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On the occasion of the 47th Mathematical Olympiad 2016 the numbers 47 and 2016 are written on the blackboard. Alice and Bob play the following game: Alice begins and in turns they choose two numbers $a$ and $b$ with $a > b$ written on the blackboard, whose difference $a-b$ is not yet written on the blackboard and write this difference additionally on the board. The game ends when no further move is possible. The winner is the player who made the last move.

Prove that Bob wins, no matter how they play.

(Richard Henner)

Csaba Daniel Farkaš
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    Hint: When the game is over, let $n$ be the least number written on the board. Of course we must have $1≤n≤47$. Prove that $n$ divides all the other numbers on the board and deduce that $n=1$. – lulu May 15 '20 at 16:35
  • If I prove that $n=1$, does it directly imply that together we will have 47+2016 numbers which is odd so Bob wins? – Csaba Daniel Farkaš May 15 '20 at 16:42
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    If $n=1$ then the numbers on the board must be all the numbers from $1$ to $2016$. As you started with two on the board, that means that the game must have taken exactly $2014$ turns to finish. – lulu May 15 '20 at 16:44

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