Difference between revisions of "2009 AMC 12A Problems/Problem 6"
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We have <math>12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{\bold{E)} P^{2n} Q^m}</math>. | We have <math>12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{\bold{E)} P^{2n} Q^m}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/T3XXMO3YvHQ | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 18:25, 7 August 2020
- The following problem is from both the 2009 AMC 12A #6 and 2009 AMC 10A #13, so both problems redirect to this page.
Contents
Problem
Suppose that and
. Which of the following is equal to
for every pair of integers
?
Solution
We have .
Video Solution
~savannahsolver
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.