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 $\boxed{\textbf{(A) }0}$.

Video Solution

https://youtu.be/V1fY0oLSHvo

~savannahsolver

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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