# 2008 AMC 10A Problems/Problem 7

## Problem

The fraction $$\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}$$ simplifies to which of the following? $\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$

## Solution

Simplifying, we get $$\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.$$ Factoring out $3^{4012}$ in the numerator and factoring out $3^{4010}$ in the denominator gives us $$\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.$$ Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.$

## Solution 2

Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes $\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$ $= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$ $= \boxed{\text{(E)}9}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 