AMC 8 · 2016 · #20

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

lcmgcdprime-factorization identify-subproblemscaseworkguess-and-check ↑ Prerequisites: prime-factorizationlcmgcd

📏 Medium solution 💡 4 insights

Problem

Three positive integers a, b, c satisfy lcm(a, b) = 12 and lcm(b, c) = 15. Find the smallest possible value of lcm(a, c).

Pick an answer.

(A)

(B)

(C)

(D)

120

(E)

180

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) is the natural fit. The two LCM conditions share the variable b, so we split the work in three pieces — first pin down what b can be (it must divide both 12 and 15), then for each choice of b find the smallest legal a and the smallest legal c separately, then take lcm(a, c). Tool #6 (Guess and Check) handles the tiny case-list — only b = 1 and b = 3 survive — so we just test both and keep the winner. We avoid Tool #13 (Algebra) here because the search space is two cases, not an equation.

1STEP 1

Factor the givens: 12 = 2² × 3 and 15 = 3 × 5, so the only primes in play are 2, 3, and 5.

12 = 2² × 3, 15 = 3 × 5

2STEP 2

b divides both 12 and 15, so b divides gcd(12, 15) = 3, leaving only b = 1 or b = 3.

b ∣ 12 and b ∣ 15 → b ∣ gcd(12, 15) = 3 → b ∈ {1, 3}

3STEP 3

Take b = 3: then a = 4 and c = 5 satisfy both lcm's, giving lcm(4, 5) = 20.

b = 3: a = 4, c = 5 → lcm(a, c) = lcm(4, 5) = 20

4STEP 4

Take b = 1: now a = 12 and c = 15 are forced, so lcm(12, 15) = 60 — much larger.

b = 1: a = 12, c = 15 → lcm(a, c) = lcm(12, 15) = 60

5STEP 5

Compare the two cases: min(20, 60) = 20, which is choice (A).

min(20, 60) = 20 → (A)

Answer

Sanity-check the answer prime by prime. lcm(a, c) must be a multiple of every prime that a or c is forced to carry. Looking at 12 = 2² × 3, the factor 2² has no home in b (since b ∣ 15 rules 2 out of b), so a must carry it — forcing 4 ∣ lcm(a, c). Similarly, 5 has no home in b, so c must carry it — forcing 5 ∣ lcm(a, c). Hence lcm(a, c) is at least lcm(4, 5) = 20. Our construction (a, b, c) = (4, 3, 5) achieves exactly 20, so 20 is both attainable and a hard lower bound. That matches (A).

💡Key takeaway

This AMC 8 problem only needs the Grade 6 GCF/LCM idea — break 12 and 15 into primes, push the shared prime into b, and the leftover 4 and 5 give the smallest answer lcm(4, 5) = 20!

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