AMC 8 · 2004 · #5
Grade 3 countingPick an answer.
AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Warm up small: 4 teams need 2 + 1 = 3 games, and 3 = 4 - 1.
Working a tiny case first shows the "teams - 1" pattern before we trust it for 16 teams.
3.OA.A.3Solve An Easier Related ProblemEvery game knocks out exactly one team, so games = eliminated teams.
A one-to-one match between games and losers is the unchanging fact that powers the whole solution.
3.OA.A.3Work BackwardsAll 16 teams but the champion lose, so eliminated teams = 16 - 1 = 15.
Subtracting the one champion from the 16 entrants gives the number of teams that lost a game.
3.OA.D.8Work BackwardsBy the invariant, total games = eliminated teams = 15 → (D).
The invariant turns a tournament-counting question into a single subtraction.
3.OA.D.8Work BackwardsEvery 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.