7. Find the number of consecutive 0 's at the end of the base 10 representation of 2006 !.
7. Ans: 500

If $p$ is a prime, then the highest power of $p$ that divides 2006 ! is
$$
f(p)=\lfloor 2006 / p\rfloor+\left\lfloor 2006 / p^{2}\right\rfloor+\left\lfloor 2006 / p^{3}\right\rfloor+\cdots .
$$
(Note that the terms in the sum are eventually 0.) The number of consecutive 0 's at the end of the base 10 representation of 2006 ! is the highest power of 10 that divides 2006!, which is $\min \{f(2), f(5)\}=f(5)=401+80+16+3=500$.