AMC 12 2006b Problem 22

Suppose $a$, $b$ and $c$ are positive integers with $a+b+c=2006$, and $a!b!c!=m\cdot 10^n$, where $m$ and $n$ are integers and $m$ is not divisible by $10$. What is the smallest possible value of $n$?

$\mathrm{(A)}\ 489 \mathrm{(B)}\ 492 \mathrm{(C)}\ 495 \mathrm{(D)}\ 498 \mathrm{(E)}\ 501$