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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> 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>
+
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>
  
 
== Solution 2 ==
 
== Solution 2 ==
  
−
Using Difference of Squares, we factor the <math>\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}{^2})}</math>
+
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}}
 
{{MAA Notice}}

Latest revision as of 16:28, 4 June 2021

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 $3^{4012}$ in the numerator and factoring out $3^{4010}$ in the denominator gives us \[\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.\] Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.$

Solution 2

Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes

$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$


$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$

$= \boxed{\text{(E)}9}$

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

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