Problem 1

Let $u_{0}, u_{1}, u_{2}, \ldots$ be integers such that

$\triangleright u_{0}=100$;

$\triangleright$ for all $k \geqslant 0$, the inequality $u_{k+2} \geqslant 2+u_{k}$ is satisfied;

$\triangleright$ for all $\ell \geqslant 0$, the inequality $u_{\ell+5} \leqslant 5+u_{\ell}$ is satisfied.

Find all possible values for the integer $u_{2023}$.

## $\S$
Solution

For all integers $n \geqslant 0$, we know that

$$
u_{n} \geqslant u_{n+5}-5 \geqslant u_{n+10}-10 \geqslant u_{n+8}-8 \geqslant u_{n+6}-6 \geqslant u_{n+4}-4 \geqslant u_{n+2}-2 \geqslant u_{n}
$$

Thus, all our inequalities are equalities, and $u_{n}=u_{n+2}-2=u_{n+5}-5$.

In particular, for all integers $n \geqslant 0$, $k \geqslant 0$, and $\ell \geqslant 0$, we have $u_{n+5 k+2 \ell}=u_{n}+5 k+2 \ell$. Here, for $n=0$, $k=1$, and $\ell=1009$, this means that $u_{2023}=u_{0}+2023=2123$. Thus, 2023 is the only potentially possible value for the integer $u_{2023}$.

Conversely, if each number $u_{n}$ is equal to $u_{n}+100$, our sequence of numbers satisfies the conditions of the statement, and $u_{2023}=2123$. In conclusion, the only possible value is indeed 2123.

## $\square$