Competition · AMC preparation · step 4 of 4

AMC 8 · 2023 · #21

Grade 5 counting
Archetype Branching State Enumeration └ Sub-archetype Rule-Governed Process Simulation
systematic-enumerationcombinations-basicset-partition caseworksystematic-enumerationtree-enumeration ↑ Prerequisites: combinations-basicmental-arithmetic
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Problem
Alina has nine cards labeled 1, 2, 3, …, 9. She wants to split them into three piles of three cards each so that all three piles have the same sum. We need to count how many different ways this split can be done.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

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.

1STEP 1

Add all nine card values

Pair the nine cards as 1+9, 2+8, 3+7, 4+6 — four tens plus the leftover 5 — so the total is 45.

1 + 2 + … + 9 = 4 × 10 + 5 = 45
2STEP 2

Split the total into thirds

Equal piles each hold a third of the total, so divide: 45 ÷ 3 = 15 is the sum every pile must reach.

45/3 = 15 ⟹ each group sums to 15
3STEP 3

List every triple summing to 15

Fix the largest card and work down: exactly seven distinct triples sum to 15, all listed at right.

&{1,5,9}, {2,4,9}, ; &{1,6,8}, {2,5,8}, {3,4,8}, ; &{2,6,7}, {3,5,7}
4STEP 4

Split into two cases

Card 9 must sit somewhere, so its pile is either {1,5,9} or {2,4,9} — two cases to check.

Case 1: {1,5,9} | Case 2: {2,4,9}
5STEP 5

Finish case 1

Case 1 takes {1,5,9}; the only sum-15 triple left with an 8 is {3,4,8}, leaving {2,6,7} — one partition.

{1,5,9}∪{3,4,8}∪{2,6,7}={1,2,3,4,5,6,7,8,9} ✓
6STEP 6

Finish case 2

Case 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.

{2,4,9}∪{1,6,8}∪{3,5,7}={1,2,3,4,5,6,7,8,9} ✓
7STEP 7

Count the partitions

Two cases, one partition each, and no other because 9 sits in just one pile: 2 ways, choice (C).

1 + 1 = 2 ⟹ (C)
Answer
2
The two partitions we found really are different — {1,5,9},{3,4,8},{2,6,7} and {1,6,8},{2,4,9},{3,5,7} share no group in common. Each row sums to 15, and across the two partitions every card 1 through 9 appears exactly once. The answer 2 also feels right intuitively: there are only seven triples that even sum to 15, and once we pin down where 9 goes, the other groups are forced — so we should expect a small number, not 3 or 4.
💡Key takeaway

This 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!

  • Add all nine card values
  • Split the total into thirds
  • List every triple summing to 15
  • Split into two cases
  • Finish case 1
  • Finish case 2
  • Count the partitions

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