Difference between revisions of "2007 AMC 10A Problems/Problem 17"
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== Solution == | == Solution == | ||
<math>3 \cdot 5^2m</math> must be a perfect cube, so each power of a prime in the factorization for <math>3 \cdot 5^2m</math> must be divisible by <math>3</math>. Thus the minimum value of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The minimum possible value for the sum of <math>m</math> and <math>n</math> is <math>60\ \mathrm{(D)}</math>. | <math>3 \cdot 5^2m</math> must be a perfect cube, so each power of a prime in the factorization for <math>3 \cdot 5^2m</math> must be divisible by <math>3</math>. Thus the minimum value of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The minimum possible value for the sum of <math>m</math> and <math>n</math> is <math>60\ \mathrm{(D)}</math>. | ||
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| + | ==Solution 2== | ||
== See also == | == See also == | ||
Revision as of 16:38, 23 October 2021
Contents
Problem
Suppose that 
and 
are positive integers such that 
. What is the minimum possible value of 
?
Solution
must be a perfect cube, so each power of a prime in the factorization for must be divisible by 
. Thus the minimum value of is 
, which makes ![$n = \sqrt[3]{3^3 \cdot 5^3} = 15$](http://latex.artofproblemsolving.com/8/b/2/8b219ec60fc52fffb3a087cd962705121a553544.png)
. The minimum possible value for the sum of and is 
.
Solution 2
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
