2008 AMC 10A Problems/Problem 7

Revision as of 19:17, 30 November 2019 by Piemax2713 (talk | contribs) (Solution)

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 $3^4012$ on the top and factoring out $3^4010$ on the bottom gives us \[\frac{(3^4-1)(3^4012)}{(3^4-1)(3^4010)}.\] Canceling out $3^4-1$ gives us $\frac{3^4012}{3^4010}=\frac{3^2}{3^0}=9.$

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AMC 10 Problems and Solutions

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