Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
Cellsecret (talk | contribs) (→Solution 3) |
Ehuang0531 (talk | contribs) (all 3 solutions are the same) |
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==Solution 1== | ==Solution 1== | ||
− | We | + | We want to find the largest prime factor of |
<cmath>16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.</cmath> | <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>. | |
− | + | ~NH14, kingofpineapplz, and Arcticturn | |
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA?t=1121 | https://youtu.be/p9_RH4s-kBA?t=1121 | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-b=7|num-a=9}} | {{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}} | {{AMC12 box|year=2021 Fall|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:36, 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 is because . What is the sum of the digits of the greatest prime number that is a divisor of ?
Solution 1
We want to find the largest prime factor of
Since is prime, our answer is .
~NH14, kingofpineapplz, and Arcticturn
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.