AMC 8 · 2023 · #21
Grade 5 countingPick an answer.
AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question "in how many ways…" with a small, finite setup is the classic trigger for Tool #2 (Make a Systematic List). Before listing, Tool #7 (Identify Subproblems) breaks the task into three cleaner pieces: (a) figure out what each group must sum to, (b) enumerate every triple of cards that hits that sum, (c) pick a triple-of-triples that uses each card exactly once. With the target sum pinned down, Tool #3 (Eliminate Possibilities) lets us anchor on the group that must contain the largest card 9, then read off the remaining groups by force — almost every candidate gets ruled out, which is why the final count is so small.
Pair the nine cards as 1+9, 2+8, 3+7, 4+6 — four tens plus the leftover 5 — so the total is 45.
Fluent addition within 100 — a Grade 2 skill — is enough to total nine small numbers.
2.NBT.B.5Identify SubproblemsEqual piles each hold a third of the total, so divide: 45 ÷ 3 = 15 is the sum every pile must reach.
Sharing a total equally among three groups is a Grade 3 division word problem.
3.OA.A.3Identify SubproblemsFix the largest card and work down: exactly seven distinct triples sum to 15, all listed at right.
Listing all the sum-decompositions of 15 into three distinct cards is the same idea as the Grade 4 "find all factor pairs" skill, applied to addition instead of multiplication.
4.OA.B.4Make A Systematic ListCard 9 must sit somewhere, so its pile is either {1,5,9} or {2,4,9} — two cases to check.
Splitting into a small number of forced cases — and following the rule "the largest leftover card must appear somewhere" — is the Grade 5 standard of generating cases from a rule.
5.OA.B.3Eliminate PossibilitiesCase 1 takes {1,5,9}; the only sum-15 triple left with an 8 is {3,4,8}, leaving {2,6,7} — one partition.
Multi-step whole-number reasoning at Grade 4 — pick a triple, subtract, check the leftover sum — finishes this case.
4.OA.A.3Eliminate PossibilitiesCase 2 takes {2,4,9}; the only sum-15 triple left with an 8 is {1,6,8}, leaving {3,5,7} — one more partition.
Same Grade 4 multi-step subtraction-and-check move as Case 1, applied to the second forced split.
4.OA.A.3Eliminate PossibilitiesTwo cases, one partition each, and no other because 9 sits in just one pile: 2 ways, choice (C).
Adding up the case counts is just Grade 2 addition.
2.NBT.B.5Make A Systematic ListThis AMC 8 problem only needs the Grade 5 habit of generating cases from a rule — once you decide where the card 9 goes, the rest of the groups are forced!
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Same archetype — closest grade level first.
