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Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!ihnp4!alberta!sask!zaphod!dkatz
From: dkatz@zaphod.UUCP (Dave Katz)
Newsgroups: net.puzzle
Subject: Re: How to win a bottle of Champagne
Message-ID: <262@zaphod.UUCP>
Date: Tue, 28-May-85 02:32:17 EDT
Article-I.D.: zaphod.262
Posted: Tue May 28 02:32:17 1985
Date-Received: Thu, 30-May-85 02:16:19 EDT
References: <179@ubvax.UUCP>
Reply-To: dkatz@zaphod.UUCP (Dave Katz)
Organization: Develcon Electronics, Saskatoon, SK
Lines: 35
Keywords: booze
Summary: 

In article <179@ubvax.UUCP> frederic@ubvax.UUCP (Frederic Bach) writes:

>   There once were two brothers herding a flock of
>sheep. One day, they decided to sell them all in
>the market. Each beast was sold for as many dollars
>as sheep there originally were in the herd.
>The total amount of money was in $10 bills, plus
>less than $10 in $1 bills. The elder brother proceded
>to share it. He took a $10 bill, then gave one to
>his brother, then took one again, and so on till
>he turned out taking the last $10 bill ; he then
>gave all of the $1 bills to the youngster, who burst
>out : "Hey, but you've got #@% dollars more than I have ".
>
>How about now working out what he really said ?

FOUR

As described, there must have been an odd number of $10's.  Since the
total value is the square of the number of sheep, look for values of
2 <= y <= 11 for which int((y^2) / 10) is odd,
(ignoring the trivial case of 1 since the problem
implies there were at least $30).  The only cases are 4^4=16, 6^6=36.
All others are even.

Further, (((n*10)+y)^2) = (n*10)^2 + 2ny + (y^2).  Since
int(((n*10)^2 + 2ny) /10 ) is always even, ((n*10)+y)^2 is odd in the
second digit iff y^2 is odd in the second digit.  This applies for all
n | 0 <= n (there being a positive number of sheep).

Hence the number of sheep must have been:
	(n*10+y | 0 <= n,  y = 4 or y = 6)
In either case, the value ends in 6 and the difference is four -

Now, where is my bottle of champagne?