Difference between revisions of "2008 AMC 10A Problems/Problem 7"
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==Problem== | ==Problem== | ||
The fraction | The fraction | ||
+ | |||
<cmath>\frac{\left(3^{2008}\right)^2-\left(3^{2006}\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}</cmath> | ||
simplifies to which of the following? | simplifies to which of the following? | ||
Line 7: | Line 8: | ||
==Solution== | ==Solution== | ||
− | + | Simplifying, we get <cmath>\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.</cmath> Factoring out <math>3^{4012}</math> in the numerator and factoring out <math>3^{4010}</math> in the denominator gives us <cmath>\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.</cmath> Canceling out <math>3^4-1</math> gives us <math>\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.</math> | |
− | <cmath>\frac{ | + | |
− | + | == Solution 2 == | |
+ | |||
+ | Using Difference of Squares, <math>\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}</math> becomes | ||
+ | |||
+ | <math>\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}</math> | ||
+ | |||
+ | |||
+ | <math> = \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}</math> | ||
+ | <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}} |
Latest revision as of 17:28, 4 June 2021
Contents
Problem
The fraction
simplifies to which of the following?
Solution
Simplifying, we get Factoring out
in the numerator and factoring out
in the denominator gives us
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.