AMC 10 · 2007 · #24

Grade 6 number-theory

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

gcddivisibility-rulescasework casework ↑ Prerequisites: gcd

📏 Long solution 💡 4 insights

Problem

Two positive whole numbers share no factor bigger than one and make a certain sum of fractions a whole number. Count the pairs.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

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

Combining gives one fraction to make whole.

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

2STEP 2

What a must divide

The first number must divide a small constant.

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

3STEP 3

What b must divide

So must the second, giving another short list.

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

4STEP 4

Narrow to a short candidate list

Together they leave only a few candidates.

a ∈ {1,2,7,14}, b ∈ {1,3,9}

5STEP 5

Test each value of b

Testing each second value kills all but one.

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

6STEP 6

Count the surviving pairs

Counting the survivors gives 4, 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