11.5. Initially, 10 ones are written on the board. Grisha and Gleb play a game, taking turns. On his turn, Grisha squares 5 of the numbers on the board. On his turn, Gleb chooses several (possibly none) of the numbers on the board and increases each of them by 1. If a number divisible by 2023 appears on the board within 10000 moves, Gleb wins; otherwise, Grisha wins. Who has a winning strategy if Grisha goes first?

(G. Nikitin)
Answer. Grisha wins.

Solution. Note that $2023=7 \cdot 17^{2}$. Grisha will divide the numbers on the board into two groups of 5 and will square the numbers from the first group and the second group alternately. It is easy to see that the squares of integers not divisible by 7, when divided by 7, can only give remainders of 1, 2, and 4. Therefore, after increasing the numbers by a maximum of 2, the numbers on the board will give remainders of $1, 2, 3, 4, 5$, and 6 when divided by 7. This means that none of the numbers will be divisible by 7, and therefore will not be divisible by 2023.

Remark. The number 2023 is incorrectly factored into prime factors, but this does not affect the solution of the problem - 1 point is deducted.

When listing the quadratic residues modulo 7, a residue different from 0 is missing - 2 points are deducted.

A reasoning is provided that solves the problem for prime numbers $p$ of the form $8k+7$, but $p$ is chosen as a composite number or a number that is not a divisor of 2023 - no more than 2 points are awarded.