Problem: What is 23+232−3+2−3\dfrac{2^{3}+2^{3}}{2^{-3}+2^{-3}}2−3+2−323+23?
Answer Choices:
A. 161616 B. 242424 C. 323232 D. 484848 E. 646464
Solution:
Note that
23+232−3+2−3=2⋅232⋅2−3=26=(E)64\dfrac{2^{3}+2^{3}}{2^{-3}+2^{-3}}=\dfrac{2 \cdot 2^{3}}{2 \cdot 2^{-3}}=2^{6}=(\text{E})\boxed{64} 2−3+2−323+23=2⋅2−32⋅23=26=(E)64
The problems on this page are the property of the MAA's American Mathematics Competitions