Difference between revisions of "2010 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
− | First, we must find the prime factorization of <math>2010</math>. <math>2010=2\cdot 3 \cdot 5 \cdot 67</math> We add the factors up to get <math>\boxed{\textbf{(C)}\ 77}</math> | + | First, we must find the prime factorization of <math>2010</math>. <math>2010=2\cdot 3 \cdot 5 \cdot 67</math>. We add the factors up to get <math>\boxed{\textbf{(C)}\ 77}</math> |
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+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/6xNkyDgIhEE?t=1236 | ||
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+ | ==Video by MathTalks== | ||
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+ | https://www.youtube.com/watch?v=6hRHZxSieKc | ||
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==See Also== | ==See Also== | ||
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{{AMC8 box|year=2010|num-b=13|num-a=15}} | {{AMC8 box|year=2010|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:55, 22 January 2023
Problem
What is the sum of the prime factors of ?
Solution
First, we must find the prime factorization of . . We add the factors up to get
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=1236
Video by MathTalks
https://www.youtube.com/watch?v=6hRHZxSieKc
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.