Difference between revisions of "2010 AMC 8 Problems/Problem 14"
(Created page with "What is the sum of the prime factors of 2010? Well to answer this question let's take the prime factorization of 2010. Which is 2,3,5, and 67. So all you do is add 2+3+5+67=77") |
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− | What is the sum of the prime factors of 2010? | + | ==Problem== |
− | + | What is the sum of the prime factors of <math>2010</math>? | |
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+ | <math> \textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210 </math> | ||
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+ | ==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> | ||
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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== | ||
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+ | {{AMC8 box|year=2010|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:55, 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.