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2019 AMC 10A Problems/Problem 2: Difference between revisions

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== Problem ==
== Problem ==
What is the hundreds digit of <math>(20!-15!)?</math>
What is the hundreds digit of <imath>(20!-15!)?</imath>


<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
<imath>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</imath>


==Solution 1==
==Solution 1==
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 <math>0</math>, this means that the difference of the hundreds digits is also <math>\boxed{\textbf{(A) }0}</math>.
Because we know that <imath>5^3</imath> is a factor of <imath>15!</imath> and <imath>20!</imath>, the last three digits of both numbers is a <imath>0</imath>, this means that the difference of the hundreds digits is also <imath>\boxed{\textbf{(A) }0}</imath>.


==Solution 2==
==Solution 2==


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


--abhinavg0627
--abhinavg0627
Line 15: Line 15:
==Solution 3 (Brute Force)==
==Solution 3 (Brute Force)==


<math>20!= 2432902008176640000</math>
<imath>20!= 2432902008176640000</imath>
<math>15!= 1307674368000</math>
and <imath>15!= 1307674368000</imath>


Then, we see that the hundreds digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>.
Then, we see that the hundreds digit is <imath>0-0=\boxed{\textbf{(A)}\ 0}</imath>.


~dragoon
~dragoon
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Note for people not used to comp. math: This is a completely reasonable way to solve this problem.
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 <imath>15!</imath> (it is easier than it seems) is <imath>2^(11) * 3^6 * 5^3 * 7^2 * 11 * 13</imath>
Notice that it includes <imath>2^3 * 5^3</imath> which is 1000
Therefore 15! and 20! are both multiples of 1000, the hundreds place is <imath>\boxed{\textbf{(A)}\ 0}</imath>.


==Video Solution by Education, the Study of Everything==
==Video Solution by Education, the Study of Everything==
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{{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}}
{{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Number Theory Problems]]

Latest revision as of 09:58, 10 November 2025

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

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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