AMC 10 Step by Step
Log inSign up
Printable

2014 AMC 10B Problem 2

Problem onlyBack to the problem
2014 AMC 10B Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

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

  1. A)

    1616

  2. B)

    2424

  3. C)

    3232

  4. D)

    4848

  5. E)

    6464

Answer

E

Key insight

Each sum is just twice a power of 2, so the fraction is 2^4 divided by 2^{-2}, which is 2^6.

Solution

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}.

Why this works

Adding a term to itself doubles it, which for powers of 22 just raises the exponent by one. Then a quotient of powers subtracts exponents, and subtracting a negative exponent adds. Keeping everything as powers of 22 avoids fraction arithmetic entirely.

The trap

Combining 2^3 + 2^3 into 2^6 (multiplying exponents when adding equal terms) instead of 2^4.

Common mistakes

  • Combining 2^3 + 2^3 into 2^6 (multiplying exponents when adding equal terms) instead of 2^4.
  • Treating 2−32^{-3} as −8-8 rather than 18\frac18, which produces a negative or nonsensical result.

Techniques

Set up the equation/formula and compute; no special trick needed