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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
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~savannahsolver
 
~savannahsolver
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==Video Solution by TheBeautyofMath==
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For AMC 10: https://youtu.be/4qgYrCYG-qw?t=795
 +
 +
For AMC 12: https://youtu.be/kuZXQYHycdk
 +
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~IceMatrix
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-b=7|num-a=9}}
 
{{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}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:35, 29 December 2022

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 largest prime factor of $16384$ is $2$ because $16384 = 2^{14}$. What is the sum of the digits of the greatest prime number that is a divisor of $16383$?

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

Solution (Difference of Squares)

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 divisible by $16383$. The sum of $127$'s digits is $1+2+7=\boxed{\textbf{(C) }10}$.

~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker

Video Solution by Interstigation

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

Video Solution

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://youtu.be/4qgYrCYG-qw?t=795

For AMC 12: https://youtu.be/kuZXQYHycdk

~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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