[ Divisibility of numbers. General properties ]

Find the smallest natural $n$ for which $(n+1)(n+2)(n+3)(n+4)$ is divisible by 1000.
For any natural $n$, the given product is divisible by 8, since among any four consecutive natural numbers, one is divisible by 4 and another by 2. Therefore, it is sufficient to find the smallest $n$ for which the given product is divisible by $125=5^{3}$. Since only one of the factors can be divisible by 5, $n$ is the smallest if the factor divisible by $125$ is the largest. Thus, $n+4=125$, which means $n=121$.

## Answer

$n=121$.