Task 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take?
Answer: 97.

Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.

For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the factors is divisible by 101. Therefore, one of the nine numbers must be at least 101, so $n+8 \geqslant 101$ and $n+4 \geqslant 97$. It is also clear that for the nine numbers $93,94, \ldots, 101$, the value $n+4=97$ is achieved, since 99 is divisible by 11, and 101 is divisible by 101.