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

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~NH14
 
~NH14
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==Video Solution by Interstigation==
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https://youtu.be/p9_RH4s-kBA?t=1121
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==See Also==
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{{AMC10 box|year=2021 Fall|ab=B|num-b=7|num-a=9}}
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{{AMC12 box|year=2021 Fall|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 02:24, 24 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 have

\[16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.\] Since $127$ is prime, our answer is $\boxed{\textbf{(C) }10}$.

~kingofpineapplz

Solution 2

Since $16384$ is $2^14$, we can consider it as $(2^7)^2$. $16383$ is $1$ less than $16384$, so it can be considered as $1$ less than a square. Therefore, it can be expressed as $(x-1)(x+1)$. Since $2^7$ is $128, 16383$ is $127 \cdot 129$. $129$ is $3 \cdot 43$, and since $127$ is larger, our answer is $\boxed {(C) 10}$.

~Arcticturn

Solution 3

We want to find the largest prime factor of $2^{14} -1 = (2^7+1)(2^7-1) = (129)(127) = 3 \cdot 43 \cdot 127.$ Thus, the largest prime factor is $127,$ which has the sum of the digits as $10.$ Thus the answer is $\boxed{\textbf{(D.)} \: 10}.$

~NH14

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

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