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}$.
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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 $(20!-15!)?$

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

Solution

The last three digits of $n!$ for all $n\geq15$ is $000$, because there are at least three 2s and three 5s in its prime factorization. Because $0-0=0$, The answer is $\boxed{\textbf{(A)}0}$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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