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2009 AMC 10B Problem 6

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2009 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$10$
B)
$12$
C)
$16$
D)
$18$
E)
$24$

Answer

Key insight

The product is k times t squared where t is the twins' age and k < t; since 128 = 2^7, only t = 8 with k = 2 works.

Solution

Let the twins each be $t$ years old and Kiana $k$ , with $k < t$ . Then $k t^2 = 128 = 2^7$ , so $t$ is a power of $2$ and $t^2$ divides $2^7$ .

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

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

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

Why this works

Twins share an age, so the product has the form $k t^2$ ; factoring $128$ as a power of $2$ makes the perfect-square part visible. The word "older" is the deciding constraint, since more than one factorization multiplies to $128$ .

The trap

Taking the twins to be 4 and Kiana 8, which multiplies to 128 but makes Kiana older than her older brothers.

Common mistakes

Taking the twins to be 4 and Kiana 8, which multiplies to 128 but makes Kiana older than her older brothers.
Forgetting that the brothers have the same age and hunting for three distinct factors of $128$ .

Techniques

Test small/specific values or special cases to find or verify the answer