The fraction \dfrac(3^2008)^2 - (3^2006)^2(3^2007)^2 - (3^2005)^2 simplifies to which of the following?

2008 AMC 10A Problem 7 — Solution

2008 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:exponent factoring

Difficulty rating: 1210

The fraction

$\dfrac{\left(3^{2008}\right)^2 - \left(3^{2006}\right)^2}{\left(3^{2007}\right)^2 - \left(3^{2005}\right)^2}$

simplifies to which of the following?

$1$

$\dfrac{9}{4}$

$3$

$\dfrac{9}{2}$

$9$

Solution:

Since $\left(3^{k}\right)^2 = 9^{k},$ the fraction is $\dfrac{9^{2008} - 9^{2006}}{9^{2007} - 9^{2005}}.$

Factoring $9^{2005}$ from each part gives $\dfrac{9^{2005}\left(9^3 - 9\right)}{9^{2005}\left(9^2 - 1\right)} = \dfrac{9\left(9^2 - 1\right)}{9^2 - 1} = 9.$

Thus, the correct answer is E.

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