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edited Mar 30, 2019 at 18:06

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The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available. This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.

The result is that

The game always ends after $34$ moves, after which the first player cannot move and loses.

The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available. This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves, after which the first player cannot move and loses.

The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available. This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves.

The result is that

The game always ends after $34$ moves, after which the first player cannot move and loses.

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created Mar 30, 2019 at 17:58

Jaap Scherphuis

55.7k
8
130
221

The numbers $25$ and $36$ are coprime. This means that if we continually replace the largest of the two numbers by the (positive) difference of the two numbers, we are essentially performing the Euclidean algorithm for finding their GCD, and will eventually get a $1$. The sequence is $36$, $25$, $36-25=11$, $25-11=14$, $14-11=3$, $11-3=8$, $8-3=5$, $5-3=2$, $3-2=1$.
Once there is a $1$ on the board, you can repeatedly subtract it to fill in any gaps and eventually produce every number from $1$ to $36$. This shows that if you have two cooperating players, all number from $1$ to $36$ can be produced.
But this also happens when the game is played between two non-cooperating players. It is impossible to prevent any of the numbers appearing. If any number in the euclidean sequence is not on the board, then there are still moves available. So eventually $1$ must be produced, and then as long as there are missing numbers between $1$ and $36$, there is at least one move available. This means that regardless of what moves are played, all numbers $1$ to $36$ will appear. We started with $2$ numbers on the board, so the game ends after $34$ moves, after which the first player cannot move and loses.