Thread: Math problem
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Old August 6th, 2003, 09:34 PM
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Default Re: Math problem

Oh wait LGM, I think I see what you are saying now. Did your program find solutions for anything between 7 and 15. The way I am looking at this now it should be possible to do for many numbers, not just powers of x.

Approaching the problem from the other end was the key if I am correct. For example, in a tournament of three man games the minimum number of players would be 3, and the number of games would be 1. The forumula should be:

x, game size
y, number of tourney games
n, total tourney field

EDIT: IGNORE EVERYTHING IN THIS POST AFTER THIS POINT. I haven't figured out what I did wrong, but this isn't right at all.

(y(x-1))+1=n

plugging in the ones we know already
(1(3-1))+1=3
(3(3-1))+1=7
(7(3-1))+1=15
(13(3-1))+1=27

We should be able to work out an n for any positive real integer X and Y fairly easy.
If x=3
y=1, n=3
y=2, n=5
y=3, n=7
y=4, n=9
.
.
y=15, n=31

So for x=3 n can be any odd number except 1.

The hard part would be actually coming up with the y number of combinations so that each player doesn't play anyone else more then once.

[ August 06, 2003, 20:44: Message edited by: geoschmo ]
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