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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==Video Solution 1==
  
The last three digits of <math>n!</math> for all <math>n\geq15</math> is <math>000</math>, because there are at least three 2s and three 5s in its prime factorization. Because <math>0-0=0</math>, The answer is <math>0</math>.
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https://youtu.be/J4Bqztwjyxw
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Education, The Study of Everything
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==Video Solution 2==
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https://youtu.be/V1fY0oLSHvo
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~savannahsolver
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== Video Solution == 3
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https://youtu.be/zfChnbMGLVQ?t=3899
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~ pi_is_3.14
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==Solution 3==
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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 0, this means that the difference of the hundreds digits is also <math>\boxed{\text{(A) }0}</math>.
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==See Also==
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{{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 15:27, 25 January 2021

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

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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