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

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<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
 
<math>\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math>
  
== Solution ==
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==Solution 1==
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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 <math>0</math>, this means that the difference of the hundreds digits is also <math>\boxed{\textbf{(A) }0}</math>.
  
The last three digits of <math>n!</math> for all <math>n\geq15</math> is <math>000</math>, because there are at least three <math>2</math>s and three <math>5</math>s in its prime factorization. Because <math>0-0=0</math>, the answer is <math>\boxed{\textbf{(A) }0}</math>.
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==Solution 2==
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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>.
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--abhinavg0627
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==Solution 3 (Brute Force)==
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<math>20!= 2432902008176640000</math>
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<math>15!= 1307674368000</math>
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Then, we see that the hundreds digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>.
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~dragoon
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Please do not do this and only use this solution as a last resort.
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==Video Solution by Education, the Study of Everything==
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https://youtu.be/J4Bqztwjyxw
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~Education, The Study of Everything
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==Video Solution by WhyMath==
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https://youtu.be/V1fY0oLSHvo
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~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/zfChnbMGLVQ?t=3899
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~pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 18:14, 4 November 2024

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)

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

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.

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 (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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