 \, 229 \, 931 \, 520$.
:After discarding all dominoes bearing a $5$ or a $6$, how many ways may the remaining $15$ dominoes be so arranged in a line?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $315$. -- Arranging the Dominoes
* : Domino Puzzles: $493$. Arranging the Dominoes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Category:Dominoes
Henry Ernest Dudeney/Puzzles and Curious Problems/315 - Arranging the Dominoes/Solution

== by : $315$ ==
;Arranging the Dominoes
== Solution ==
There are $126 \, 720$ ways to arrange $15$ dominoes as described.
== Proof ==
== Sources ==
* : Solutions: $315$. -- Arranging the Dominoes
* : Answers: $493$. Arranging the Dominoes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/316 - Queer Golf

== by : $316$ ==
;Queer Golf
:A certain links had nine holes, $300$, $250$, $200$, $325$, $275$, $350$, $225$, $375$, and $400$ yards apart.
:If a man could always strike the ball in a perfectly straight line and send it exactly one of two distances,
::so that it would either go towards the hole, pass over it, or drop into it,
::what would those two distances be that would carry him in the least number of strokes round the whole course?
:Two very good distances are $125$ and $75$, which carry you round in $28$ strokes,
::but this is not the correct answer.
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $316$. -- Queer Golf
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $466$. Queer Golf
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/316 - Queer Golf/Solution

== by : $316$ ==
;Queer Golf
== Solution ==
The two best distances are:
:$100$ yards (known as "the appproach")
and:
:$125$ yards (known as "the drive").
Hence:
:Hole $1$ can be reached in $3$ approaches
:Hole $2$ can be reached in $2$ drives
:Hole $3$ can be reached in $2$ approaches
:Hole $4$ can be reached in $2$ approaches and $1$ drive
:Hole $5$ can be reached in $3$ drives and $1$ backward approach
:Hole $6$ can be reached in $2$ drives and $1$ approach
:Hole $7$ can be reached in $1$ drive and $3$ approaches
:Hole $8$ can be reached in $3$ drives
:Hole $9$ can be reached in $4$ approaches
which is $26$ strokes in all.
== Sources ==
* : Solutions: $316$. -- Queer Golf
* : Answers: $466$. Queer Golf
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/317 - The Archery Match

== by : $317$ ==
;The Archery Match
:On a target on which the scoring was $40$ for the bull's-eye, and $39$, $24$, $23$, $17$ and $16$ respectively for the rings from the centre outwards, as shown in the diagram,
::three players had a match with six arrows each.
:
:The result was:
::Miss Dora Talbot: $120$ points;
::Reggie Watson, $110$ points;
::Mrs. Finch, $100$ points.
:Every arrow scored, and the bull's-eye was only once hit.
:Can you, from these facts, determine the exact six hits made by each competitor?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $317$. -- The Archery Match
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $467$. The Archery Match
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/317 - The Archery Match/Solution

== by : $317$ ==
;The Archery Match
== Solution ==
Mrs. Finch scored $100$ with four $17$s and two $16$s.
Reggie Watson scored $110$ with two $23$s and four $16$s.
Miss Dora Talbot scored $120$ with one $40$ and five $16$s.
There are other ways to make $120$, but only one which uses the bull's-eye.
This is the only way the bull's-eye can be used and still fit the constraints of the question.
== Sources ==
* : Solutions: $317$. -- The Archery Match
* : Answers: $467$. The Archery Match
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/318 - Target Practice

== by : $318$ ==
;Target Practice
:Three people in an archery competition had each had six shots at a target, and the result is shown in the diagram,
::where they all hit the target every time.
:
:The bull's-eye scores $50$, then the scores are $25$, $20$, $10$, $5$, $3$, $2$, $1$ for the rings from the centre outwards respectively.
:It is seen that the hits on target are $1$ bull's-eye, two $25$s, three $20$s, three $10$s, three $1$s and two hits in each of the other rings.
:The three men tied with an equal score.
:Can you work out who hit what?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $318$. -- Target Practice
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $468$. Target Practice
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/318 - Target Practice/Solution

== by : $318$ ==
;Target Practice
== Solution ==
The total score was $213$, so each contestant scored $71$.
This can be done as:
:one man scores $50$, $10$, $5$, $3$, $2$, $1$
:one man scores $25$, $20$, $20$, $3$, $2$, $1$
:one man scores $25$, $20$, $10$, $10$, $5$, $1$
== Sources ==
* : Solutions: $318$. -- Target Practice
* : Answers: $468$. Target Practice
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/319 - The Ten Cards

== by : $319$ ==
;The Ten Cards
:Place any ten playing cards in a row face up.
:There are two players.
:The first player may turn face down any single card he chooses.
:Then the second player can turn face down any single card or any $2$ adjacent cards.
:And so on.
:Thus the first player must turn face down a single, but afterwards either player may turn down either a single or two adjacent cards.
:The player who turns down the last card wins.
:Should the first or second player win?
=== Click here for solution ===
== Sources ==
* : Crossing River Problem, and Problems Concerning Games and Puzzle Games: $319$. -- The Ten Cards
* : Game Puzzles: $478$. The Ten Cards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/319 - The Ten Cards/Solution

== by : $319$ ==
;The Ten Cards
== Solution ==
The first player can always win.
== Proof ==
Let $A$ denote the person who plays first, and $B$ denote the person who plays second.
Let $O$ denote a card turned up, and $\text X$ denote a card turned down.
There are $3$ ways $A$ can win.
;Third Card
$A$ turns down the $3$rd from either end.
This leaves:
:$00 \text X 0000000$
Whatever happens next, $A$ can always leave one of the following:
:$000 \text X 000$
:$00 \text X 00 \text X 0 \text X 0$
:$0 \text X 00 \text X 000$
The order does not matter.
In the first case, $A$ copies in one triplet what $B$ does in the other triplet, until he gets the last card.
In the second case, $A$ similarly copies $B$ until he gets the last card.
In the third case, whatever $B$ does, $A$ can leave:
:$0 \text X 0$
:$0 \text X 0 \text X 0 \text X 0$
:$00 \text X 00$
and again the win is apparent.
;Second Card
$A$ turns down the $2$nd from either end.
This leaves:
:$0 \text X 00000000$
;Fourth Card
$A$ turns down the $4$th card from either end.
This leaves the cards with $3$ turned up, one turned down, and $6$ turned up.
:$000 \text X 000000$
== Historical Note ==
== Sources ==
* : Solutions: $319$. -- The Ten Cards
* : Answers: $478$. The Ten Cards
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/31 - A Queer Settling Up

== by : $31$ ==
;A Queer Settling Up
:Person 1: "Here is my purse, give me just as much money as you find in it."
:Person 2, having done that: "If you give me as much as I have left of my own, we shall be square."
:After Person 2 has done that, Person 1 find his purse contains three shillings and sixpence,
::while Person 2 has three shillings.
:How much did each possess at the start?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $31$. -- A Queer Settling Up
* : Arithmetical and Algebraical Problems: Money Puzzles: $28$. A Queer Settling Up
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/31 - A Queer Settling Up/Solution

== by : $31$ ==
;A Queer Settling Up
== Solution ==
:Person $1$ started with $2 \shillings 6 \oldpence$
:Person $2$ started with $4 \shillings$
== Proof ==
Before the second transaction:
:person $2$ has half of three shillings, that is, $1 \shillings 6 \oldpence$
:person $1$ has ${3 \shillings 6 \oldpence} + {1 \shillings 6 \oldpence}$, that is, $5 \shillings$
So before the first transaction:
:person $1$ has half of $5 \shillings$, that is, $2 \shillings 6 \oldpence$
:person $2$ has ${1 \shillings 6 \oldpence} + {2 \shillings 6 \oldpence}$, that is, $4 \shillings$
Hence the result.
== Sources ==
* : Solutions: $31$. -- A Queer Settling Up
* : Answers: $28$. A Queer Settling Up
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/320 - An Awkward Time

== by : $320$ ==
;An Awkward Time
:When I told someone the other morning that I had to catch the $12:50$ train, he told me it was a very awkward time for a train to start.
:I asked him to explain why.
:Can you guess his answer?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $320$. -- An Awkward Time
* : Unclassified Puzzles: $523$. An Awkward Time
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/320 - An Awkward Time/Solution

== by : $320$ ==
;An Awkward Time
== Solution ==
:Because it's ten to one if you catch it.
A betting joke.
== Sources ==
* : Solutions: $320$. -- An Awkward Time
* : Answers: $523$. An Awkward Time
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/321 - Cryptic Addition

== by : $321$ ==
;Cryptic Addition
:Can you prove that the following addition sum is correct?
:
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $321$. -- Cryptic Addition
* : Unclassified Puzzles: $524$. Cryptic Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/321 - Cryptic Addition/Solution

== by : $321$ ==
;Cryptic Addition
== Solution ==
:Look at it upside down and you will see it says that one, nine, one and eight added together make nineteen.
== Sources ==
* : Solutions: $321$. -- Cryptic Addition
* : Answers: $524$. Cryptic Addition
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/322 - The New Gun

== by : $322$ ==
;The New Gun
:An inventor undertook that a new gun which he had manufactured, which when once loaded,
::would fire fifteen shots at the rate of a shot a minute.
:A series of tests were made, and the gun certainly fired fifteen shots in a quarter of an hour.
:However, the Government refused to buy the gun, on the grounds that it did not do the job as advertised.
:Why?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $322$. -- The New Gun
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/322 - The New Gun/Solution

== by : $322$ ==
;The New Gun
== Solution ==
For the same reason they did not buy the gun in the puzzle published as $99$ - The Rejected Gun in 's $1926$ collection , of which this is a retread:
== Sources ==
* : Solutions: $322$. -- The New Gun
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/323 - Cats and Mice

== by : $323$ ==
;Cats and Mice
:A number of cats (more than one) killed between them $999 \, 919$ mice, and every cat killed an equal number of mice (more than one).
:Each cat killed more mice than there were cats.
:How many cats were there?
's original question was wordy to the extreme.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $323$. -- Cats and Mice
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $251$. Cats and Mice
Category:Henry Ernest Dudeney/Puzzles and Curious Problems
Henry Ernest Dudeney/Puzzles and Curious Problems/323 - Cats and Mice/Solution

== by : $323$ ==
;Cats and Mice
== Solution ==
:$991$ cats.
== Proof ==
:$999 \, 919$ is a semiprime whose proper divisors are $991$ and $1009$.
== Sources ==
* : Solutions: $323$. -- Cats and Mice
* : Answers: $251$. Cats and Mice
Category:Henry Ernest Dudeney/Puzzles and Curious Problems/Solutions
Henry Ernest Dudeney/Puzzles and Curious Problems/324 - The Two Snakes

== by : $324$ ==
;The Two Snakes
:Suppose two snakes started swallowing one another simultaneously,
::each getting the tail of the other in its mouth,
::so that the circle formed by the snakes becomes smaller and smaller.
:What will eventually happen?
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $324$. -- The Two Snakes
* : Unclassified Puzzles: $525$. The Two Snakes
Category:Henry Ernest Dudeney/Puzzles and Curious Problems