2009 AMC 10B Problems/Problem 6

The following problem is from both the 2009 AMC 10B #6 and 2009 AMC 12B #5, so both problems redirect to this page.

Problem

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

$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$

Solution

The age of each person is a factor of $128 = 2^7$. So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$. The answer is $\mathrm{(D)}$.

Video Solution

https://youtu.be/F8k7r3LDXoA

~savannahsolver

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2009 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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