9.5. Varya and Miron are playing the following game. There are 10 piles of 10 stones each on the table. The players take turns, with Varya starting first. On their turn, a player divides any pile that has more than one stone into several equal piles. The player who cannot make a move (when there is exactly one stone in each pile before their turn) wins. Who will win if both players play optimally?
Answer: Miron.

Solution. We will call a pile with one stone "unitary," and a pile with a prime number of stones - "prime." A prime pile of $p$ stones can only be divided into $p$ unitary piles.

For Miron to win, it is sufficient for him to repeat Varya's moves until only one pile of 10 stones remains. At this point, apart from it, there will be some number (possibly zero) $n$ of prime piles on the table. If $n$ is even, Miron needs to divide the last full pile into 5 piles of 2 stones each; if $n$ is odd, then into 2 piles of 5 stones each or 10 piles of 1 stone each. Now all non-unitary piles are prime, and before Varya's move, their number is odd, while before Miron's move, it is even. On each move, the number of prime piles decreases by one, so Varya will divide the last prime pile, and Miron will win.

Remark. For Miron to win, he does not necessarily have to exactly repeat Varya's moves. He needs to ensure that after each of his moves, an even number of full piles remain. For this, he should, after Varya divides a full pile, divide another full pile into such a number of primes that the total number of primes becomes even. This should be done until only one full pile remains. From there, he should act as described in the solution. Other ways of playing for Miron lead to defeat.

Comment. Proof of the statement that the one who divides the last full pile wins - 2 points.