smc

We first observe that since there will be more 2s than 5s in $1005!$, we are looking for the largest $n$ such that $5^n$ divides $1005!$. We will use the fact that: $$n=\lfloor \frac{1005}{5^1}\rfloor +\lfloor \frac{1005}{5^2}\rfloor +\lfloor \frac{1005}{5^3}\rfloor \cdots$$ (This is an application of Legendre's formula). From $k=5$ and onwards, $\lfloor \frac{1005}{5^k}\rfloor =0$. Thus, our calculation becomes $$n=\lfloor \frac{1005}{5^1}\rfloor +\lfloor \frac{1005}{5^2}\rfloor +\lfloor \frac{1005}{5^3}\rfloor +\lfloor \frac{1005}{5^4}\rfloor$$ $$n=\lfloor \frac{1005}{5}\rfloor +\lfloor \frac{1005}{25}\rfloor +\lfloor \frac{1005}{125}\rfloor +\lfloor \frac{1005}{625}\rfloor$$ $$n=201+40+8+1=\boxed{250}$$ Since none of the answer choices equal 250, the answer is $\boxed{\text{none of the above}}$.