2019 AMC 10A Problems/Problem 2

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 1

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 $\boxed{\textbf{(A) }0}$ .

Solution 2

We can clearly see that $20! \equiv 15! \equiv 0 \pmod{1000}$ , so $20! - 15! \equiv 0 \pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $0$ , or $\boxed{\textbf{(A)}\ 0}$ .

--abhinavg0627

Solution 3 (Brute Force)

$20!= 2432902008176640000$ $15!= 1307674368000$

Then, we see that the hundred digit is $0-0=\boxed{\textbf{(A)}\ 0}$ .

~dragoon

Video Solution by Education, the Study of Everything

https://youtu.be/J4Bqztwjyxw

~Education, The Study of Everything

Video Solution 2

https://youtu.be/V1fY0oLSHvo

~savannahsolver

Video Solution 3 by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=3899

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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