# Difference between revisions of "2014 AMC 10B Problems/Problem 2"

## Problem

What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$?

$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$

## Solution

We can synchronously multiply ${2^3}$ to the polynomials both above and below the fraction bar. Thus, $$\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.$$ Hence, the fraction equals to $\boxed{{64 (\textbf{E})}}$.

## Solution 2

We have $$\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,$$ so our answer is $\boxed{{(\textbf{E})}}$.

~savannahsolver