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

Revision as of 01:24, 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 ~NH14
+==Video Solution by Interstigation==
+https://youtu.be/p9_RH4s-kBA?t=1121
+==See Also==
+{{AMC10 box|year=2021 Fall|ab=B|num-b=7|num-a=9}}
+{{AMC12 box|year=2021 Fall|ab=B|num-b=5|num-a=7}}
+{{MAA Notice}}

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