Difference between revisions of "2019 AMC 10A Problems/Problem 2"

Revision as of 05:40, 1 January 2021

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

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

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2019 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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 <math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
-== Solution ==
-The last three digits of <math>n!</math> for all <math>n\geq15</math> are <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>.
 ==Video Solution 1==