Difference between revisions of "2008 AMC 10A Problems/Problem 7"
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<math> = \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}</math> | <math> = \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}</math> | ||
| − | <math> = \boxed | + | <math> = \boxed{\text{(E)}9}</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 10:44, 25 December 2019
Contents
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
Simplifying, we get ![\[\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.\]](http://latex.artofproblemsolving.com/c/b/1/cb1decaca126604949ba6ae3013d1d5716cbe164.png)
Factoring out 
on the top and factoring out 
on the bottom gives us ![\[\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.\]](http://latex.artofproblemsolving.com/4/5/d/45ddc8ab77e26848dcfe1758bdc31dbcead4dcb8.png)
Canceling out 
gives us 
Solution 2
Using Difference of Squares, 
becomes
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. 
