2009 AMC 12B Problem 5

Attempt Problem 5 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Kiana has two older twin brothers. The product of their three ages is

128.

What is the sum of their three ages?

10

12

16

18

24

128 = 2^7,

so every age is a power of

2

If each twin is

t

years old, then

t^2

times Kiana’s age is

128,

and Kiana is younger

Since

128 = 2^7,

each age is a power of

2.

The twins share an age

t,

so Kiana’s age is

\dfrac{128}{t^2}.

Taking

t = 8

gives Kiana

\dfrac{128}{64} = 2,

who is younger than the twins. (Smaller twins would make Kiana older, which is not allowed.) The sum is

8 + 8 + 2 = 18.

Thus, the correct answer is D.

Problem 5 in other years