2019 AMC 10A Problems/Problem 2

Revision as of 17:33, 9 February 2019 by Brendanb4321 (talk | contribs) (Problem)

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 $n\geq15$ is $000$, because there are at least three 2s and three 5s in its prime factorization. Because $0-0=0$, The answer is $0$.

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