2019 AMC 10A Problems/Problem 2

Revision as of 11:06, 26 December 2023 by Leonidastheconquerer (talk | contribs) (Solution 4)

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

Video Solution 3 by OmegaLearn

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

~ pi_is_3.14

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

Solution 4

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 5 (Brute Force)

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

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

this was the indended solution for this question

-dragoon

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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