Difference between revisions of "2013 AMC 10A Problems/Problem 8"

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Cancelling out the <math>2^{2012}</math> from the numerator and denominator, we see that it simplifies to <math>\frac{5}{3}</math>, <math>\textbf{(C)}</math>.
 
Cancelling out the <math>2^{2012}</math> from the numerator and denominator, we see that it simplifies to <math>\frac{5}{3}</math>, <math>\textbf{(C)}</math>.
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==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=7|num-a=9}}
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{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}

Revision as of 21:58, 7 February 2013

Problem

What is the value of $\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?$


$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024}$


Solution

Factoring out, we get: $\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)} ?$


Cancelling out the $2^{2012}$ from the numerator and denominator, we see that it simplifies to $\frac{5}{3}$, $\textbf{(C)}$.

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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
2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 12 Problems and Solutions