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

(Solution 2 is wrong since divisibility by 100 doesn't imply the hundreds digit of their difference is 0.)
(Solution 4)
 
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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> are <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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https://youtu.be/J4Bqztwjyxw
  
==Video Solution==
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Education, The Study of Everything
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==Video Solution 2==
 
https://youtu.be/V1fY0oLSHvo
 
https://youtu.be/V1fY0oLSHvo
  
 
~savannahsolver
 
~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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==Solution 4==
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We can clearly see that <math>20! \equiv 15! \equiv 0 \pmod{100}</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{\text{(A)}}</math>.
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--abhinavg0627
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==Solution 5==
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<math>20!= 2432902008176640000</math>
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<math>15!= 1307674368000</math>
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Subtract
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-dragoon
  
 
==See Also==
 
==See Also==

Latest revision as of 14:12, 13 August 2022

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


Solution 4

We can clearly see that $20! \equiv 15! \equiv 0 \pmod{100}$, so $20! - 15! \equiv 0 \pmod{100}$ meaning that the last two digits are equal to $00$ and the hundreds digit is $0$, or $\boxed{\text{(A)}}$.

--abhinavg0627


Solution 5

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

Subtract

-dragoon

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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