Competition · AMC preparation · step 4 of 4
AMC 8 · 2016 · #20
Grade 6 number-theoryPick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Factor 12 and 15 into primes
Factor the givens: 12 = 2² × 3 and 15 = 3 × 5, so the only primes in play are 2, 3, and 5.
Splitting a number into its prime building blocks is the first move whenever the question is about lcm or gcd — each prime can be handled on its own.
6.EE.A.1Identify SubproblemsNarrow down b
b divides both 12 and 15, so b divides gcd(12, 15) = 3, leaving only b = 1 or b = 3.
If b tried to use the prime 2 it would force lcm(b, c) to be even, but 15 is odd. The same logic rules 5 out of b via 12.
The number b, shared by both conditions, must be either 1 or 3.
▸ Why?
b divides 12: the least common multiple of a and b is 12, a number both a and b share as a multiple, so 12 is a whole number of copies of b with none left over.
▸ Why?
b divides 15: the least common multiple of b and c is 15, again a shared multiple of b, so 15 is a whole number of copies of b with none left over.
▸ Why?
No whole number larger than 3 divides both 12 and 15, so b cannot exceed 3, leaving only 1 and 3.
▸ Why?
12 is built from the primes 2, 2, 3 and 15 from the primes 3, 5, so the only prime they share is a single 3; a number dividing both can be assembled only from that shared prime, which yields just 1 and 3.
Test b equal to 3
Take b = 3: then a = 4 and c = 5 satisfy both lcm's, giving lcm(4, 5) = 20.
Loading b with every shared prime (3 in this case) lets a and c stay as small as possible, and lcm(a, c) stays small with them.
6.NS.B.4Guess And CheckTest b equal to 1
Take b = 1: now a = 12 and c = 15 are forced, so lcm(12, 15) = 60 — much larger.
When b shares nothing, every prime gets pushed entirely into a or c, blowing up lcm(a, c).
6.NS.B.4Guess And CheckPick the smaller result
Compare the two cases: min(20, 60) = 20, which is choice (A).
Only two cases survived the b-filter, so the brute compare is just "pick the smaller of two numbers".
4.OA.B.4Identify SubproblemsThis 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!
- Factor 12 and 15 into primes
- Narrow down b
- Test b equal to 3
- Test b equal to 1
- Pick the smaller result
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