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

--abhinavg0627

Solution 3 (Brute Force)

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

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

~dragoon

Please do not do this and only use this solution as a last resort.

Note for people not used to comp. math: This is a completely reasonable way to solve this problem.

Solution 4 (Solution 1 but simpler)

The prime factorization of $15!$ (it is easier than it seems) is $2^(11) * 3^6 * 5^3 * 7^2 * 11 * 13$

Notice that it includes $2^3 * 5^3$ which is 1000

Therefore 15! and 20! are both multiples of 1000, the hundreds place is $\boxed{\textbf{(A)}\ 0}$ .

Video Solution by Education, the Study of Everything

https://youtu.be/J4Bqztwjyxw

~Education, The Study of Everything

Video Solution by WhyMath

https://youtu.be/V1fY0oLSHvo

~savannahsolver

Video Solution 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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