Difference between revisions of "2007 AMC 8 Problems/Problem 3"
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<math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math> | <math>\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 12</math> | ||
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| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=272 | ||
== Solution == | == Solution == | ||
| − | + | The prime factorization of <math>250</math> is <math>2 \cdot 5^3</math>. The smallest two are <math>2</math> and <math>5</math>. <math>2+5 = \boxed{\text{(C) }7}</math>. | |
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| − | The two | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=2|num-a=4}} | {{AMC8 box|year=2007|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:59, 29 December 2022
Problem
What is the sum of the two smallest prime factors of 
?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=272
Solution
The prime factorization of is 
. The smallest two are 
and 
. 
.
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
