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

Revision as of 12:23, 27 January 2015

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

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

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 $\boxed{\textbf{(C) }\frac{5}{3}}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

Revision as of 11:06, 4 July 2013 (view source) Nathan wailes (talk \| contribs) ← Older edit		Revision as of 12:23, 27 January 2015 (view source) Equationcrunchor (talk \| contribs) m (→‎Solution: ?) Newer edit →
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−	Factoring out, we get: <math>\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)} ?</math>	+	Factoring out, we get: <math>\frac{2^{2012}(2^2 + 1)}{2^{2012}(2^2-1)}</math>.
−

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