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

(Solution)
(Solution (Difference of Squares))
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& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\
 
& = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\
 
& = 129 \cdot 127 \\
 
& = 129 \cdot 127 \\
& = 3 \cdot 43 \cdot 127.
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
Therefore, the greatest prime divisor of <math>16383</math> is <math>127.</math> The sum of its digits is <math>1+2+7=\boxed{\textbf{(C)} \: 10}.</math>
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Since <math>129</math> is composite, <math>127</math> is the largest prime divisible by <math>16383</math>. The sum of <math>127</math>'s digits is <math>1+2+7=\boxed{\textbf{(C) }10}</math>.
  
~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn
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~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==

Revision as of 10:05, 17 August 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

See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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