9.2 In the fishing, 11 experienced fishermen and $n$ children participated. Together they caught $n^{2}+$ $5 n+22$ fish, with all experienced fishermen catching the same amount, and all children catching the same amount, but each 10 less than an experienced fisherman. Who was there more of at the fishing - experienced fishermen or children?
Solution: Let each child catch $m$ fish. Then $n m+11(m+10)=n^{2}+5 n+22$. From this, $(n+11) m=n^{2}+3 n-88$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+5 n-88=$ $(n+11)(n-6)-22$, so 22 is divisible by $n+11$. The only divisor of 22 greater than 11 is 22 itself, so $n+11=22, n=11$. Therefore, the experienced fishermen and children were equal in number.

## Criteria:

- No points are deducted for the absence of an example for $n=11$.
- It is shown that $n=11$ works, but it is not proven that there are no other options - 1 point.