Difference between revisions of "2008 AMC 10A Problems/Problem 7"
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<math>\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}</math> | <math>\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}</math> | ||
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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> | ||
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+ | <math> = \frac{3^{2006} \sout {9+1} \cdot 3^{2006} \sout {9-1}}{3^2006} \sout {9+1} \cdot 3^{2006} \sout {9-1}</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 10:39, 25 December 2019
Contents
Problem
The fraction simplifies to which of the following?
Solution
Simplifying, we get Factoring out on the top and factoring out on the bottom gives us Canceling out gives us
Solution 2
Using Difference of Squares, becomes
$= \frac{3^{2006} \sout {9+1} \cdot 3^{2006} \sout {9-1}}{3^2006} \sout {9+1} \cdot 3^{2006} \sout {9-1}$ (Error compiling LaTeX. Unknown error_msg)
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.