Difference between revisions of "2013 AMC 8 Problems/Problem 10"
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For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. <math>3^2 \times 2</math> = 18. | For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. <math>3^2 \times 2</math> = 18. | ||
| − | Thus the answer = <math>\frac{5940}18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math> | + | Thus the answer = <math>\frac{5940}{18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=9|num-a=11}} | {{AMC8 box|year=2013|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:47, 27 November 2013
Problem
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
Solution
This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.
The prime factorization of 180 = 
The prime factorization of 594 = 
Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is 
). Multiply all of these to get 5940.
For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. 
= 18.
Thus the answer = 
= 
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
