2009 AMC 10B Problem 6 — Solution | LIVE by Po-Shen Loh

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2009 AMC 10B Problem 6

Below is the professionally curated solution for Problem 6 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Concepts:prime factorization power of 2 ages

Difficulty rating: 960

6.

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

$10$

$12$

$16$

$18$

$24$

Solution:

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 $k\lt t.$

Taking $t=8$ gives $k=2,$ which works. The sum is $8+8+2=18.$

Thus, the correct answer is D.

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Problem 6 in Other Years

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