In a single-elimination tournament, each game is between two players. Only the winner of each game advances to the next round. In a particular such tournament there are 256 players. How many individual games must be played to determine the champion?
The "Think About It" name usually hints at an elusive trick to the question. There doesn't appear to be a trick anywhere; here's my best attempt:
If we arrange the players into two-player groups, we'll cut the players in half. 128 games have been played.
Here is the player to game progression:
(256,0)
(128,128)
(64,128 + 64)
(32, 128 + 64 + 32)
(16, 128 + 64 + 32 + 16)
(8, 128 + 64 + 32 + 16 + 8)
(4, 128 + 64 + 32 + 16 + 8 + 4)
(2, 128 + 64 + 32 + 16 + 8 + 4 + 2)
(1, 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1)
Fun fact: sum of powers of 2 starting from 2^0 up to 2^n = 2^n-1
Proof: our answer is 255.
Oh man... now I get the trick (think about it).
Fun fact: sum of powers of 2 starting from 2^0 up to 2^n = 2^n-1
It’s probably not fun if it’s not a fact.... But it’s definitely NOT a fact.
The formula for the sum of powers of (2) is \(2^{(n+1)}-1 \text { where (n) is an integer}\)
... now I get the trick (think about it).
OK, I’ve thought about it...
Your formula (2^n – 1) gives the correct number of games played when (n=8), but how does the problem solver find the (8)? ...And after finding it, how does it help in solving the problem?
This is a binary question. There are 256 Players. Each game eliminates one player. How many players must be eliminated until only one remains? Answer: 255. Because 256 – 255 = 1
For this kind of question, the answer is always one less than the number of players.
After thinking about it, I’m nominating your solution for a Fields MeTal. Category: Rube Goldburg solution methods (Overly complex solutions for simple questions).
You’re not likely to win. This is the best one https://web2.0calc.com/questions/help-algebra_116 so far....
GA
--. .-