Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Since <math>129</math> is composite, <math>127</math> is the largest prime | + | Since <math>129</math> is composite, <math>127</math> is the largest prime which can divide <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 ~MrThinker | + | ~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker ~abed_nadir(youtube.com/@indianmathguy) |
==Video Solution by Interstigation== | ==Video Solution by Interstigation== |
Latest revision as of 04:28, 12 November 2023
- 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.
Contents
Problem
The greatest prime number that is a divisor of is
because
. What is the sum of the digits of the greatest prime number that is a divisor of
?
Solution
We have
Since is composite,
is the largest prime which can divide
. The sum of
's digits is
.
~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker ~abed_nadir(youtube.com/@indianmathguy)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=1121
Video Solution (Just 1 min!)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
For AMC 10: https://www.youtube.com/watch?v=RyN-fKNtd3A&t=797
For AMC 12: https://www.youtube.com/watch?v=4qgYrCYG-qw
~IceMatrix
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.