AMC 10 · 2007 · #25

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

gcddivisibility-rulescasework casework ↑ Prerequisites: gcd

📏 Long solution 💡 4 insights

Problem

An ordered pair of positive whole numbers (a,b) shares no common factor bigger than 1, and the expression

\frac{a}{b}

\frac{14b}{9a}

must come out to a whole number. Find how many such pairs there are.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Two separate fractions are hard to judge, so first combine them into one fraction (9a²+14b²)/9ab; then "is an integer" simply means the bottom divides the top. Tool #7 (Identify Subproblems) splits that one divisibility question into smaller pieces: what must a divide, what must b divide, and how does the shared factor of 3 behave. The coprime rule turns each piece into a short divisor list, so a and b each have only a handful of possible values. Tool #2 (Make a Systematic List) then tests those few candidates, and Tool #3 (Eliminate Possibilities) throws out the ones that fail, leaving the exact count.

1STEP 1

Combine into one fraction

Over the common bottom 9ab the sum is $\frac{9a^2+14b^2}{9ab}$ , so being a whole number just means 9ab divides 9a²+14b².

a/b+14b/9a=9a²/9ab+14b²/9ab=(9a²+14b²)/9ab

2STEP 2

What a must divide

The bottom's a must divide 9a²+14b²; it already divides 9a² and shares nothing with b, so a divides 14: a is 1, 2, 7, or 14.

a ∣ 9a²+14b² → a ∣ 14b² → a ∣ 14 → a ∈ {1,2,7,14}

3STEP 3

What b must divide

Likewise b must divide 9a²+14b²; it already divides 14b² and shares nothing with a, so b divides 9: b is 1, 3, or 9.

b ∣ 9a²+14b² → b ∣ 9a² → b ∣ 9 → b ∈ {1,3,9}

4STEP 4

Narrow to a short candidate list

Divisor lists are necessary, not sufficient: the bottom still holds a 9 beside b, so test the three cases b = 1, 3, 9.

a ∈ {1,2,7,14}, b ∈ {1,3,9}

5STEP 5

Test each value of b

b = 1 would need 9a to divide 14 — never; b = 9 needs 9 to divide a²+126, which fails — only b = 3 works, leaving $\frac{a^2+14}{3a}$ .

b=1: 9a ∣ 14 (never); b=9: 9 ∤ a²+126; b=3: (a²+14)/3a (always)

6STEP 6

Count the surviving pairs

With b = 3 every allowed a works: (1,3), (2,3), (7,3), (14,3) — 4 pairs, choice (A).

(1,3),(2,3),(7,3),(14,3) → 4 pairs → (A)

Answer

The four pairs were each plugged back in and gave the whole numbers 5, 3, 3, 5, so all four truly work. No pair was double-counted, and the case analysis on b proved that b=1 and b=9 contribute nothing, so nothing was missed — this rules out (E) infinitely many and the larger counts (B) 6, (C) 9, (D) 12. Exactly 4 pairs remain, matching (A).

💡Key takeaway

Glue the fractions into one, then let the no-shared-factor rule force each letter onto a tiny divisor list — after that you only have a handful of cases to check.

Combine into one fraction
What a must divide
What b must divide
Narrow to a short candidate list
Test each value of b
Count the surviving pairs