AMC 8 · 2012 · #14
Easy mode Grade 5Problem
A football league has some teams in it. During the season, every team plays every other team exactly once.
By the end of the season, a total of 21 games have been played.
How many teams are in the league?
Pick an answer.
AMC 8 2012 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: In the BIG N middle school football conference, every team plays every other team exactly once. A total of $21$ games were played in the $2012$ season. How many teams are in the conference?
Givens: Every pair of teams plays exactly one game against each other (round-robin); Total number of games played $= 21$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The number of teams $n$ in the conference
Understand
Restated: In the BIG N middle school football conference, every team plays every other team exactly once. A total of $21$ games were played in the $2012$ season. How many teams are in the conference?
Givens: Every pair of teams plays exactly one game against each other (round-robin); Total number of games played $= 21$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #6 Guess and Check, #5 Look for a Pattern
A game is a pair of teams, so the question is really "how many pairs can $n$ teams make?" Tool #2 (Make a Systematic List) lets us count those pairs directly for small $n$ — for $3$ teams the pairs are AB, AC, BC ($3$ games); for $4$ teams the pairs are AB, AC, AD, BC, BD, CD ($6$ games). Tool #5 (Look for a Pattern) turns those counts into the rule "new team $n$ adds $n-1$ new games," giving the running totals $1, 3, 6, 10, 15, 21, \ldots$. Tool #6 (Guess and Check) then matches $21$ to a choice without any algebra.
Execute — Answer: B
4.OA.A.3 Step 1 - Start with the smallest cases.
- With $2$ teams there is $1$ game (AB).
- Add a third team C; team C plays both A and B, so $2$ new games (AC, BC) — total $1 + 2 = 3$.
- The new team always plays every earlier team once.
💡 Listing pairs in order (A first, then B, then C) is the systematic-list move — no pair gets missed or double-counted.
5.OA.A.2 Step 2 - Spot the pattern.
- Each time we add a team, the new team plays every previous team exactly once, so adding the $n$-th team adds $n-1$ new games.
- The total number of games for $n$ teams is the sum $1 + 2 + 3 + \cdots + (n-1)$.
💡 Writing the count as a numerical expression (a sum of consecutive whole numbers) is a Grade 5 "express a calculation as an expression" move.
4.OA.A.3 Step 3 - Extend the list of running totals until we hit $21$.
- Keep a small table of (teams, games).
💡 Each row adds the next whole number: $10 + 5 = 15$, then $15 + 6 = 21$ — there is the $21$ we need.
4.OA.A.3 Step 4 - Check the answer with Guess and Check on the listed choices.
- We want the choice $n$ for which the running total is exactly $21$.
💡 Choice (B) $n = 7$ is the only one that produces exactly $21$ games — the answer is $\textbf{(B)}$.
4.OA.A.3 Start with the smallest cases. With $2$ teams there is $1$ game (AB). Add a thir 5.OA.A.2 Spot the pattern. Each time we add a team, the new team plays every previous tea 4.OA.A.3 Extend the list of running totals until we hit $21$. Keep a small table of (team 4.OA.A.3 Check the answer with Guess and Check on the listed choices. We want the choice Review
Reasonableness: Sanity check by a different counting argument. Each of the $7$ teams plays the other $6$ teams once, which seems like $7 \times 6 = 42$ game-slots. But every game is counted twice (once for each team), so the real game count is $42 \div 2 = 21$. That matches the given total exactly, confirming $n = 7$.
Alternative: Tool #13 (Convert to Algebra) gives the closed form $\binom{n}{2} = \dfrac{n(n-1)}{2} = 21$, so $n(n-1) = 42$. The two consecutive whole numbers whose product is $42$ are $6$ and $7$, so $n = 7$. The systematic-list path above reaches the same answer without needing the variable equation, which is why we preferred it for younger solvers.
CCSS standards used (min grade 5)
4.OA.A.3Solve multi-step word problems with whole numbers using the four operations (Counting games for small numbers of teams ($1, 3, 6, 10, 15, 21$) and matching the running total to the choices — a multi-step whole-number word problem.)5.OA.A.2Write and interpret numerical expressions (Writing the total number of games for $n$ teams as the expression $1 + 2 + 3 + \cdots + (n-1)$ and evaluating it without solving an equation.)
⭐ You can solve this AMC 8 problem just by listing how many games show up when you add one team at a time — no algebra needed.
⭐ You can solve this AMC 8 problem just by listing how many games show up when you add one team at a time — no algebra needed.
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