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2013 AMC 10A Problem 8

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2013 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

What is the value of $\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?$

A)
$-1$
B)
$1$
C)
$\frac{5}{3}$
D)
$2013$
E)
$2^{4024}$

Answer

Key insight

Factor 2^2012 out of the numerator and denominator; what remains is (4+1)/(4-1).

Solution

Write $2^{2014} = 2^{2012} \cdot 2^2 = 4 \cdot 2^{2012}$ . Then both the numerator and the denominator have the common factor $2^{2012}$ :
$\frac{2^{2014} + 2^{2012}}{2^{2014} - 2^{2012}} = \frac{2^{2012}(4 + 1)}{2^{2012}(4 - 1)} = \frac{5}{3}.$

The answer is $\boxed{\textbf{(C)}\ \frac{5}{3}}$ .

Why this works

Huge exponents are a signal that the numbers themselves never matter; only their ratio does. Pull out the smallest power present, and the expression collapses to small integers. The general lesson: in a sum or difference of powers of the same base, factor out the lowest power before doing anything else.

Alternative approach

Replace $2012$ by a small exponent to see the pattern: $\dfrac{2^{4} + 2^{2}}{2^{4} - 2^{2}} = \dfrac{16 + 4}{16 - 4} = \dfrac{20}{12} = \dfrac{5}{3}$ . The value does not depend on the exponent as long as the two exponents differ by $2$ .

The trap

Cancelling the 2^2014 terms or the 2^2012 terms individually across the fraction, as if they were factors rather than summands.

Common mistakes

Cancelling the 2^2014 terms or the 2^2012 terms individually across the fraction, as if they were factors rather than summands.
Subtracting exponents as though the fraction were $2^{2014} / 2^{2012}$ and answering $4$ or similar.

Techniques

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