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\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 \boxed{(A) 0} | + | 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>\boxed{\textbf{(A)}0}</math>. |
==See Also== | ==See Also== |
Revision as of 18:42, 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 .
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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All AMC 10 Problems and Solutions |
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