AMC 10 Step by Step
Log inSign up
Printable

2008 AMC 10A Problem 7

Problem onlyBack to the problem
2008 AMC 10A Problem 7AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 2 of 5 · about 2 min · position P1-10

The fraction

(32008)2−(32006)2(32007)2−(32005)2\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2} simplifies to which of the following?

  1. A)

    11

  2. B)

    94\frac{9}{4}

  3. C)

    33

  4. D)

    92\frac{9}{2}

  5. E)

    99

Answer

E

Key insight

Numerator and denominator have the same shape; the numerator is the denominator with every exponent raised by 2, i.e. multiplied by 3^2 = 9.

Solution

Rewrite each square: (32008)2=34016(3^{2008})^2 = 3^{4016}, (32006)2=34012(3^{2006})^2 = 3^{4012}, (32007)2=34014(3^{2007})^2 = 3^{4014}, (32005)2=34010(3^{2005})^2 = 3^{4010}.

Factor the smallest power out of each difference:
34012(34−1)34010(34−1).\frac{3^{4012}\left(3^{4} - 1\right)}{3^{4010}\left(3^{4} - 1\right)}.

The factors 34−1=803^4 - 1 = 80 cancel, leaving 3401234010=32=9\frac{3^{4012}}{3^{4010}} = 3^2 = 9.

The answer is (E) 9\boxed{\textbf{(E)}\ 9}.

Why this works

Both differences are "a power of 33 times (81−1)(81 - 1)"; the only difference between top and bottom is the power pulled out. Whenever an expression has the same structure with all exponents shifted by a constant, the ratio is just the base raised to that shift. Factoring the common power is the step that makes this visible.

Alternative approach

Substitute x=32005x = 3^{2005}. The numerator is (27x)2−(3x)2=720x2(27x)^2 - (3x)^2 = 720x^2 and the denominator is (9x)2−x2=80x2(9x)^2 - x^2 = 80x^2; the ratio is 99.

The trap

Cancelling term by term (3^2008 over 3^2007 and so on) as if the fraction were a product, or forgetting that squaring doubles the exponent.

Common mistakes

  • Cancelling term by term (3^2008 over 3^2007 and so on) as if the fraction were a product, or forgetting that squaring doubles the exponent.
  • Concluding 33 because "the exponents differ by one," without accounting for the squaring.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta