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

(Solution 2 is wrong since divisibility by 100 doesn't imply the hundreds digit of their difference is 0.)
(Video Solution)
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The last three digits of <math>n!</math> for all <math>n\geq15</math> are <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>.
 
The last three digits of <math>n!</math> for all <math>n\geq15</math> are <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>.
  
==Video Solution==
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==Video Solution 1==
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https://youtu.be/J4Bqztwjyxw
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Education, The Study of Everything
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==Video Solution 2==
 
https://youtu.be/V1fY0oLSHvo
 
https://youtu.be/V1fY0oLSHvo
  

Revision as of 19:31, 20 November 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 $\boxed{\textbf{(A) }0}$.

Video Solution 1

https://youtu.be/J4Bqztwjyxw

Education, The Study of Everything


Video Solution 2

https://youtu.be/V1fY0oLSHvo

~savannahsolver

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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