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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
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== Problem ==
 
== Problem ==
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The largest prime factor of <math>16384</math> is <math>2</math> because <math>16384 = 2^{14}</math>. What is the sum of the digits of the greatest prime number that is a divisor of <math>16383</math>?
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The greatest prime number that is a divisor of <math>16{,}384</math> is <math>2</math> because <math>16{,}384 = 2^{14}</math>. What is the sum of the digits of the greatest prime number that is a divisor of <math>16{,}383</math>?
  
 
<math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math>
 
<math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math>
  
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==Solution 1==
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== Solution==
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We want to find the largest prime factor of
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We have
 +
<cmath>\begin{align*}
 +
16383 & = 2^{14} - 1 \\
 +
& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\
 +
& = 129 \cdot 127 \\
 +
\end{align*}</cmath>
  
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<cmath>16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.</cmath>
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Since <math>129</math> is composite, <math>127</math> is the largest prime which can divide <math>16383</math>. The sum of <math>127</math>'s digits is <math>1+2+7=\boxed{\textbf{(C) }10}</math>.
  
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Since <math>127</math> is prime, our answer is <math>1+2+7=\boxed{\textbf{(C) }10}</math>.
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~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker ~abed_nadir(youtube.com/@indianmathguy)
  
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~NH14, kingofpineapplz, and Arcticturn
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==Video Solution by Interstigation==
 +
https://youtu.be/p9_RH4s-kBA?t=1121
  
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== Solution 2 ==
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==Video Solution (Just 1 min!)==
−
We have
+
https://youtu.be/NB6CamKgDaw
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<cmath>
+
 
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\begin{align*}
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~Education, the Study of Everything
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16,383 & = 2^{14} - 1 \\
 
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& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\
 
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& = 129 \cdot 127 \\
 
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& = 3 \cdot 43 \cdot 127 .
 
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\end{align*}
 
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</cmath>
 
  
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Therefore, the greatest prime divisor of 16.383 is 127.
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==Video Solution by WhyMath==
 +
https://youtu.be/JBNbjKsw_tU
  
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Therefore, the answer is <math>\boxed{\textbf{(C) }10}</math>.
+
~savannahsolver
 +
==Video Solution by TheBeautyofMath==
 +
For AMC 10: https://www.youtube.com/watch?v=RyN-fKNtd3A&t=797
  
−
~Steven Chen (www.professorchenedu.com)
+
For AMC 12: https://www.youtube.com/watch?v=4qgYrCYG-qw
  
−
==Video Solution by Interstigation==
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~IceMatrix
−
https://youtu.be/p9_RH4s-kBA?t=1121
 
  
 
==See Also==
 
==See Also==

Revision as of 03:28, 12 November 2023

The following problem is from both the 2021 Fall AMC 10B #8 and 2021 Fall AMC 12B #6, so both problems redirect to this page.

Problem

The greatest prime number that is a divisor of $16{,}384$ is $2$ because $16{,}384 = 2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of $16{,}383$?

$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$

Solution

We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}

Since $129$ is composite, $127$ is the largest prime which can divide $16383$. The sum of $127$'s digits is $1+2+7=\boxed{\textbf{(C) }10}$.

~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker ~abed_nadir(youtube.com/@indianmathguy)

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=1121

Video Solution (Just 1 min!)

https://youtu.be/NB6CamKgDaw

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/JBNbjKsw_tU

~savannahsolver

Video Solution by TheBeautyofMath

For AMC 10: https://www.youtube.com/watch?v=RyN-fKNtd3A&t=797

For AMC 12: https://www.youtube.com/watch?v=4qgYrCYG-qw

~IceMatrix

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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