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

Notice that $9$ can be factored out of the numerator: \[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}=\frac{9\left(3^{2007}\right)^2-9\left(3^{2005}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}=9\cdot\frac{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] Thus, the expression is equal to $9$, and the answer is $\mathrm{(E)}$.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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