# 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

## See Also

 2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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