AMC 8 · 2015 · #22
Grade 6 number-theoryPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each day's arrangement corresponds to one divisor of N, so "12 days but no 13th" simply says N has exactly 12 divisors. That gives us a single sharp test. We then use Tool #3 (Eliminate Possibilities) on the five answer choices: knock out any that fail the divisibility constraints (15 ∣ N and 6 ∣ N, i.e. 30 ∣ N) and any whose divisor count isn't 12. To check the divisor count we use Tool #2 (Make a Systematic List) — list every divisor of each surviving candidate in order from smallest to largest and count them. Working from the smallest choice upward, the first one that passes both tests is the answer.
Each day's row size k satisfies N = r × k, so k is a divisor of N — the daily arrangements are exactly the divisors of N.
Rectangular arrays match factor pairs — exactly the Grade 4 "factors and multiples" idea.
4.OA.B.4Eliminate PossibilitiesNew arrangements for 12 days then none on the 13th means N has exactly 12 divisors.
"12 work, 13th doesn't" is the most direct way the problem tells us the divisor count.
4.OA.B.4Eliminate PossibilitiesSince 15 and 6 both divide N, so does lcm(15, 6) = 30; (A) 21 isn't a multiple of 30, so cross it out.
lcm is the Grade 6 way to combine two divisibility requirements into one.
6.NS.B.4Eliminate PossibilitiesList the divisors of 30 in pairs — 1·30, 2·15, 3·10, 5·6 — only 8 of them, not 12, so cross out (B).
Pairing k with is the systematic way to list divisors without missing any.
4.OA.B.4Make A Systematic ListList the divisors of 60 in pairs — six pairs give exactly 12 divisors, and both 15 and 6 appear.
Six factor pairs → exactly 12 divisors — and 15 and 6 are both on the list.
4.OA.B.4Make A Systematic List60 clears both tests and is smaller than 90 and 1080, so it is the smallest valid choice.
Once a candidate passes from the smallest end of the list, larger candidates are no longer minimal.
4.OA.B.4Eliminate PossibilitiesThis AMC 8 problem only needs Grade 6 factor-and-multiple reasoning: turn "rows" into divisors, then test the choices!