CareerCup

For n matches the number of tournaments needs to play is n-1

Ex:
teams-tournaments
2-1
3-2
4-3
5-4
6-5
7-6

this questions seems to bear a lot of ambiguity.. plz clear the procedure of promoting the odd i.e 9th team in upper levels.if the promotion is to the level right above it then the answer is 11.
gullu do reply to this post ..

8 matches .. as from 9 teams only 1 will win , rest 8 teams must lose , in knockout tournament 1 team goes out in 1 match .

When the number of teams is not a square of 2, then we need to have byes for some teams. (this is how tournament brackets are constructed).
A binary tree needs to be built and then the number of non-leaf nodes needs to calculated - that will be the number of matches.
Find x where x is the nearest power of 2 less than n.
n-x teams are extra, choose another (n-x) teams from the pool. Remaining teams = n - 2(n-x) = 2x - n. So, 2(n-x) teams will play in the first round while the remaining get a bye. After round 1, we will get 2*(n-x)/2 + (2x-n) = x number of teams. Since x is a perfect square of 2, we can build the binary tree from up there.
The goal is to get to a square of 2 number of teams after round 1.

Number of matches: (n-x) in the first round + (ln(x) + 1)
(Number of non-leaf nodes in a full binary tree is ln(x) + 1 where x is the number of leaves)

Given this, if n = 9, x = 8. So, number of matches = (9-8) + (ln 8 + 1) = 1 + (3 + 1) = 5.

Since we are interested in the number of losers....the answer should be 8.