2009 AMC 10B Problem 6
Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
Show solution without answering
Key insight
The twins share an age, so 128 is Kiana's age times the twins' age twice. Try the few possible twin ages and keep the one where Kiana is younger.
Solution
The twins have the same age, so the product is (Kiana's age) (twins' age) (twins' age). Try each possible twins' age.
- , so every age must be a power of :
- List the ways to pick the twins' age and find Kiana's age from :
| Twins' age | Twins multiplied | Kiana's age | Kiana younger? |
|---|---|---|---|
| no | |||
| no | |||
| no | |||
| yes |
(A twins' age of is too big, since is more than .)
3. Only the last row fits the word "older": the twins are and Kiana is .
4. The sum is .
The answer is .
Why this works
Twins give a repeated factor, so the product has a squared part. Writing as a product of 's shows there are only a few possible ages to test. Several choices multiply to , so the word "older" is what picks the right one; always use every condition in the problem.
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Let the twins each be years old and Kiana , with . Then , so is a power of and divides .
- : , and . Valid.
- : , but then Kiana is older. Invalid.
- or : or , far older. Invalid.
So the ages are and the sum is .
The answer is .
Review in the AoPS books
- Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
- Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)
- Introduction to Algebra, Chapter 20: Special Functions (§20.3)
- Introduction to Number Theory, Chapter 13: Divisibility Rules (§13.2)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.