AMC 8 · 2006 · #23
Grade 6 number-theoryPick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each remainder condition pins N to a clear arithmetic sequence — "start at 4, step by 6" for the first, "start at 3, step by 5" for the second — so Tool #2 (Systematic List) writes both sequences cleanly and the first shared value is the smallest N. Tool #5 (Look for a Pattern) sharpens the search: the gap to the next multiple of 6 is 2, and the gap to the next multiple of 5 is also 2, so N + 2 must be a common multiple of 5 and 6. That pattern means we only need the smallest common multiple of 5 and 6, which is 30, and then N = 30 - 2 = 28. Both tools land on the same N, after which the final remainder by 7 is one division.
The numbers leaving remainder 4 when divided by 6 start at 4 and grow by 6.
Grade 4 "generate a number pattern from a rule" — the rule here is +6, starting at 4.
4.OA.C.5Make A Systematic ListThe numbers leaving remainder 3 when divided by 5 start at 3 and grow by 5.
Same Grade 4 pattern move with a different rule: +5, starting at 3.
4.OA.C.5Make A Systematic ListScanning both lists in order, the first shared value is 28, so that is the smallest N.
Intersecting two short lists is the cleanest Grade 4 way to pin down the smallest number that fits two multiple-style rules at once.
4.OA.B.4Make A Systematic ListDividing 28 by 7 gives exactly 4 with remainder 0.
Grade 4 division-with-remainder closes the problem: 7 divides 28 evenly, so nothing is left over.
4.NBT.B.6Make A Systematic ListWhen a count leaves two awkward remainders, two short arithmetic lists usually meet within a few terms — and spotting that both shortfalls are the same number turns the search into a quick LCM.