Problem 16 · AMC 8 Stretch
Core
Counting & Probability
Logic & Word Problems
seeking-complementsasking-key-questions
A tournament has \(36\) players. One loss knocks you out. How many games must be played to crown a single champion? Then: how would the answer change if it took TWO losses to be knocked out?
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Answer: 35 games (single elimination); 70 or 71 games if two losses are needed
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Hint 1 of 3
Don't count games from the winner's side — that's hard. Ask the complement question: how many LOSERS are there, and how does a loss relate to a game?
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Hint 2 of 3
Each game produces exactly one loser. So the number of games equals the total number of losses handed out.
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Hint 3 of 3
Everyone except the one champion gets eliminated. For one-loss: \(35\) players must be eliminated, so \(35\) losses, so \(35\) games. For two-loss: each eliminated player needs \(2\) losses.
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Approach: Seeking complements — count losses, not wins
- Ask the complement question: count losses, not wins. Every game makes exactly one loser, so the number of games equals the number of losses.
- One loss to eliminate: out of \(36\) players, exactly \(1\) becomes champion and the other \(35\) are each eliminated by \(1\) loss. That is \(35\) losses, so \(35\) games.
- Two losses to eliminate: each of the \(35\) eliminated players must collect \(2\) losses, giving \(35 \times 2 = 70\) losses. The champion might also pick up a loss along the way (one is allowed), so the total is \(70\) games if the champion never lost, or \(71\) if the champion lost exactly once before winning.
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