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

(all 3 solutions are the same)
Line 14: Line 14:
  
 
~NH14, kingofpineapplz, and Arcticturn
 
~NH14, kingofpineapplz, and Arcticturn
 +
 +
== Solution 2 ==
 +
We have
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Therefore, the greatest prime divisor of 16.383 is 127.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }10}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==

Revision as of 21:22, 25 November 2021

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 1

We want to find the largest prime factor of

\[16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.\]

Since $127$ is prime, our answer is $1+2+7=\boxed{\textbf{(C) }10}$.

~NH14, kingofpineapplz, and Arcticturn

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png