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>
  
==Video 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>.
  
https://youtu.be/J4Bqztwjyxw
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==Solution 2==
  
Education, The Study of Everything
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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>.
  
 +
--abhinavg0627
  
 +
==Solution 3 (Brute Force)==
  
==Video Solution 2==
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<math>20!= 2432902008176640000</math>
https://youtu.be/V1fY0oLSHvo
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<math>15!= 1307674368000</math>
  
~savannahsolver
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Then, we see that the hundreds digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>.
  
== Video Solution 3 by OmegaLearn ==
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~dragoon
https://youtu.be/zfChnbMGLVQ?t=3899
 
  
~ pi_is_3.14
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Please do not do this and only use this solution as a last resort.
  
==Solution 3==
+
==Video Solution by Education, the Study of Everything==
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{\textbf{(A) }0}</math>.
 
  
==Solution 4==
+
https://youtu.be/J4Bqztwjyxw
  
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>0</math>, or <math>\boxed{\textbf{(A)}\ 0}</math>.
+
~Education, The Study of Everything
  
--abhinavg0627
+
==Video Solution by WhyMath==
 +
 
 +
https://youtu.be/V1fY0oLSHvo
  
==Solution 5 (Brute Force)==
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~savannahsolver
  
<math>20!= 2432902008176640000</math>
+
==Video Solution by OmegaLearn==
<math>15!= 1307674368000</math>
 
  
Then, we see that the hundred digit is <math>0-0=\boxed{\textbf{(A)}\ 0}</math>.
+
https://youtu.be/zfChnbMGLVQ?t=3899
  
~dragoon
+
~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
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All AMC 10 Problems and Solutions

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