9. [7] Let $n$ be an integer and
$$
m=(n-1001)(n-2001)(n-2002)(n-3001)(n-3002)(n-3003) \text {. }
$$

Given that $m$ is positive, find the minimum number of digits of $m$.
Answer:
11

Solution: One can show that if $m>0$, then we must either have $n>3003$ or $n1000^{5}$, meaning that $m$ has at least 16 digits. However, if $n>3003$, it is clear that the minimal $m$ is achieved at $n=3004$, which makes
$$
m=2002 \cdot 1002 \cdot 1001 \cdot 3 \cdot 2 \cdot 1=12 \cdot 1001 \cdot 1001 \cdot 1002,
$$
which is about $12 \cdot 10^{9}$ and thus has 11 digits.