Difference between revisions of "2019 AMC 10A Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | The last three digits of <math>n!</math> for all <math>n | + | The last three digits of <math>n!</math> for all <math>n\geq15</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 16:29, 9 February 2019
Problem
What is the hundreds digit of 
Solution
The last three digits of 
for all 
is 
, because there are at least three 2s and three 5s in its prime factorization. Because 
, The answer is 
.
