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Path: utzoo!watmath!watnot!water!emneufeld
From: emneufeld@water.UUCP (Eric Neufeld)
Newsgroups: net.puzzle
Subject: Re: 1/3 of the Lottery, real sad spoiler
Message-ID: <93@water.UUCP>
Date: Fri, 29-Nov-85 15:25:59 EST
Article-I.D.: water.93
Posted: Fri Nov 29 15:25:59 1985
Date-Received: Sat, 30-Nov-85 00:38:41 EST
Distribution: net
Organization: U of Waterloo, Ontario
Lines: 19

Another goof on my part!

Two ways to derive the solution 42 as the number of tickets:
	1: Use block design theory
	2. Compute C(36,2)/C(6,2)

But in order to take advantage of this minimum number, somehow, you must
be able to figure out what to write on the tickets you buy.

That is, there must be someway of writing the (36 choose 2 ) pairs on
42 tickets which each allow (6 choose 2) pairs, and making sure
each pair appears on exactly one ticket.

This turns out to be impossible, see Corollary 17.2, Section VII of
Street and Wallis, "Combinatorial Theory: an introduction", or in any
introudction to block design theory, what is known as the Bruck-Ryser-Chowla
theorem.

So you have to add a little redundancy to your tickets, that is upcoming.