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>? | |
| + | <math> \textbf{(A)}\ 67\qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 77\qquad\textbf{(D)}\ 201\qquad\textbf{(E)}\ 210 </math> | ||
| + | |||
| + | ==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> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2010|num-b=13|num-a=15}} | ||
Revision as of 17:34, 5 November 2012
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 | ||
