AMC 8 · 2005 · #14
Easy mode Grade 5Problem
Picture a basketball league called the Little Twelve. It has two divisions, and each division has 6 teams.
Here are the rules for the schedule:
- Each team plays every other team in its own division twice.
- Each team plays every team in the other division once.
How many games are scheduled in the whole league?
Pick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Two divisions, $6$ teams each ($12$ teams total); Intra-division: each pair of teams in the same division plays $2$ games; Inter-division: each pair (one team from each division) plays $1$ game; Answer choices: (A) $80$, (B) $96$, (C) $100$, (D) $108$, (E) $192$
Unknowns: The total number of games scheduled across the whole conference
Understand
Restated: 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?
Givens: Two divisions, $6$ teams each ($12$ teams total); Intra-division: each pair of teams in the same division plays $2$ games; Inter-division: each pair (one team from each division) plays $1$ game; Answer choices: (A) $80$, (B) $96$, (C) $100$, (D) $108$, (E) $192$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List
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 $\binom{n}{2}$ formula because grade-5 arithmetic is enough.
Execute — Answer: B
4.OA.A.3 Step 1 - Subproblem 1: count the unordered pairs of teams inside one division using a systematic list.
- Label the $6$ teams $1, 2, 3, 4, 5, 6$.
- For each team, list only the partners with a larger label so no pair is counted twice.
💡 Team $1$ has $5$ new partners, team $2$ has $4$ new ones (it already paired with team $1$), and so on. This is the same as the "handshake" count.
4.OA.A.1 Step 2 - Each of those $15$ pairs plays twice, and there are two divisions that follow the same rule.
- So the intra-division total multiplies the pairs-per-division by $2$ games and by $2$ divisions.
💡 "Plays twice" is a multiplicative comparison: it doubles the count. Two identical divisions double it again.
5.OA.A.2 Step 3 - Subproblem 2: count inter-division games.
- Each of the $6$ teams in Division A plays each of the $6$ teams in Division B exactly once, so this is a simple multiplication.
💡 Pairing every Division A team with every Division B team is a $6 \times 6$ grid of matchups.
4.OA.A.3 Step 4 Add the two subproblem totals to get the full schedule, then match to a choice.
💡 Intra-division and inter-division games never overlap, so the totals just add.
4.OA.A.3 Subproblem 1: count the unordered pairs of teams inside one division using a sys 4.OA.A.1 Each of those $15$ pairs plays twice, and there are two divisions that follow th 5.OA.A.2 Subproblem 2: count inter-division games. Each of the $6$ teams in Division A pl 4.OA.A.3 Add the two subproblem totals to get the full schedule, then match to a choice. Review
Reasonableness: 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 $12 \times 11 = 132$ 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 $\times 2$), $192$ is $96 \times 2$ (counting each game from both teams' point of view), and $80$ is $40 + 40$ (forgetting the second division's intra games).
Alternative: Tool #9 (Solve an Easier Related Problem): shrink each division to $2$ teams. Intra-division: each division has $1$ pair playing twice, so $1 \times 2 \times 2 = 4$ games. Inter-division: $2 \times 2 = 4$ games. Total $8$. With $3$ teams each: intra $= 3 \times 2 \times 2 = 12$, inter $= 3 \times 3 = 9$, total $21$. The pattern "intra $= \binom{n}{2} \cdot 2 \cdot 2$ and inter $= n \cdot n$" holds; plugging $n = 6$ gives $60 + 36 = 96$, confirming (B).
CCSS standards used (min grade 5)
4.OA.A.1Interpret a multiplication equation as a comparison (Reading "each pair plays twice" as a multiplicative comparison that doubles the intra-division count.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Listing the $15$ pairs per division and combining the intra- and inter-division subtotals into the final sum.)5.OA.A.2Write simple expressions that record calculations with numbers (Recording the inter-division count as $6 \times 6$ and the intra-division count as $15 \times 2 \times 2$ before evaluating.)
⭐ 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.
⭐ 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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