Two players play the following game: from $923^{k}$ pebbles, they take turns removing 9 or 2 or 3 pebbles. The one who can no longer do so loses. Does either player have a winning strategy?
Consider the following two sets:

$$
S=\{2,3,4,7,8,9,10\}, \quad R=\{0,1,5,6\}
$$

Let the number of remaining pebbles after the $t$-th step be $n_{t}$, and the remainder of $n_{t}$ when divided by 11 be $m_{t}$.

Consider the case when $m_{t} \in S$. If $m_{t} \in\{2,3,7,8\}$, then $n_{t} \geq 2$. By removing 2 pebbles from the set, we get that $m_{t+1} \in R$.

If $m_{t} \in\{4,9\}$, then $n_{t} \geq 4$, so by removing 3 pebbles from the set, we get that $m_{t+1} \in R$. The only case left is $m_{t}=10$. By removing 9 pebbles from the set, we get that $m_{t+1} \in R$. Therefore, if $m_{t} \in S$, we can always make a move such that $m_{t+1} \in R$. It is easy to see that if $m_{t} \in R$, then no matter what move we make, $m_{t+1} \in S$. In summary: if $m_{0} \in S$, the first player can always play such that the remainder of the number of pebbles modulo 11 is in $S$ before each of their moves. Thus, they can always make a move, so they win in this case. By similar reasoning, if $m_{0} \in R$, the second player wins. In our case:

$$
m_{0} \equiv(923)^{k} \equiv(-1)^{k} \quad(\bmod 11)
$$

Therefore, if $k$ is odd, the first player has a winning strategy; otherwise, the second player has a winning strategy.

Based on the work of Székelyhidi Gábor (Kuwait, New English School, 11th grade)