2013 AMC 12B Problems/Problem 22
Contents
Problem
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Let and
be integers. Suppose that the product of the solutions for
of the equation
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is the smallest possible integer. What is ?
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Solution
− Rearranging logs, the original equation becomes
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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
.
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It remains to minimize the integer value of . Since
, we can check that
and
work. Thus the answer is
.
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Video Solution
− For those who prefer a video solution: https://www.youtube.com/watch?v=vX0y9lRv9OM&t=312s
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See also
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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 |
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.