Problem 4

Find all integers $n \geqslant 0$ such that $20 n+2$ divides $2023 n+210$.

## $\S$
Solution $\mathbf{n}^{\circ} 1$

Let $n \geqslant 0$ be any solution. The integer $20 n+2$ divides

$$
20 \times(2023 n+210)-2023 \times(20 n+2)=4200-4046=154=2 \times 7 \times 11
$$

whose divisors are $1,2,7,11,14,22,77$ and 154. Thus, $20 n+2$ is 2 or 22, so $n$ is 0 or 1.

Conversely, $n=0$ is a solution, because 210 is even, but $n=1$ is not a solution, because $2023+210$ is odd while $20+2$ is even. In conclusion, $n=0$ is the only solution.

## $\S$