10. Agnese, Beatrice, Claudio, and Dario are playing with 53 stacks of coins. No matter which two stacks are chosen, they have a different number of coins. On each turn, a player chooses a stack and removes one coin from it. The player loses if, by removing a coin from a stack, they make that stack the same height as another stack on the table.

A stack can have 0 coins, and two stacks with 0 coins are considered equal. Agnese starts, followed by Beatrice, Claudio, and Dario, after which it is Agnese's turn again, and the order continues.

If at the beginning of the game there are 2020 coins in total and if all players play optimally, who loses?
(A) Agnese
(B) Beatrice
(C) Claudio
(D) Dario
(E) It is not possible to determine with the given information
10. The answer is (C). Let $a_{1}, a_{2}, \ldots, a_{53}$ be the piles ordered in increasing height.

We observe that the game ends. With each move, the total number of tokens decreases by 1, and if there are 51 tokens in total, the game is already over because there are at least two piles with 0 tokens.

Certainly, if the game has not already been lost, $a_{i}$ will have fewer tokens than $a_{j}$ if $i<j$: if this were not the case, since the number of tokens decreases by 1 with each move, there is a moment when the two piles have the same number of tokens, but this implies that the game is over, which is absurd.

Moreover, the only configuration in which each move leads the player to lose (let's call it the decisive configuration) is the one where the heights of the piles are $0,1,2, \ldots, 52$, in some order. Indeed, if the height difference between $a_{i+1}$ and $a_{i}$ were at least 2 for some $i$, then we could remove a token from $a_{i+1}$; if $a_{1}$ had more than 0 tokens, then we could remove a token from this pile.

Thus, the pile $a_{i}$ at the end of the game will have $i-1$ tokens for every $i \in\{1,2, \ldots, 53\}$. The number $N$ of tokens that will be removed to reach the decisive configuration is equal to the difference between the number of tokens initially present and the number of tokens in the final configuration, that is,

$$
N=2020-(0+1+\cdots+52)=2020-\frac{52 \cdot 53}{2}=642
$$

Since we can write $N=4 \cdot 160+2$ and there are 4 players, when the decisive configuration is reached, it will be Claudio's turn, so Claudio loses.