2002 AMC 10B Problems/Problem 1: Difference between revisions

Revision as of 10:22, 4 July 2013

The ratio $\frac{2^{2001}\cdot3^{2003}}{6^{2002}}$ is:

$\mathrm{(A) \ } 1/6\qquad \mathrm{(B) \ } 1/3\qquad \mathrm{(C) \ } 1/2\qquad \mathrm{(D) \ } 2/3\qquad \mathrm{(E) \ } 3/2$

$\frac{2^{2001}\cdot3^{2003}}{6^{2002}}=\frac{6^{2001}\cdot 3^2}{6^{2002}}=\frac{9}{6}=\frac{3}{2}$ or $\mathrm{ (E) \ }$

2002 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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	[[Category:Introductory Number Theory Problems]]		[[Category:Introductory Number Theory Problems]]
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