Difference between revisions of "2021 Fall AMC 12B Problems/Problem 6"
R00tsofunity (talk | contribs) (IceMatrix had wrong link for the AMC 12 video -- I fixed that. Maybe he confused 2021 Fall with 2021 Feb.) |
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== Problem == | == Problem == | ||
− | The | + | The greatest prime number that is a divisor of <math>16\text{,}384</math> is <math>2</math> because <math>16\text{,}384 = 2^{14}</math>. What is the sum of the digits of the greatest prime number that is a divisor of <math>16\text{,}383</math>? |
<math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math> | <math>\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22</math> |
Revision as of 00:57, 30 December 2022
- 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 (Difference of Squares)
We have
Since is composite, is the largest prime divisible by . The sum of 's digits is .
~Steven Chen (www.professorchenedu.com) ~NH14 ~kingofpineapplz ~Arcticturn ~MrThinker
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=1121
Video Solution
~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.