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

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== Solution ==
 
== Solution ==
  
The last three digit of <math>n!</math> for all <math>n>=15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.
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The last three digits of <math>n!</math> for all <math>n>=15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.

Revision as of 16:14, 9 February 2019

Problem

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


Solution

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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