Problem 6 · AMC 8 Stretch
Core
Counting & Probability
Logic & Word Problems
count-the-complementconsider-extreme-cases
A tennis tournament has 61 players. In any round with an odd number of players, one player gets a 'bye' (they skip that round and advance without playing). Every match is played until someone wins; the loser is knocked out. How many matches are played in all before one champion is left undefeated?
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Answer: 60 matches
Show hints
Hint 1 of 4
First make sure you understand a 'bye': it just means a player moves on to the next round without playing, because there was an odd number of players.
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Hint 2 of 4
You COULD draw the whole bracket round by round and add up the matches. That works, but it's a lot of bookkeeping.
Still stuck? Show hint 3 →
Hint 3 of 4
Try a smarter idea: instead of counting winners, count LOSERS. Every match knocks out exactly one player.
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Approach: Count losers, not winners
- Bracket way (the long way): Round 1 has 61 players (one bye), 30 matches, 31 left; Round 2: 15 matches, 16 left; Round 3: 8 matches, 8 left; Round 4: 4 matches; Round 5: 2 matches; Round 6: 1 match. Total \(30+15+8+4+2+1=60\).
- The clever way: the champion is the only player who never loses, and everyone loses exactly once, by being knocked out in one match.
- A bye is not a match and knocks no one out. Each match knocks out exactly one player, and we must knock out everyone except the champion.
- So the number of matches is \(61-1=60\). Counting losers skips all the messy bye bookkeeping.
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