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18. February 2004, 23:48:14
Kevin 
Subject: Membership Competition
Since I have to renew my membership soon anyways, i figured i'd have another membership competition so i could pay for them together. I do not want to simply set up a tournament because i want any pawn to be able to enter regardless of their current tournament status, and do not want to give it simply to the person who is the best at a certain game.

So I have decided to have the following competition:

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Any player at the pawn or knight membership level can enter. To enter, you must send me either a private message (to Kevin) or an email (to Kevin_P86@hotmail.com or kevin_p861@yahoo.com) which contains as many english words of a certain length that you can find. However, each word in your list must be the same number of letters and the winners will be decided based on total number of letters in the list of words. All the words must either be found at www.dictionary.com or the URL must be provided to the online dictionary in which it was found. No proper nouns, slang or abbreviations allowed (including those found at www.dictionary.com). The prizes will be a number of 1 year Brain Rook memberships - the number depends on the number of entrants. The deadline for entries will be Friday, March 19, 2004 (the last day of winter). Only one entry per player will be accepted.

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I will allow Rooks to enter as well provided that they give me the name of a pawn or knight that will receive the membership, should they win.

Good luck everyone!
Kevin

18. February 2004, 23:53:35
Linda J 
Subject: Re: Membership Competition
That sounds like a fun idea. Good luck with it Kevin. :•)

22. February 2004, 01:54:22
Kevin 
Subject: Re: Membership Competition
So far i have only gotten 4 entries - one with three-letter words, two with five-letter words and one with eight-letter words.

There is just under a month left to enter, but the more entrants the more winners, so go ahead and enter!

The details of this competition can be found at This page or message#87012 on this board.

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