AMC 10 · 2018 · #19

Grade 6 rate-ratio
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
divisor-countmultiplesdigit-sum convert-to-algebrapattern-recognition ↑ Prerequisites: divisor-count
📏 Medium solution 💡 3 insights
Problem
Joey, Chloe, and their daughter Zoe share a birthday. Joey is 1 year older than Chloe, and Zoe is exactly 1 today. Today is the first of exactly 9 birthdays on which Chloe's age is a whole-number multiple of Zoe's age. Find the sum of the two digits of Joey's age the next time his age is a whole-number multiple of Zoe's age.

Pick an answer.

(A)
7
(B)
8
(C)
9
(D)
10
(E)
11

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Tool #16 (Change Focus / Count the Complement): instead of tracking the messy ratio of two growing ages, focus on the gap between them. Two people who share a birthday keep the same age difference forever, and that fixed difference is the key — Zoe's age is a multiple-maker exactly when it divides that constant gap. Tool #13 (Convert to Algebra): name Chloe's age and write the ages after t years so "multiple" becomes a clean divisibility statement. Tool #5 (Look for a Pattern): because Zoe's age runs through 1, 2, 3, … in order, the multiple-birthdays line up exactly with the divisors of the fixed gap, so counting birthdays becomes counting divisors.

1STEP 1

Name the ages

Let Chloe be C today, so Joey is C+1 and Zoe is 1; then '1+t divides C+t' after t years is exactly what 'multiple' means.

Chloe = C+t, Zoe = 1+t, Joey = C+1+t
2STEP 2

Look at the fixed gap

Since Chloe and Zoe share a birthday, their gap (C+t)-(1+t) = C-1 is fixed, so 1+t divides C+t exactly when it divides C-1.

(C+t)-(1+t) = C-1 → (1+t) ∣ (C-1)
3STEP 3

Birthdays equal divisors

As t grows, Zoe's age 1+t hits every whole number in turn, so multiple-birthdays match the divisors of C-1 - and there are 9.

#{multiple-birthdays} = #{divisors of C-1} = 9
4STEP 4

Find the gap with nine divisors

Nine divisors comes from p⁸ or p²q²; the smallest realistic gap is C-1 = 36 = 2²·3², so Chloe is 37 and Joey is 38 today.

C-1 = 2²· 3² = 36 → C = 37, Joey = 38
5STEP 5

Do the same for Joey

Joey's gap over Zoe is C = 37, a prime, so the next multiple-birthday is Joey = 74; digits 7+4 = 11, answer (E).

37 prime → next Zoe = 37, Joey = 74, 7+4 = 11 → (E)
Answer
11
Confirm 36 truly has nine divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36 — that is exactly 9, matching the nine Chloe-birthdays, and the first one (Zoe =1, today) is included as promised. With Chloe 37 and Joey 38, Joey's gap over Zoe is 37, a prime, so the next multiple-birthday is Zoe =37, Joey =74, and indeed 74 = 2 × 37 is a clean multiple. The digit sum 7+4 = 11 is answer (E). The alternative C-1 = 256 would make Chloe 257 — impossible for a parent of a 1-year-old — so 36 is the only sensible gap, confirming the result.
💡Key takeaway

People who share a birthday keep the same age gap forever, so "one age is a multiple of another" just means the small age divides that fixed gap — count the gap's divisors and you count the birthdays.

  • Name the ages
  • Look at the fixed gap
  • Birthdays equal divisors
  • Find the gap with nine divisors
  • Do the same for Joey

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