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

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== Solution 2 ==
 
== Solution 2 ==
  
20 and 15 are both greater than 10, they are divisible by 100, and therefore, the hundreds digit is <math>\boxed{\textbf{(A) }0}</math>. ~ Peppapig_
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20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is <math>\boxed{\textbf{(A) }0}</math>. ~ Peppapig_
  
 
==See Also==
 
==See Also==

Revision as of 20:59, 1 April 2020

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

Solution 2

20 and 15 are both greater than 10, therefore they are divisible by 100 because of powers of 5 and powers of 2, so the hundreds digit is $\boxed{\textbf{(A) }0}$. ~ Peppapig_

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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