AMC 10 · 2009 · #5

Grade 4 number-theory
prime-factorizationfactorslogical-deduction systematic-enumeration ↑ Prerequisites: prime-factorization
📏 Medium solution 💡 2 insights
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Problem
Two of three ages are equal and both larger than the third. The three ages multiply to a given number. Find their sum.

Pick an answer.

(A)
10
(B)
12
(C)
16
(D)
18
(E)
24
How to solve
Strategy Introduce a Variable

Name the twin age t and Kiana's age k. Since the twins share an age, the product becomes t × t × k = 128, so t shows up twice. That lets me list only the whole-number twin ages whose square divides 128, then throw out the ones that break the 'older than Kiana' rule.

1STEP 1

Name the ages

The equal pair means the product holds a square.

t × t × k = 128
2STEP 2

List the twin ages that fit

Listing them gives four candidates.

1{·}1{·}128, 2{·}2{·}32, 4{·}4{·}8, 8{·}8{·}2
3STEP 3

Keep only older twins

The older condition leaves only one.

8 > 2 ✓
4STEP 4

Add the three ages

Adding gives 18, choice (D).

8 + 8 + 2 = 18
Answer
18
Check the survivor directly: 8 × 8 × 2 = 128 matches the product, and both brothers (8) are older than Kiana (2), so every condition holds. The sum 8 + 8 + 2 = 18 is choice (D).
💡Key takeaway

When two people are the same age, that age gets used twice in the product, so only ages whose square fits the number can work.

  • Name the ages
  • List the twin ages that fit
  • Keep only older twins
  • Add the three ages