AMC 10 Step by Step

2009 AMC 10B Problem 6

P1-10~2 minPrint

Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

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Key insight

The twins share an age, so 128 is Kiana's age times the twins' age twice. Try the few possible twin ages and keep the one where Kiana is younger.

Solution

The twins have the same age, so the product is (Kiana's age) ×\times (twins' age) ×\times (twins' age). Try each possible twins' age.

  1. 128=2×2×2×2×2×2×2128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2, so every age must be a power of 22: 1,2,4,8,…1, 2, 4, 8, \ldots
  2. List the ways to pick the twins' age tt and find Kiana's age from 128÷(t×t)128 \div (t \times t):
Twins' ageTwins multipliedKiana's ageKiana younger?
1111128128no
22443232no
44161688no
88646422yes

(A twins' age of 1616 is too big, since 16×16=25616 \times 16 = 256 is more than 128128.)
3. Only the last row fits the word "older": the twins are 88 and Kiana is 22.
4. The sum is 8+8+2=188 + 8 + 2 = 18.

The answer is (D) 18\boxed{\textbf{(D)}\ 18}.

Why this works

Twins give a repeated factor, so the product has a squared part. Writing 128128 as a product of 22's shows there are only a few possible ages to test. Several choices multiply to 128128, so the word "older" is what picks the right one; always use every condition in the problem.

Concise solution

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

Let the twins each be tt years old and Kiana kk, with k<tk < t. Then kt2=128=27k t^2 = 128 = 2^7, so tt is a power of 22 and t2t^2 divides 272^7.

  • t=8t = 8: k=128/64=2k = 128/64 = 2, and 2<82 < 8. Valid.
  • t=4t = 4: k=128/16=8k = 128/16 = 8, but then Kiana is older. Invalid.
  • t=2t = 2 or t=1t = 1: k=32k = 32 or 128128, far older. Invalid.

So the ages are 2,8,82, 8, 8 and the sum is 2+8+8=182 + 8 + 8 = 18.

The answer is (D) 18\boxed{\textbf{(D)}\ 18}.

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.