Problem 4. There are three piles of 40 stones each. Petya and Vasya take turns, Petya starts. On a turn, one must combine two piles, then divide these stones into four piles. The player who cannot make a move loses. Which of the players (Petya or Vasya) can win, no matter how the opponent plays?

$[6$ points] (A.V. Shapovalov)
Answer: Vasya.

Comment: In fact, this is a joke game: Vasya wins regardless of the players' actions.

Solution: In one move, two piles are replaced by four, i.e., the number of piles increases by 2. So, how many piles will there be when the game ends? Initially, the number of piles was odd, so, increasing by two, it will always remain odd.

If a move cannot be made, then the two largest piles together contain no more than 3 stones. Then in all other piles, there is 1 stone each, and there are no fewer than $120-3=117$ such piles. That is, there should be (at least) 119 piles in total.

To get 119 piles, you need to make $(119-3): 2=$ $=58$ moves. This number is even, meaning the last move was made by Vasya (and he won).