Competition · AMC preparation · step 4 of 4
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.
Pull out the constant remainder
"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
Find 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.
The smallest positive whole number that 3, 4, 5, and 6 all divide evenly is 60.
▸ Why?
Sixty really is divided evenly by all four, since 60 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10, so each divisor goes in a whole number of times with nothing left over.
▸ Why?
No positive number below 60 works, because any number all four divide is a common multiple of 3, 4, 5, and 6, and the smallest such common multiple is their least common multiple, which is 60.
▸ Why?
The numbers that 3, 4, 5, and 6 all divide are exactly the multiples of one single smallest value — their least common multiple — so that value is the first shared number and nothing below it can be divided evenly by all four.
▸ Why?
That least common multiple comes out to 60: a number all four divide must carry two 2s (to be a multiple of 4), a 3, and a 5, and the smallest number built from exactly those primes is 2 × 2 × 3 × 5 = 60; the 6 forces nothing new because 6 = 2 × 3 is already present.
▸ Why?
A multiple of 4 must contain 2 × 2, a multiple of 3 must contain a 3, and a multiple of 5 must contain a 5, because a multiple is that divisor taken a whole number of times and so carries all of the divisor's own factors.
▸ Why?
Taking each demanded prime the fewest times forced, and adding no spare factor, gives the one smallest number holding them all, because every whole number is a product of primes in just one way, so a smaller number would have to drop a required prime and fail.
List the candidates in order
Write 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 PatternMatch 62 to the range
Check 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!
- Pull out the constant remainder
- Find the least common multiple
- List the candidates in order
- Match 62 to the range
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