Competition · AMC preparation · step 4 of 4
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.
Restate the story as divisors
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.
An arrangement with k students in each equal row is possible exactly when k is a divisor of the total number of students N.
▸ Why?
Standing the students in equal rows of k splits the whole group into some whole number r of rows, each holding exactly k students, so the total is r rows of k.
▸ Why?
r rows that each hold k students are r equal groups of k, and r equal groups of k count up to r × k altogether.
▸ Why?
The rows together hold every student once, with nobody left over and nobody counted twice, so the row contents add back up to the whole group N.
▸ Why?
So a working arrangement needs a whole number of rows r with N = r × k, and such an r exists exactly when N ÷ k comes out to a whole number — which is what it means for k to divide N.
▸ Why?
Getting r back from N = r × k means undoing the multiplication by k, that is r = N ÷ k, and this r lands on a whole number exactly when k divides N with nothing left over.
Read off the divisor count
New 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 PossibilitiesCross out by divisibility
Since 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
List 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 N / k is the systematic way to list divisors without missing any.
4.OA.B.4Make A Systematic ListList the divisors of 60
List 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 ListPick the smallest survivor
60 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!
- Restate the story as divisors
- Read off the divisor count
- Cross out by divisibility
- List the divisors of 30
- List the divisors of 60
- Pick the smallest survivor
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