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Difference between revisions of "2002 AMC 10B Problems/Problem 1"

(New page: == Problem == The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is: <math> \mathrm{(A) \ } 1/6\qquad \mathrm{(B) \ } 1/3\qquad \mathrm{(C) \ } 1/2\qquad \mathrm{(D) \ } 2/3\q...)
(No difference)

Revision as of 01:21, 27 December 2008

Problem

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

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

Solution

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

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