+204 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?
+133 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).
+2511 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
--. .-
Give yourself a point 
