Difference between revisions of "2016 AMC 8 Problems/Problem 9"
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The prime factorization is <math>2016=2^5\times3^2\times7</math>. Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>. Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>. | The prime factorization is <math>2016=2^5\times3^2\times7</math>. Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>. Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>. | ||
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| + | ==Video Solution== | ||
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| + | https://youtu.be/I_jevKp3Kyg?si=qKDYbUmjFkioeIvE | ||
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| + | A solution so simple that a 12-year-old made it! | ||
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| + | ~Elijahman~ | ||
==Video Solution (CREATIVE THINKING!!!)== | ==Video Solution (CREATIVE THINKING!!!)== | ||
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:51, 20 June 2024
Contents
Problem
What is the sum of the distinct prime integer divisors of 
?
Solutions
Solution 1
The prime factorization is 
. Since the problem is only asking us for the distinct prime factors, we have 
. Their desired sum is then 
.
Video Solution
https://youtu.be/I_jevKp3Kyg?si=qKDYbUmjFkioeIvE
A solution so simple that a 12-year-old made it!
~Elijahman~
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
