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

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~savannahsolver
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==Solution 3==
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Because we know that <math>5^3</math> is a factor of <math>15!</math> and <math>20!</math>, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also <math>0</math>.
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==See Also==
 
==See Also==

Revision as of 21:41, 6 January 2021

Problem

What is the hundreds digit of $(20!-15!)?$

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

Video Solution 1

https://youtu.be/J4Bqztwjyxw

Education, The Study of Everything


Video Solution 2

https://youtu.be/V1fY0oLSHvo

~savannahsolver

Solution 3

Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also $0$.



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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