2019 AMC 10A Problems/Problem 2

Revision as of 16:14, 9 February 2019 by Mathwizard07 (talk | contribs) (Solution)


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


The last three digits of $n!$ for all $n>=15$ 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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