AMC 8 · 2005 · #14

Grade 5 counting

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

combinations-basicmulti-digit-arithmeticset-partition identify-subproblemscasework ↑ Prerequisites: combinations-basic

📏 Short solution 💡 2 insights

Problem

The Little Twelve Basketball Conference has two divisions of 6 teams each. Every team plays every other team in its own division twice, and every team in the other division once. How many conference games are scheduled in total?

Pick an answer.

(A)

(B)

(C)

100

(D)

108

(E)

192

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The total schedule has two kinds of games with different rules, so Tool #7 (Identify Subproblems) splits the count into a clean sum: intra-division games (both divisions) plus inter-division games. Each subproblem is a short multiplication. To count the unordered team pairings inside one division, Tool #2 (Make a Systematic List) is the kid-friendly way to get 15 without a combination formula — just list each team's new opponents in order so no pair gets counted twice. We avoid Tool #13 (Algebra) and the C(n, 2) formula because grade-5 arithmetic is enough.

1STEP 1

Subproblem 1: pair each of a division's 6 teams only with higher-numbered teams — this handshake count gives 15 pairs per division.

5 + 4 + 3 + 2 + 1 + 0 = 15 pairs per division

2STEP 2

Each of the 15 pairs plays twice and there are two divisions, so the intra-division total is 60 games.

intra = 15 × 2 × 2 = 60 games

3STEP 3

Subproblem 2: each of Division A's 6 teams plays each of Division B's 6 teams once — a 6 × 6 grid gives 36 games.

inter = 6 × 6 = 36 games

4STEP 4

Intra and inter games never overlap, so add them: 60 + 36 = 96 games, choice (B).

total = 60 + 36 = 96 → (B)

Answer

Sanity check the size. A loose upper bound is to imagine every one of the 12 teams playing every other team twice: that would be 12 × 11 = 132 ordered pairs, or 66 unordered pairs, so at most 132 games if every pair played twice. Our intra-division share (60) plus the once-only inter-division share (36) gives 96, which sits comfortably below that bound. Trap choices come from common slips: 100 is 60 + 40 (over-counting inter-division), 108 is 72 + 36 (treating both intra and inter as × 2), 192 is 96 × 2 (counting each game from both teams' point of view), and 80 is 40 + 40 (forgetting the second division's intra games).

💡Key takeaway

Split the schedule into "same-division" and "other-division" games, count each piece with a quick multiplication, then add. With that split, this AMC 8 problem becomes a Grade 5 multistep arithmetic exercise.

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