jmc

We have $4000=2^5\cdot 5^3$. Any factor of 4000 is in the form $2^a\cdot 5^b$ for $0\le a\le 5$ and $0\le b\le 3$. Since $100=2^2\cdot 5^2$, we must count the factors of 4000 that have $a\ge 2$ and $b\ge 2$. That gives $(5-2+1)(3-2+1)=4\cdot 2=\boxed{8}$ factors.