Difference between revisions of "2008 AMC 10A Problems/Problem 7"
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Notice that <math>9</math> can be factored out of the numerator: | Notice that <math>9</math> can be factored out of the numerator: | ||
<cmath>\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}</cmath> | <cmath>\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}</cmath> | ||
| − | Thus, the expression is | + | Thus, the expression is equal to <math>9</math>, and the answer is <math>\mathrm{(E)}</math>. |
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2008|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:19, 10 February 2016
Problem
The fraction
![\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\]](http://latex.artofproblemsolving.com/2/6/e/26e933ebb23b926f939dea3997c79761c2d177d5.png)
simplifies to which of the following?
Solution
Notice that 
can be factored out of the numerator:
![\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}=\frac{9\left(3^{2007}\right)^2-9\left(3^{2005}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}=9\cdot\frac{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\]](http://latex.artofproblemsolving.com/3/6/8/368380e08dd7d0dd2735baee6e0715d6e8b68c7a.png)
Thus, the expression is equal to , and the answer is 
.
See also
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
