7.5. There are 11 empty boxes. In one move, you can place one coin in any 10 of them. Two players take turns. The winner is the one who, after their move, first has 21 coins in one of the boxes. Who wins with correct play?
# Solution.

Number the boxes: $1, \ldots$, 11 and denote the move by the number of the box where we did not put a coin. We can assume that the first player started the game with move 1. To win, the second player needs to, regardless of the first player's moves, make moves $2, \ldots, 11$. These ten moves, together with the first player's move, will result in 10 coins in each box. Additionally, there will be a box (let's call it $A$) into which the first player put a coin with each of their moves from 2 to 11. Thus, after the first player's 11th move, box $A$ will have 20 coins, and no other box will have more. The second player's 11th move should be to place a coin in box $A$. This way, the second player wins.

Answer. The second player wins.