AMC 8 · 2012 · #15
Grade 6 number-theoryPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem packs four divisibility conditions into one question. Tool #7 (Identify Subproblems) turns that into one clean step: subtract the common remainder of 2, and the question becomes "what is the smallest positive number that is divisible by 3, 4, 5, and 6 at the same time?" — the definition of the least common multiple. Tool #5 (Look for a Pattern) handles the leftover comparison: every working N has the form 60k + 2, so we just pick the smallest k and walk the pattern. Tool #3 (Eliminate Possibilities) then maps the resulting N against the five range choices and rules out the four that miss. We avoid tool #13 (Algebra) and modular-arithmetic notation because the "shift by 2" subproblem move makes them unnecessary.
"Remainder 2 by d" means N - 2 is a multiple of d; apply it to all four divisors, so you need the smallest N - 2 divisible by 3, 4, 5, 6.
Stripping the shared remainder is the Tool #7 subproblem move — one harder question becomes one easier question about plain multiples.
4.OA.B.4Identify SubproblemsFind the least common multiple of 3, 4, 5, 6 by prime factorization: taking 2², 3, and 5 gives LCM = 60.
The 6 is "free" because 6 = 2 × 3 is already covered by the 4 and 3 — no new prime is added.
6.NS.B.4Identify SubproblemsWrite N - 2 = 60k for k = 1, 2, 3, …; the smallest N greater than 2 comes from k = 1, giving N = 62.
Listing the first few terms of the "add 60" pattern makes the smallest one — and the gap to the next one — obvious.
4.OA.C.5Look For A PatternCheck 62 against the ranges — 40–50, 51–55, 56–60, 61–65, 66–99 — and only 61–65 contains it, so the answer is (D).
Comparing 62 against each range is just the Tool #3 "eliminate possibilities" move — four ranges miss, one fits.
1.NBT.B.3Eliminate PossibilitiesThis AMC 8 problem boils down to one Grade 6 idea — finding the least common multiple of 3, 4, 5, 6 — once you spot that subtracting the shared remainder of 2 makes the four conditions collapse into one!