AMC 8 · 2004 · #5
Easy mode Grade 3Problem
Imagine a basketball tournament with 16 teams. In each game, two teams play and the loser is out of the tournament. The winner moves on to play the next game.
The tournament ends when only one team is left. That team is the champion.
How many games are played in all?
Pick an answer.
AMC 8 2004 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: Sixteen teams play a single-elimination tournament: every game eliminates the losing team, and the tournament ends when one champion remains. How many games are played in total?
Givens: $16$ teams enter the tournament; Single elimination: the losing team of each game is out; The tournament ends when one champion is left; Answer choices: (A) $4$, (B) $7$, (C) $8$, (D) $15$, (E) $16$
Unknowns: The total number of games played
Understand
Restated: Sixteen teams play a single-elimination tournament: every game eliminates the losing team, and the tournament ends when one champion remains. How many games are played in total?
Givens: $16$ teams enter the tournament; Single elimination: the losing team of each game is out; The tournament ends when one champion is left; Answer choices: (A) $4$, (B) $7$, (C) $8$, (D) $15$, (E) $16$
Plan
Primary tool: #11 Find an Invariant
Secondary: #9 Solve an Easier Problem
Tool #11 (Find an Invariant) is the cleanest path here. Across the whole tournament, the relation "each game eliminates exactly one team" never changes — that one-loser-per-game rule is the invariant. Since every team except the champion must be eliminated, the number of games equals the number of eliminated teams: $16 - 1 = 15$. Tool #9 (Solve an Easier Problem) lets us first try a small bracket (like $4$ teams) to see why the rule "games $=$ teams $-$ 1" must always hold, before applying it to $16$ teams.
Execute — Answer: D
3.OA.A.3 Step 1 - Warm up with a smaller bracket.
- Try $4$ teams in a single-elimination tournament.
- Round 1 has $2$ games (4 teams paired), leaving $2$ winners.
- Round 2 has $1$ game (the final).
- Total games: $2 + 1 = 3$, and $3 = 4 - 1$.
💡 Working a tiny case first shows the "teams $-$ 1" pattern before we trust it for $16$ teams.
3.OA.A.3 Step 2 - State the invariant.
- In every game, exactly one team wins and exactly one team is eliminated.
- That "one loser per game" rule never changes, no matter how many teams remain.
💡 A one-to-one match between games and losers is the unchanging fact that powers the whole solution.
3.OA.D.8 Step 3 - Count the eliminated teams.
- There are $16$ teams to start.
- The tournament ends with exactly one champion, who is the only team never eliminated.
- So the number of eliminated teams is $16 - 1 = 15$.
💡 Subtracting the one champion from the $16$ entrants gives the number of teams that lost a game.
3.OA.D.8 Step 4 - Apply the invariant.
- Since each game eliminates exactly one team, the number of games equals the number of eliminated teams.
💡 The invariant turns a tournament-counting question into a single subtraction.
3.OA.A.3 Warm up with a smaller bracket. Try $4$ teams in a single-elimination tournament 3.OA.A.3 State the invariant. In every game, exactly one team wins and exactly one team i 3.OA.D.8 Count the eliminated teams. There are $16$ teams to start. The tournament ends w 3.OA.D.8 Apply the invariant. Since each game eliminates exactly one team, the number of Review
Reasonableness: Double-check by adding up the rounds: with $16$ teams, Round 1 has $16 \div 2 = 8$ games, Round 2 has $8 \div 2 = 4$ games, Round 3 (semifinals) has $4 \div 2 = 2$ games, Round 4 (final) has $1$ game. Total: $8 + 4 + 2 + 1 = 15$ games. That matches choice (D). The smaller choices (A) $4$, (B) $7$, (C) $8$ count only some rounds (for example, $8$ is just Round 1), so they undershoot. Choice (E) $16$ would mean even the champion lost a game, which contradicts "the champion never loses."
Alternative: Tool #5 (Look for a Pattern): list games per round as the bracket halves. Round 1: $8$. Round 2: $4$. Round 3: $2$. Round 4: $1$. The pattern is a halving sequence, and its sum is $8 + 4 + 2 + 1 = 15$. Same answer (D), reached by summing the rounds instead of counting eliminations.
CCSS standards used (min grade 3)
3.OA.A.3Use multiplication and division within $100$ to solve word problems (Pairing teams into games (dividing by $2$ each round in the warm-up case) to see why the "games $=$ teams $-$ 1" rule appears.)3.OA.D.8Solve two-step word problems using the four operations (Subtracting the one champion from $16$ teams to get $15$ eliminated teams, then identifying that count as the total number of games.)
⭐ Every game in a single-elimination tournament knocks out exactly one team. So the number of games equals the number of teams that have to be eliminated — which is everyone except the champion. For $16$ teams, that's $16 - 1 = 15$ games, no bracket drawing required.
⭐ Every game in a single-elimination tournament knocks out exactly one team. So the number of games equals the number of teams that have to be eliminated — which is everyone except the champion. For $16$ teams, that's $16 - 1 = 15$ games, no bracket drawing required.
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