nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/193 - A Domino Square

== by : $193$ ==
;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.
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $193$. -- A Domino Square
* : Domino Puzzles: $482$. A Domino Square
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/194 - A Domino Star

== by : $194$ ==
;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.
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $194$. -- A Domino Star
* : Domino Puzzles: $483$. A Domino Star
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/194 - A Domino Star/Solution

== by : $194$ ==
;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.
== Solution ==
:
== Sources ==
* : Solutions: $194$. -- A Domino Star
* : Answers: $483$. A Domino Star
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/195 - Domino Groups

== by : $195$ ==
;Domino Groups
:I wonder how many of my readers know that if you lay out the $28$ dominoes in line according to the ordinary rules --
::$6$ against $6$, $2$ against $2$, blank against blank, and so on --
::the last number must always be the same as the first, so that they will really always form a circle.
:It is a very ancient trick to conceal one domino (but do not take out a double) and then ask him to arrange all the others in line without your seeing.
:It will astonish him when you tell him, after he has succeeded, what the two end numbers are.
:They must be those on the domino that you have withdrawn, for that domino completes the circle.
: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?
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $195$. -- Domino Groups
* : Domino Puzzles: $484$. Domino Groups
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/196 - Les Quadrilles

== by : $196$ ==
;Les Quadrilles
:This old French puzzle will, I think, be found very interesting.
: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?
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $196$. -- Les Quadrilles
* : Domino Puzzles: $485$. Les Quadrilles
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/196 - Les Quadrilles/Historical Note

== Historical Note on by : $196$ - Les Quadrilles ==
Domino problems of this type, called quadrilles, date back some way.
 discussed them in his of $1894$.
More recently, they appeared in an issue of of $1964$.
== Sources ==
* 
* 
* : Answers: $485$. Les Quadrilles
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/196 - Les Quadrilles/Solution

== by : $196$ ==
;Les Quadrilles
== Solution ==
:Here is one solution:
:
== Historical Note ==
== Sources ==
* : Solutions: $196$. -- Les Quadrilles
* : Answers: $485$. Les Quadrilles
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/197 - A Puzzle with Cards

== by : $197$ ==
;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?
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $197$. -- A Puzzle with Cards
* : Unclassified Puzzles: $511$. A Puzzle with Cards
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/197 - A Puzzle with Cards/Solution

== by : $197$ ==
;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?
== Solution ==
Arrange the pack in the following order face down with $9 \, \clubsuit$ at the top and $5 \, \spadesuit$ at the bottom.
Reading across:
:$\begin{array}{ccccccccccccc}
9 \, \clubsuit & \mathrm J \, \diamondsuit & 5 \, \clubsuit & \mathrm A \, \diamondsuit & \mathrm K \, \heartsuit & \mathrm K \, \spadesuit & 7 \, \heartsuit & 2 \, \diamondsuit & 6 \, \spadesuit & \mathrm Q \, \diamondsuit & 10 \, \spadesuit & \mathrm A \, \spadesuit & 3 \, \clubsuit \\
3 \, \diamondsuit & 8 \, \clubsuit & \mathrm K \, \diamondsuit & 8 \, \heartsuit & 7 \, \clubsuit & 4 \, \diamondsuit & 2 \, \spadesuit & \mathrm A \, \heartsuit & \mathrm A \, \clubsuit & 7 \, \spadesuit & 5 \, \diamondsuit & 9 \, \heartsuit & 2 \, \heartsuit\\
\mathrm J \, \spadesuit & 6 \, \diamondsuit & \mathrm Q \, \clubsuit & 6 \, \clubsuit & 10 \, \heartsuit & 3 \, \spadesuit & 3 \, \heartsuit & 7 \, \diamondsuit & 4 \, \clubsuit & 2 \, \clubsuit & 8 \, \spadesuit & \mathrm J \, \heartsuit & 4 \, \heartsuit\\
8 \, \diamondsuit & \mathrm J \, \clubsuit & 4 \, \spadesuit & \mathrm Q \, \spadesuit & \mathrm K \, \clubsuit & 9 \, \diamondsuit & 5 \, \heartsuit & 10 \, \clubsuit & \mathrm Q \, \heartsuit & 10 \, \diamondsuit & 9 \, \spadesuit & 6 \, \heartsuit & 5 \, \spadesuit & \end{array}$
Puzzles of this nature are straightforward to solve as follows:
Start with the last card to be spelled, and perform all operations in the reverse order.
You will end up with the complete set of cards in the correct order.
== Sources ==
* : Solutions: $197$. -- A Puzzle with Cards
* : Answers: $511$. A Puzzle with Cards
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/198 - A Card Trick

== by : $198$ ==
;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.
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $198$. -- A Card Trick
* : Arithmetical and Algebraical Problems: Miscellaneous Puzzles: $183$. A Card Trick
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/198 - A Card Trick/Solution

== by : $198$ ==
;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.
== Solution ==
Let there be $n$ piles.
Let $p_k$ denote the number of cards in pile $k$.
Let $b_k$ denote the value of the bottom card.
Let $m$ be the number of cards left over.
We have that:
:$p_k = 13 - b_k$
for all $1 \le k \le n$.
That is, the sum of all the bottom cards in the piles equals the number of piles minus $4$, all times $13$, plus the ones left over.
== Sources ==
* : Solutions: $198$. -- A Card Trick
* : Answers: $183$. A Card Trick
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/199 - Golf Competition Puzzle

== by : $199$ ==
;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}$
:For such a small number it is easy but interesting.
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $199$. -- Golf Competition Puzzle
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $451$. Golf Competition Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/199 - Golf Competition Puzzle/Solution

== by : $199$ ==
;Golf Competition Puzzle
== Solution ==
Here is one arrangement:
$\qquad \begin {array} {r | c |}
\text {Rounds} & 1 & 2 & 3 & 4 & 5 \\
\hline \text{$1$st Links} & CE & DE & DF & CF & AB \\
\hline \text{$2$nd Links} & AF & BF & AE & BE & CD \\
\hline \text{$3$rd Links} & BD & AC & BC & AD & EF \\ 
\hline \end{array}$
== Derivation ==
Fill up the final column with any pairings.
Then each of the remaining four columns of each row consists of both possible pairings of the other four players.
The first row can again be assigned arbitrarily.
The second and third row follow after some straightforward trial and error.
== Sources ==
* : Solutions: $199$. -- Golf Competition Puzzle
* : Answers: $451$. Golf Competition Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/19 - Market Transactions

== by : $19$ ==
;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?
:Most people will solve this, if they succeed at all, by more or less laborious trial,
:but there are several direct ways of getting the solution.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $19$. -- Market Transactions
* : Arithmetical and Algebraical Problems: Money Puzzles: $11$. Market Transactions
Category:Henry Ernest Dudeney/Modern Puzzles
Category:One Hundred Fowls Problems
Henry Ernest Dudeney/Modern Puzzles/19 - Market Transactions/Solution

== by : $19$ ==
;Market Transactions
== Solution ==
:$19$ cows, $1$ sheep and $80$ rabbits.
== Proof ==
Recall:
:$1$ pound sterling ($\pounds 1$) is $20$ shillings ($20 \shillings$)
Let all prices be expressed, therefore, in shillings.
Thus:
:the price of cows is $5 \times 20 = 100 \shillings$ each
:the price of sheep is $20 \shillings$ each
:the price of rabbits is $1 \shillings$ each
and:
:the total amount of money to spend is $100 \times 20 = 2000 \shillings$.
Let $c$, $s$ and $r$ denote the number of cows, sheep and rabbits respectively.
We have:
Note that both $c$ and $s$ need to be positive.
We need to find possible values of $c$ such that $1900 - 99 c$ is divisible by $19$.
This can happen only when $c$ itself is divisible by $19$.
It is implicit that there are at least some cows are bought, so the solution:
:$c = 0, s = 100, r = 0$
is usually ruled out.
Hence we have:
:$c = 19, s = 1, r = 80$
== Sources ==
* : Solutions: $19$. -- Market Transactions
* : Answers: $11$. Market Transactions
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/1 - Concerning a Cheque

== by : $1$ ==
;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.
=== Click here for solution ===
=== Variant ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $1$. -- Concerning a Cheque
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/1 - Concerning a Cheque/Solution

== by : $1$ ==
;Concerning a Cheque
== Solution ==
As puts it:
:If you set to work under the notion that there were only pounds and shillings -- no pence --- in the amount, a solution is impossible.
The amount on the cheque was $\pounds 5 \ 11 \shillings 6 \oldpence$
He received $\pounds 11 \ 5 \shillings 6 \oldpence$
After spending half a crown, which is $2 \shillings 6 \oldpence$, he had $\pounds 11 \ 3 \shillings$
This is twice as much as $\pounds 5 \ 11 \shillings 6 \oldpence$
== Proof ==
We recall the conversion factors:
:$12$ pennies make one shilling
:$20$ shillings make one pound.
:$1$ half crown consists of $2$ shillings and $6$ pence.
Let $C$ be the amount written on the cheque.
Let $C'$ be the amount paid out to the man by the bank.
Let $P$ be the final amount in the man's pocket.
Suppose $C$ were an integral number of shillings.
Then $C'$ would also be an integral number of shillings.
But $P$, which is $C'$ less $2 \tfrac 1 2$ shillings is then not an integral number of shillings.
That means $P$ could not be twice $C$. 
Hence 's warning.
Let $C$ consist of $C_l$ pounds, $C_s$ shillings and $C_d$ pennies.
Let $P$ consist of $P_l$ pounds, $P_s$ shillings and $P_d$ pennies.
But what we can say about the pennies is that:
:$(1): \quad 2 \times \paren {C_d - 6} = P_d \pmod {12}$
because $2 \paren {C - 2 \shillings 6 \oldpence} = P$
and also that:
:$(2): \quad C_d - 6 = P_d \pmod {12}$
because $P_d$ is the actual number of pennies.
from which it follows that $C_d = 6$.
Let $D_1$ be the value of the cheque in pennies.
Let $D_2$ be the money the man left the bank with in pennies.
Let $D_3$ be the money the man arrived home with in pennies.
We have:
:$D_1 = 240 C_l + 12 C_s + 6$
After coming out of the bank, the man has:
:$D_2 = 240 C_s + 12 C_l + 6$
After arriving home, the man has:
:$D_3 = 240 C_s + 12 \paren {C_l - 2}$
But we have:
:$D_1 \times 2 = D_3$
which leads us to:
By inspection, we arrive at:
:$13 \times 5 = 65 = 6 \times 11 - 1$
and the answer follows.
== Sources ==
* : Solutions: $1$. -- Concerning a Cheque
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/200 - Cricket Scores

== by : $200$ ==
;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?
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $200$. -- Cricket Scores
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/200 - Cricket Scores/Solution

== by : $200$ ==
;Cricket Scores
== Solution ==
== Proof ==
Let $a$, $b$, $c$ and $d$ be Great Muddleton's first innings' score, Little Wurzleford's first innings' score, Great Muddleton's second innings' score and Little Wurzleford's second innings' score respectively.
Then:
The rest of the scores fall to simple arithmetic.
== Sources ==
* : Solutions: $200$. -- Cricket Scores
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/201 - Football Results

== by : $201$ ==
;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?
=== Click here for solution ===
== Sources ==
* : Problems Concerning Games: $201$. -- Football Results
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $452$. Football Results
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/201 - Football Results/Solution

== by : $201$ ==
;Football Results
== Solution ==
The final scores:
:Scotland $3$, England $0$
:Scotland $2$, Wales $1$
:Scotland $2$, Ireland $0$
:England $2$, Ireland $0$
:England $0$, Wales $0$
:Wales $2$, Ireland $1$
== Proof ==
We are given that England conceded $3$ goals to Scotland.
Thus must have been the only $3$ goals they conceded.
Therefore the draw with Wales (which was the only drawn match) must have been goalless.
England's only other match was against Ireland, which they won by $2$ goals against $0$.
Wales must have lost against Scotland and won against Ireland.
The goal conceded by Scotland was pushed in by either Wales or Ireland.
Suppose Wales conceded $1$ against Scotland.
Then Ireland must have scored $2$ against Wales, which could not happen.
So Wales did not concede $1$ against Scotland
Suppose Wales conceded $3$ against Scotland.
Then Ireland's goal must have been against Scotland.
But then Scotland only put one past Ireland.
But that would mean the Scotland-Ireland match was a $1-1$ draw, which it could not have been.
The only other possibility is that Wales conceded $2$ against Scotland.
Then Wales also conceded $1$ against Ireland.
Hence Ireland scored no goals against Scotland.
Then Wales must have scored the only goal against Scotland.
This means they must have scored $2$ against Ireland.
Hence Scotland also scored $2$ against Ireland, and the table is complete.
== Sources ==
* : Solutions: $201$. -- Football Results
* : Answers: $452$. Football Results
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/202 - Noughts and Crosses

== by : $202$ ==
;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, ,
::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?
=== Click here for solution ===
== Linguistic Note ==
== Sources ==
* : Puzzle Games: $202$. -- Noughts and Crosses
* : Game Puzzles: $471$. Tic Tac Toe
Category:Henry Ernest Dudeney/Modern Puzzles