Henry Ernest Dudeney/Modern Puzzles/206 - The 37 Puzzle Game/Solution

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Modern Puzzles by Henry Ernest Dudeney: $206$

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.


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.


Dudeney describes the above by providing sample games.

$\blacksquare$


Sources

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