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2008 AMC 10A Problem 7

Problem 7 of 25EasierAlgebra

The fraction (32008)2−(32006)2(32007)2−(32005)2 \dfrac{\left(3^{2008}\right)^2 - \left(3^{2006}\right)^2}{\left(3^{2007}\right)^2 - \left(3^{2005}\right)^2} simplifies to which of the following?

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Solution

Since (3k)2=9k,\left(3^{k}\right)^2 = 9^{k}, the fraction is 92008−9200692007−92005.\dfrac{9^{2008} - 9^{2006}}{9^{2007} - 9^{2005}}. Factoring 920059^{2005} from each part gives 92005(93−9)92005(92−1)=9(92−1)92−1=9. \dfrac{9^{2005}\left(9^3 - 9\right)}{9^{2005}\left(9^2 - 1\right)} = \dfrac{9\left(9^2 - 1\right)}{9^2 - 1} = 9. Thus, the correct answer is E.

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Concepts: exponent · factoring

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