Difference between revisions of "2009 AMC 12A Problems/Problem 6"

Latest revision as of 17:25, 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.

Suppose that $P = 2^m$ and $Q = 3^n$ . Which of the following is equal to $12^{mn}$ for every pair of integers $(m,n)$ ?

$\textbf{(A)}\ P^2Q \qquad \textbf{(B)}\ P^nQ^m \qquad \textbf{(C)}\ P^nQ^{2m} \qquad \textbf{(D)}\ P^{2m}Q^n \qquad \textbf{(E)}\ P^{2n}Q^m$

We have $12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{\bold{E)} P^{2n} Q^m}$ .

~savannahsolver

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #6]] and [[2009 AMC 10A Problems|2009 AMC 10A #13]]}}
 == Problem ==
 Suppose that <math>P = 2^m</math> and <math>Q = 3^n</math>. Which of the following is equal to <math>12^{mn}</math> for every pair of integers <math>(m,n)</math>?
@@ Line 6: / Line 8: @@
 == Solution ==
-We have <math>12^{mn} = (2\cdot 2\cdot 3)^{mn} = 2^{2mn} \cdot 3^{mn} = (2^m)^{2n} \cdot (3^n)^m = \boxed{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 ==
 {{AMC12 box|year=2009|ab=A|num-b=5|num-a=7}}
+{{AMC10 box|year=2009|ab=A|num-b=12|num-a=14}}
+{{MAA Notice}}