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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
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<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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We want to find the largest prime factor of
 
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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>127</math> is prime, our answer is <math>1+2+7=\boxed{\textbf{(C) }10}</math>.
 
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~NH14, kingofpineapplz, and Arcticturn
 
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 01:52, 29 January 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 2

We have \begin{align*} 16,383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ & = 3 \cdot 43 \cdot 127 . \end{align*}

Therefore, the greatest prime divisor of 16.383 is 127.

Therefore, the answer is $\boxed{\textbf{(C) }10}$.

~Steven Chen (www.professorchenedu.com)

Video Solution by Interstigation

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

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

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