32. A pile contains 1992 stones. Two players play the following game: on each turn, they can take from the pile any number of stones that is a divisor of the number of stones the opponent took on their previous turn. The first player can take any number of stones on their first move, but not all at once. The player who takes the last stone wins. Who wins with correct play?
92.32. The first player wins. With his first move, he takes eight stones, and 1984 stones remain in the pile, which is $1984=64 \cdot 31$. After this, the first player repeats all the moves of the second player. Since each move involves taking one, two, four, or eight stones from the pile, and since 1984 is divisible by 16, an even number of moves (not counting the first move) will be made, and thus the first player will make the last move.