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

Revision as of 00:41, 30 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://www.youtube.com/watch?v=RyN-fKNtd3A&t=797

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

~IceMatrix

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.

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