# Difference between revisions of "1977 AHSME Problems/Problem 25"

## Problem 25

Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$.

$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$

### Solution

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:

$$n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots$$

(This is an application of Legendre's formula).

From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$. Thus, our calculation becomes

$$n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor$$

$$n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor$$

$$n = 201 + 40 + 8 + 1 = \boxed{250}$$

Since none of the answer choices equal 250, the answer is $\boxed{\textbf{(E)}}$.

- mako17