AMC 10 Step by Step

2014 AMC 10B Problem 2

P1-10~1 minPrint

What is 23+232−3+2−3\frac{2^3 + 2^3}{2^{-3} + 2^{-3}} ?

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Key insight

The top is 8 + 8 = 16 and the bottom is 1/8 + 1/8 = 1/4; dividing by 1/4 is multiplying by 4.

Solution

The fraction has a top and a bottom. Each is small enough to work out as ordinary numbers, so do that first.

1. Top. 23=82^3 = 8, so 23+23=8+8=162^3 + 2^3 = 8 + 8 = 16.
2. Bottom. A negative exponent means "one over": 2−3=123=182^{-3} = \frac{1}{2^3} = \frac18. So 2−3+2−3=18+18=28=142^{-3} + 2^{-3} = \frac18 + \frac18 = \frac28 = \frac14.
3. Divide. Dividing by 14\frac14 is the same as multiplying by 44:
161/4=16×4=64.\frac{16}{1/4} = 16 \times 4 = 64.

The answer is (E) 64\boxed{\textbf{(E)}\ 64}.

Why this works

Adding a number to itself doubles it, and doubling a power of 22 just raises the exponent by one: 8+8=168 + 8 = 16 is 23+23=242^3 + 2^3 = 2^4. The mistake to avoid is adding the exponents when you add the numbers; exponents add only when you multiply. When the numbers are small, turning each power into an ordinary number is the safest route.

Alternative approach

Stay with powers of 22 the whole time. The top is 2⋅23=242 \cdot 2^3 = 2^4 and the bottom is 2⋅2−3=2−22 \cdot 2^{-3} = 2^{-2}. Dividing powers of the same base subtracts the exponents: 24−(−2)=26=642^{4 - (-2)} = 2^{6} = 64.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The numerator is 23+23=8+8=162^3 + 2^3 = 8 + 8 = 16. The denominator is 2−3+2−3=18+18=142^{-3} + 2^{-3} = \frac18 + \frac18 = \frac14.

Dividing, 161/4=16⋅4=64\frac{16}{1/4} = 16 \cdot 4 = 64.

Equivalently, 23+23=2⋅23=242^3 + 2^3 = 2 \cdot 2^3 = 2^4 and 2−3+2−3=2−22^{-3} + 2^{-3} = 2^{-2}, so the quotient is 24−(−2)=26=642^{4 - (-2)} = 2^6 = 64.

The answer is (E) 64\boxed{\textbf{(E)}\ 64}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.