Difference between revisions of "2013 AMC 12A Problems/Problem 4"

Revision as of 17:31, 22 February 2013

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

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

$\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}$

We can factor a ${2^{2012}}$ out of the numerator and denominator to obtain

$\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}$

The ${2^{2012}}$ cancels, so we get

$\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}$ , which is $C$

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

@@ Line 1: / Line 1: @@
+== Problem ==
+What is the value of <cmath>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?</cmath>
+<math> \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} </math>
+==Solution==
 <math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math>
@@ Line 8: / Line 17: @@
 <math>\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}</math>, which is <math>C</math>
+== See also ==
+{{AMC12 box|year=2013|ab=A|num-b=3|num-a=5}}