Folklore

Find the smallest natural value of $n$ for which the number $n!$ is divisible by 990.
The number $11!=1 \cdot 2 \cdot \ldots \cdot 9 \cdot 10 \cdot 11$ is divisible by $9 \cdot 10 \cdot 11=990$.

For $n!<11$, the number $n!$ is not even divisible by 11.

## Answer

$n=11$.