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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"

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{{duplicate|[[2021 Fall AMC 10B Problems#Problem 8|2021 Fall AMC 10B #8]] and [[2021 Fall AMC 12B Problems#Problem 6|2021 Fall AMC 12B #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 ==
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== Solution==
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We want to find the largest prime factor of <math>2^{14} -1 = (2^7+1)(2^7-1) = (129)(127) = 3 \cdot 43 \cdot 127.</math> Thus, the largest prime factor is <math>127,</math> which has the sum of the digits as <math>10.</math> Thus the answer is <math>\boxed{\textbf{(D.)} \: 10}.</math>
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We have
 +
<cmath>\begin{align*}
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16383 & = 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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\end{align*}</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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~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker ~abed_nadir(youtube.com/@indianmathguy)
 +
~PaperMath
 +
 
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=== Note ===
 +
Note that you can quickly tell that <math>2^7 -1</math> is prime because it is a [[Mersenne prime|Mersenne prime]].
 +
 
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~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 +
 
 +
==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
  
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~NH14
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==See Also==
 +
{{AMC10 box|year=2021 Fall|ab=B|num-b=7|num-a=9}}
 +
{{AMC12 box|year=2021 Fall|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 00:54, 3 November 2024

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) ~PaperMath

Note

Note that you can quickly tell that $2^7 -1$ is prime because it is a Mersenne prime.

~South

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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