9.2 In Pokémon hunting, 11 adults and $n$ children participated. Together, they caught $n^{2}+3 n-2$ Pokémon, with all adults catching the same number, and all children catching the same number, but each child catching 6 fewer than an adult. How many children participated in the game?
Solution: Let each child catch $m$ pokemons. Then $n m+11(m+6)=n^{2}+3 n-2$. From this, $(n+11) m=n^{2}+3 n-68$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+3 n-68=$ $n(n+11)-8(n+11)+20$, so 20 is divisible by $n+11$. The only divisor of 20 greater than 10 is 20 itself, so $n+11=20, n=9$.

## Criteria:

- No points are deducted for the lack of an example for $n=9$.
- It is shown that $n=9$ works, but it is not proven that there are no other options -1 point.