Difference between revisions of "2013 AMC 12B Problems/Problem 22"
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It remains to minimize the integer value of <math>\sqrt[8]{m^7n^6}</math>. Since <math>m, n>1</math>, we can check that <math>m = 2^2</math> and <math>n = 2^3</math> work. Thus the answer is <math>4+8 = \boxed{\textbf{(A)}\ 12}</math>. | It remains to minimize the integer value of <math>\sqrt[8]{m^7n^6}</math>. Since <math>m, n>1</math>, we can check that <math>m = 2^2</math> and <math>n = 2^3</math> work. Thus the answer is <math>4+8 = \boxed{\textbf{(A)}\ 12}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}} | {{AMC12 box|year=2013|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:27, 28 September 2020
Contents
Problem
Let and be integers. Suppose that the product of the solutions for of the equation is the smallest possible integer. What is ?
Solution
Rearranging logs, the original equation becomes
By Vieta's Theorem, the sum of the possible values of is . But the sum of the possible values of is the logarithm of the product of the possible values of . Thus the product of the possible values of is equal to .
It remains to minimize the integer value of . Since , we can check that and work. Thus the answer is .
Video Solution
For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.