2009 AMC 10B Problem 6

Question

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

Every age divides

128=2^7,

and the twins share one age

If the twins are

t

and Kiana is

k,

then

t^2 k=128

with

k\lt t

Since

128=2^7,

every age is a power of

2.

Writing the twins’ common age as

t

and Kiana’s as

k,

we need

t^2k=128

with

k\lt t.

Write

t=2^j.

Then

k=2^{7-2j}.

For

k

to be a positive integer age we need

7-2j\ge0,

so

j\le3.

Because Kiana is younger than the twins,

7-2j\lt j,

so

j\ge3.

Therefore

j=3,

giving

t=8

and

k=2.

The sum is

8+8+2=18.

Thus, the correct answer is D.

Problem 6 in other years