nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/203 - The Horse-Shoe Game

== by : $203$ ==
;The Horse-Shoe Game
:This little game is an interesting companion to our "Noughts and Crosses".
:There are two players.
:One has two white counters, the other two black.
:Playing alternately, each places a counter on a vacant point, where he leaves it.
:When all are played, you slide only, and the player is beaten who is so blocked that he cannot move.
:In the example, Black has just placed his lower counter.
:
:White now slides his lower one to the centre, and wins.
:Black should have played to the centre himself, and won.
:Now, which player ought to win at this game?
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $203$. -- The Horse-Shoe Game
* : Game Puzzles: $472$. The Horseshoe Game
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/203 - The Horse-Shoe Game/Solution

== by : $203$ ==
;The Horse-Shoe Game
== Solution ==
Every game should end up as a draw.
The only way one player can win is through a blunder made by his opponent.
== Proof ==
== Sources ==
* 
* : Solutions: $203$. -- The Horse-Shoe Game
* : Answers: $472$. The Horseshoe Game
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/204 - Turning the Die

== by : $204$ ==
;Turning the Die
:This is played with a single die.
:The first player calls any number he chooses, from $1$ to $6$, and the second player throws the die at hazard.
:Then they take it in turns to roll over the die in any direction they choose, but never giving it more than a quarter turn.
:The score increases as they proceed, and the player wins who manages to score $25$ or forces his opponent to score beyond $25$.
:I will give an example game.
:Player $A$ calls $6$, and $B$ happens to throw $3$, making the score $9$.
:Now $A$ decides to turn up $1$, scoring $10$;
::$B$ turns up $3$, scoring $13$;
::$A$ turns up $6$, scoring $19$;
::$B$ turns up $3$, scoring $22$;
::$A$ turns up $1$, scoring $23$;
::and $B$ turns up $2$, scoring $25$ and winning.
:What call should $A$ make in order to have the best chance at winning?
:Remember that the numbers on opposite sides of a correct die always sum to $7$, that is, $1 - 6$, $2 - 5$, $3 - 4$.
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $204$. -- Turning the Die
* : Game Puzzles: $473$. Turning the Die
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/204 - Turning the Die/Solution

== by : $204$ ==
;Turning the Die
:This is played with a single die.
:The first player calls any number he chooses, from $1$ to $6$, and the second player throws the die at hazard.
:Then they take it in turns to roll over the die in any direction they choose, but never giving it more than a quarter turn.
:The score increases as they proceed, and the player wins who manages to score $25$ or forces his opponent to score beyond $25$.
:I will give an example game.
:Player $A$ calls $6$, and $B$ happens to throw $3$, making the score $9$.
:Now $A$ decides to turn up $1$, scoring $10$;
::$B$ turns up $3$, scoring $13$;
::$A$ turns up $6$, scoring $19$;
::$B$ turns up $3$, scoring $22$;
::$A$ turns up $1$, scoring $23$;
::and $B$ turns up $2$, scoring $25$ and winning.
:What call should $A$ make in order to have the best chance at winning?
:Remember that the numbers on opposite sides of a correct die always sum to $7$, that is, $1 - 6$, $2 - 5$, $3 - 4$.
== Solution ==
The best call is either $2$ or $3$.
In either case, only one specific throw will defeat him.
If he calls $1$ then either $3$ or $6$ defeats him.
If he calls $2$, then only $5$ defeats him.
If he calls $3$, then only $4$ defeats him.
If he calls $4$, then $3$ or $4$ defeats him.
If he calls $5$, then $2$ or $3$ defeats him.
If he calls $6$, then $1$ or $5$ defeats him.
== Proof ==
The basic idea is that if at any time you score $5$, $6$, $9$, $10$, $14$, $15$, $18$, $19$ or $23$ with the die any side up, you should lose.
If you score $7$ or $16$ with any side up, you win.
The chances of winning with any other score depends on how the die lies.
== Sources ==
* : Solutions: $204$. -- Turning the Die
* : Answers: $473$. Turning the Die
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/205 - The Three Dice

== by : $205$ ==
;The Three Dice
:Mason and Jackson were playing with three dice.
:The player won whenever the numbers thrown added up to one of the two numbers he selected at the beginning of the game.
:As a matter of fact, Mason selected $7$ and $13$, and one of his winning throws was $6$, $4$, $3$.
:What were his chances of winning a throw?
:And what two other numbers should Jackson have selected for his own throws to make his chances of winning exactly equal?
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $205$. -- The Three Dice
* : Game Puzzles: $474$. The Three Dice
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/205 - The Three Dice/Solution

== by : $205$ ==
;The Three Dice
== Solution ==
Mason's chances of winning were $1$ in $6$.
If Jackson selected $8$ and $14$, his chances were the same.
== Proof ==
To make $7$, you need to make the throws:
:$1 \ 1 \ 5$
:$1 \ 2 \ 4$
:$1 \ 3 \ 3$
:$1 \ 4 \ 2$
:$1 \ 5 \ 1$
:$2 \ 1 \ 4$
:$2 \ 2 \ 3$
:$2 \ 3 \ 2$
:$2 \ 4 \ 1$
:$3 \ 1 \ 3$
:$3 \ 2 \ 2$
:$3 \ 3 \ 1$
:$4 \ 1 \ 2$
:$4 \ 2 \ 1$
:$5 \ 1 \ 1$
a total of $15$ throws.
To make $13$, you need to make the throws:
:$1 \ 6 \ 6$
:$2 \ 5 \ 6$
:$2 \ 6 \ 5$
:$3 \ 4 \ 6$
:$3 \ 5 \ 5$
:$3 \ 6 \ 4$
:$4 \ 3 \ 6$
:$4 \ 4 \ 5$
:$4 \ 5 \ 4$
:$4 \ 6 \ 3$
:$5 \ 2 \ 6$
:$5 \ 3 \ 5$
:$5 \ 4 \ 4$
:$5 \ 5 \ 3$
:$5 \ 6 \ 2$
:$6 \ 6 \ 1$
:$6 \ 5 \ 2$
:$6 \ 4 \ 3$
:$6 \ 3 \ 4$
:$6 \ 2 \ 5$
:$6 \ 6 \ 1$
a total of $21$ throws.
Hence Mason wins with a total of $36$ throws.
For each of throw $a \ b \ c$ there exists the complementary throw $\paren {7 - a} \ \paren {7 - b} \ \paren {7 - c}$ which makes a total of $21 - \paren {a + b + c}$.
Hence there is an equal chance of making $21 - \paren {a + b + c}$ as there is of making $a + b + c$.
Hence Jackson's numbers are $21 - 13$ and $21 - 7$, giving $8$ and $14$.
== Sources ==
* : Solutions: $205$. -- The Three Dice
* : Answers: $474$. The Three Dice
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/206 - The 37 Puzzle Game

== by : $206$ ==
;The $37$ Puzzle Game
:Here is a beautiful new puzzle game, absurdly simple to play but quite fascinating.
:To most people it will seem to be practically a game of chance -- equal for both players --
::but there are pretty subtleties in it, and I will show how to win with certainty.
:Place the five dominoes $1$, $2$, $3$, $4$, $5$, on the table.
:There are two players, who play alternately.
:The first player places a coin on any domino, say the $5$, which scores $5$;
::then the second player removes the coin to another domino, say to the $3$,
::and adds that domino, scoring $8$;
::then the first player removes the coin again, say to the $1$, scoring $9$; and so on.
:The player who scores $37$, or forces his opponent to score more than $37$, wins.
:Remember, the coin must be removed to a different domino at each play.
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $206$. -- The $37$ Puzzle Game
* : Game Puzzles: $475$. The $37$ Puzzle Game
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dominoes
Henry Ernest Dudeney/Modern Puzzles/206 - The 37 Puzzle Game/Solution

== by : $206$ ==
;The $37$ Puzzle Game
== Solution ==
Player $A$ can always win.
== Proof ==
The first player leads with $4$.
The winning scores which must be secured during play are $4$, $11$, $17$, $24$, $30$, $37$.
Let us refer to these as the milestones.
It is noted that there are gaps of $7$, $6$, $7$, $6$ and $7$ between them.
Once player $A$ secures one of these numbers, it is not possible for player $B$ to prevent player $A$ to reach the next one or the one after that.
Suppose player $A$ is on milestone $P$ such that there are $6$ points to the next milestone $Q$.
Let $A$ be placed on point $n$.
Let $B$ then move to point $m$ such that $m \ne 3$.
Then $A$ moves to $6 - m$ and so achieves the next milestone.
Now if $B$ did move to the $3$, $A$ cannot move to the $3$, but instead moves to the $5$.
There are then $5$ points to the next milestone, but as $A$ is already on the $5$, $B$ cannot move to the $5$.
So if $B$ plays $n$, $A$ then playes $5 - n$ which cannot equal $n$.
Thus $A$ reaches the milestone $13$ points after $P$.
Suppose player $B$ is on milestone $P$ such that there are $7$ points to the next milestone $Q$.
Let $A$ play any $n$ such that $n > 1$.
Then $B$ plays $7 - n$ and has then achieved $Q$.
So, instead, let $A$ play $1$.
Then there are $6$ points to $Q$.
$B$ plays $3$, which is bound to be empty because $A$ just played $1$.
But $A$ now cannot play $3$.
If he plays $n$ less than $3$, $B$ plays $3 - n$ and achieves milestone $Q$.
So, let $A$ play $4$ or $5$, to take him $1$ point beyond $Q$.
If $Q - 37$, he has just lost the game.
Otherwise there are $5$ or $4$ points to the next milestone, neither of which point $A$ is actually on.
So $A$ achieves the next milestone after $Q$ after all -- or has won the game.
 describes the above by providing sample games.
== Sources ==
* : Solutions: $206$. -- The $37$ Puzzle Game
* : Answers: $475$. The $37$ Puzzle Game
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/207 - The Twenty-Two Game

== by : $207$ ==
;The Twenty-Two Game
:Here is a variation of our little "Thirty-one Game" (: No. $79$).
:Lay out the $16$ cards as shown.
$\qquad \begin{matrix}
\boxed {A \heartsuit} & \boxed {A \spadesuit} & \boxed {A \diamondsuit} & \boxed {A \clubsuit} \\
\boxed {2 \heartsuit} & \boxed {2 \spadesuit} & \boxed {2 \diamondsuit} & \boxed {2 \clubsuit} \\
\boxed {3 \heartsuit} & \boxed {3 \spadesuit} & \boxed {3 \diamondsuit} & \boxed {3 \clubsuit} \\
\boxed {4 \heartsuit} & \boxed {4 \spadesuit} & \boxed {4 \diamondsuit} & \boxed {4 \clubsuit} \\
\end{matrix}$
:Two players alternately turn down a card and add it to the common score,
::and the player who makes the score of $22$, or forces his opponent to go beyond that number, wins.
:For example, $A$ turns down a $4$, $B$ turns down $3$ (counting $7$), $A$ turns down a $4$ (counting $11$),
::$B$ plays a $2$ (counting $13$), $A$ plays $1$ (counting $14$), $B$ plays $3$ ($17$), and whatever $A$ does, $B$ scores the winning $22$ next play.
:Now, which player should always win, and how?
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $207$. -- The Twenty-Two Game
* : Game Puzzles: $476$. The $22$ Game
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/207 - The Twenty-Two Game/Solution

== by : $207$ ==
;The Twenty-Two Game
== Solution ==
The first player can always win by playing $1$.
== Proof ==
Apart from the cards running out, the winning series is:
:$7, 12, 17, 22$
If you can score $17$ and leave at least one pair adding up to $5$ of both kinds: $1 - 4$ and $3 - 2$, you must win.
If you can score $12$ and leave at least two pairs adding up to $5$ of both kinds, you must win.
If you can score $7$ and leave at least $3$ pairs adding up to $5$ of both kinds, you must win.
Call the first player $A$ and the second player $B$.
So, if the $A$ plays $3$ or $4$, $B$ plays $4$ and $3$, gets $7$, and wins.
Nothing now stops $B$ scoring $12$, then $17$, then $22$.
If $A$ plays $2$, then $B$ replies with either $3$ or $2$.
So, for example: $2 - 3$, $2 - 3$, $2 - 3$, $2 - 3$ making $20$, and there is no $2$ left to make $22$.
$A$ must then play $1$ leaving another $1$ for $B$ to play and win.
Similarly: $2 - 3$, $1 - 3$, 3 - 2$, 3 - 2$ giving $19$ and again $B$ wins, as there is no $3$ left to make $22$.
And so on.
So, $A$ plays $1$.
Examples:
:$1 - 1$, $4 - 1$, $4 - 1$, $4$ making $16$ and $A$ wins.
:$1 - 3$, $1 - 2$, $4 - 1$, $4 - 1$, $4$ making $21$ and $A$ wins.
:$1 - 4$, $2$ making $7$ and $A$ wins.
:$1 - 2$, $4$ making $7$ and $A$ wins.
== Sources ==
* : Solutions: $207$. -- The Twenty-Two Game
* : Answers: $476$. The $22$ Game
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/208 - The Nine Squares Game

== by : $208$ ==
;The Nine Squares Game
:Make the simple square diagram show and provide a box of matches.
:The side of the large square is three matches in length.
:The game is, playing one match at a time alternately, to enclose more of those small squares than your opponent.
:For every small square that you enclose, you not only score one point, but you play again.
:This illustration shows an illustrative game in progress.
:
:Twelve matches are placed, my opponent and myself having made six plays each, and, as I had first play, it is now my turn to play a match.
:What is my best line of play in order to win most squares?
:If I play $FG$ my opponent will play $BF$ and score one point.
:Then, as he has the right to play again, he will score another with $EF$ and again with $IJ$, and still again with $IJ$, and still again with $GK$.
:If he now plays $CD$, I have nothing better than $DH$ (scoring one), but, as I have to play again, I am compelled, whatever I do, to give him all the rest.
:So he will win by $8$ to $1$ -- a bad defeat for me.
:Now, what should I have played instead of that disastrous $FG$?
:There is room for a lot of skilful play in the game, and it can never end in a draw.
=== Click here for solution ===
== Sources ==
* : Puzzle Games: $208$. -- The Nine Squares Game
* : Game Puzzles: $477$. The Nine Squares Game
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Dots and Boxes Game
Henry Ernest Dudeney/Modern Puzzles/208 - The Nine Squares Game/Solution

== by : $208$ ==
;The Nine Squares Game
== Solution ==
There are $2$ best plays:
:$MN$, which is that suggested by 
:$HL$
== Proof ==
Neither of $MN$ or $HL$ allow any boxes to be created immediately.
Playing:
:$(A): \quad MN$
leaves the plays $HL$, $CD$ or $DH$ for the opponent.
Any other reply is a blunder which gives you $7$ boxes immediately, and the win.
After:
:$(B): \quad HL$
you play $CD$ or $DH$, allowing his play of either $DH$ or $CD$, claiming $1$, and forcing a further play which gives you the remaining $8$ boxes.
After:
:$(B): \quad CD$ or $DH$
you play $DH$ or $CD$, claiming $1$, then $HL$.
$B$'s next move gives you the rest.
Playing:
:$(A): \quad HL$
leaves the plays $MN$, $CD$ or $DH$ for the opponent, and the same analysis is used, exchanging $HL$ and $MN$.
== Sources ==
* : Solutions: $208$. -- The Nine Squares Game
* : Answers: $477$. The Nine Squares Game
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/209 - A Wheel Fallacy

== by : $209$ ==
;A Wheel Fallacy
:Here is a curious fallacy that I have found to be very perplexing to many people.
:The wheel shown in the diagram makes one complete revolution in passing from $A$ to $B$.
:It is therefore obvious that the line $AB$ is exactly equal in length to the circumference of the wheel.
:
:Now the inner circle (the large hub in the diagram) also makes one complete revolution along the dotted line $CD$ and,
::since the line CD is equal to the line $AB$, the circumference of the larger and smaller circles are the same.
:This is clearly not true.
:Wherein lies the fallacy?
=== Click here for solution ===
== Sources ==
* : Wheel Paradox Problems: $209$. -- A Wheel Fallacy
* : Geometrical Problems: Circle Puzzles: $293$. A Wheel Fallacy
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Fallacies
Henry Ernest Dudeney/Modern Puzzles/209 - A Wheel Fallacy/Solution

== by : $209$ ==
;A Wheel Fallacy
:Here is a curious fallacy that I have found to be very perplexing to many people.
== Solution ==
Fairly obviously, the inner wheel does not only roll along $CD$ but slides along it as well.
Suppose the inner wheel rolls along $CD$ without sliding.
Then the bottom of the upper wheel move backwards in the same way as the bottom of the flange on a railway wheel, as analysed in $211$ - Another Wheel Paradox in this collection.
== Sources ==
* : Solutions: $209$. -- A Wheel Fallacy
* : Answers: $293$. A Wheel Fallacy
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/20 - The Seven Applewomen

== by : $20$ ==
;The Seven Applewomen
:Here is an old puzzle that people are frequently writing to me about.
: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?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $20$. -- The Seven Applewomen
* : Arithmetical and Algebraical Problems: Money Puzzles: $12$. The Seven Applewomen
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/20 - The Seven Applewomen/Solution

== by : $20$ ==
;The Seven Applewomen
== Solution ==
Each woman sold her apples at:
:$7$ for $1 \oldpence$
:$3 \oldpence$ for the odd ones left over.
Thus each received the same amount::
:$1 \shillings 8 \oldpence$
As put it:
:Without questioning the ingenuity of the thing, I have always thought the solution unsatisfactory,
:because really indeterminate, even if we admit that such an eccentric way of selling may be fairly termed a "price".
:It would seem just as fair if they sold them at different rates and afterwards divided the money;
:or sold different kinds of apples at different values;
:or sold by weight, the apples being of different sizes;
:or sold by rates diminishing with the age of the apples;
:and so on.
:That is why I have never held a high opinion of this old puzzle.
:In a general way, we can say that $n$ women, possessing $a n + \paren {n - 1}$, $\paren {a + b} n + \paren {n - 2}$, $\paren {a + 2 b} n + \paren {n - 3}$, $\ldots$, $\paren {a + \paren {n - 1} b} n$ apples respectively,
:can sell at $n$ for the penny and $b$ pence for each odd one left over,
:and each receive $a + b \paren {n - 1}$ pence.
:In the case of our puzzle $a = 2$, $b = 3$, and $n = 7$.
== Sources ==
* : Solutions: $20$. -- The Seven Applewomen
* : Answers: $12$. The Seven Applewomen
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/210 - A Famous Paradox

== by : $210$ ==
;A Famous Paradox
:There is a question that one is perpetually hearing asked, but to which I have never heard or read an answer that was satisfactory or really convincing to the ordinary person.
:It is this.
:When a bicycle is in motion, does the upper part of each wheel move faster than the bottom part near the ground?
 embarks on a prolix discussion on why he believes some people may think otherwise, which is just too tedious to repeat.
=== Click here for solution ===
== Sources ==
* : Wheel Paradox Problems: $210$. -- A Famous Paradox
* : Geometrical Problems: Circle Puzzles: $294$. A Famous Paradox
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Veridical Paradoxes
Henry Ernest Dudeney/Modern Puzzles/210 - A Famous Paradox/Solution

== by : $210$ ==
;A Famous Paradox
== Solution ==
Obviously. Du-uh.
== Proof ==
The part of the wheel that is in contact with the ground moves at the same speed relative to the ground that the ground does.
The conclusion follows in all its bletheringly obvious glory.
How this can be considered "paradoxical" completely confounds me.
== Sources ==
* : Solutions: $210$. -- A Famous Paradox
* : Answers: $294$. A Famous Paradox
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/211 - Another Wheel Paradox

== by : $211$ ==
;Another Wheel Paradox
:When a railway train is in motion, it is always the case that some parts of the train are travelling backwards.
:Explain why.
 couches this paradox in a tediously long-winded conversation between a pair of cyclists leaning on a bridge.
=== Click here for solution ===
== Sources ==
* : Wheel Paradox Problems: $211$. -- Another Wheel Paradox
* : Geometrical Problems: Circle Puzzles: $295$. Another Wheel Paradox
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Veridical Paradoxes
Henry Ernest Dudeney/Modern Puzzles/211 - Another Wheel Paradox/Solution

== by : $211$ ==
;Another Wheel Paradox
:When a railway train is in motion, it is always the case that some parts of the train are travelling backwards.
:Explain why.
 couches this paradox in a tediously long-winded conversation between a pair of cyclists leaning on a bridge.
== Solution ==
Let us direct our attention to the wheels of the train.
A train wheel has a flange on it, so as to make sure it stays on the rail.
This flange descends below the level of the rail a short but measurable distance.
As the train rolls along the track, the part of the flange that is at any one time below the level of the rail is demonstrably moving backwards.
:
The graph of the motion of a point on this flange can be illustrated as:
== Sources ==
* : Solutions: $211$. -- Another Wheel Paradox
* : Answers: $295$. Another Wheel Paradox
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/212 - A Chain Puzzle

== by : $212$ ==
;A Chain Puzzle
:A man has $80$ links of old chain in $13$ fragments, as shown in the diagram.
:
:It will cost him $1 \oldpence$ to open a link and $2 \oldpence$ to weld one together again.
:What is the lowest price it must cost him to join all the pieces together so as to form an endless chain?
:A new chain will cost him $3 \shillings$ (that is, $36 \oldpence$)
:What is the cheapest method of procedure?
:Remember that the small and large links must run alternately.
=== Click here for solution ===
== Sources ==
* : Unclassified Problems: $212$. -- A Chain Puzzle
* : Unclassified Puzzles: $513$. A Chain Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/212 - A Chain Puzzle/Solution

== by : $212$ ==
;A Chain Puzzle
== Solution ==
Open all the links of the piece of chain with $4$ links, on the right hand end of the middle row.
Open all the links of the piece of chain with $3$ links, in the middle of the middle row.
Open all the links of one of the remaining pieces of chain with $3$ links, for example from the middle of the bottom row.
We now have $10$ open links: $5$ large and $5$ small.
At the ends of the remaining $10$ pieces, there are $10$ large links and $10$ small links in total.
Each of the large open links can be used to connect $2$ of the small open links.
Similarly, each of the small open links can be used to connect $2$ of the large open links.
This costs $2 \shillings 6 \oldpence$
== Proof ==
To open a link and rejoin it costs $3 \oldpence$
By opening one link at the end of each of the $13$ pieces will cost $39 \oldpence$, that is, $3 \shillings 3 \oldpence$
Using taht strategy it would be cheaper to buy a new chain.
The best that can be done is to open all the links of $3$ pieces containing $10$ links of the appropriate nature so as to use them to join the remaining $10$ pieces at a cost of $30 \oldpence$, that is $2 \shillings 6 \oldpence$
== Sources ==
* : Solutions: $212$. -- A Chain Puzzle
* : Answers: $513$. A Chain Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/213 - The Six Pennies

== by : $213$ ==
;The Six Pennies
:Lay six pennies on the table, and arrange them as shown by the $6$ white circles in the diagram,
::so that a seventh penny can be dropped into the centre and touch each of the other $6$.
:
:It is required to get it exact, without any dependence on the eye.
:In this case it is not allowed to lift any penny off the table, nor can any measuring or marking be employed.
:However, you are allowed to slide the pennies after they have been placed on the table.
:You require only the six pennies.
=== Click here for solution ===
== Also see ==
* by : $329$ - The Four Pennies
== Sources ==
* : Unclassified Problems: $213$. -- The Six Pennies
* : Geometrical Problems: Moving Counter Puzzles: $383$. The Six Pennies
Category:Henry Ernest Dudeney/Modern Puzzles