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> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=13|num-a=15}} | {{AMC8 box|year=2010|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:33, 24 October 2017
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 
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. 
