# 2019 AMC 10A Problems/Problem 2

## Problem

What is the hundreds digit of $(20!-15!)?$ $\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

## Video Solution 1

Education, The Study of Everything

~savannahsolver

~ pi_is_3.14

## Solution 3

Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also $\boxed{\textbf{(A) }0}$.

## Solution 4

We can clearly see that $20! \equiv 15! \equiv 0 \pmod{100}$, so $20! - 15! \equiv 0 \pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $0$, or $\boxed{\textbf{(A)}\ 0}$.

--abhinavg0627

## Solution 5 (Brute Force) $20!= 2432902008176640000$ $15!= 1307674368000$

Then, we see that the hundred digit is $0-0=\boxed{\textbf{(A)}\ 0}$.

-dragoon

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 