# 2013 AMC 12A Problems/Problem 4

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

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 $C$

## Solution 2

Suppose that $$x=2^{2012}.$$ Then the given expression is equal to $$\frac{4x+x}{4x-x}=\frac{5x}{3x}=\boxed{\frac{5}{3}\textbf{(C)}}.$$

~sugar_rush

## Solution 3 (Elimination)

Choice $A$ cannot be true, because $2^{2014}+2^{2012}$ is clearly larger than $2^{2014}-2^{2012}$. We can apply our same logic to choice $B$ and eliminate it as well. $2013$ and $2^{4024}$ are all whole numbers, but $2^{2014}+2^{2012}$ is not a multiple of $2^{2014}-2^{2012}$, so we can eliminate choices $D$ and $E$ too. This leaves us with choice $\boxed{C}$ as our final answer.

~dbnl

## Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=545 (solution for #4 starts at 9:05)

~sugar_rush

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 