Difference between revisions of "2008 AMC 10A Problems/Problem 7"

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==Solution==
 
==Solution==
Simplifying, we get <cmath>\frac{3^4016-3^4012}{3^4014-3^4010}.</cmath> factoring out <math>3^4012</math> on the top and factoring out <math>3^4010</math> on the bottom 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.</math>
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Simplifying, we get <cmath>\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.</cmath> factoring out <math>3^{4012}</math> on the top and factoring out <math>3^{4010}</math> on the bottom 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.</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 19:18, 30 November 2019

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

Simplifying, we get \[\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.\] factoring out $3^{4012}$ on the top and factoring out $3^{4010}$ on the bottom gives us \[\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.\] Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9.$

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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