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 1

We can synchronously multiply ${2^3}$ to the expresions 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 $\boxed{{\textbf{(E) }64}}$.

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 $\boxed{{(\textbf{E) }64}}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/NwWf1uYFZqw

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Video Solution

https://youtu.be/vLFULrT_7yk

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See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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