Problem

Source:

Tags: combinatorics



Two players, $A$ and $B,$ alternatively take stones from a pile of $n \geq 2$ stones. $A$ plays first and in his first move he must take at least one stone and at most $n-1$ stones. Then each player must take at least one stone and at most as many stones as his opponent took in the previous move. The player who takes the last stone wins. Which player has a winning strategy?