# Difference between revisions of "2019 AMC 10A Problems/Problem 2"

## Problem

What is the hundreds digit of $(20!-15!)?$

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

## Solution

The last three digits of $n!$ for all $n\geq15$ are $000$, because there are at least three $2$s and three $5$s in its prime factorization. Because $0-0=0$, the answer is $\boxed{\textbf{(A) }0}$.

## Solution 2

20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is $\boxed{\textbf{(A) }0}$. ~peppapig_

~savannahsolver