Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/5/84; site luke.UUCP
Path: utzoo!linus!philabs!cmcl2!seismo!hao!hplabs!oliveb!bene!luke!sml
From: sml@luke.UUCP (Steven List)
Newsgroups: net.puzzle
Subject: Re: Re: How to win a bottle of Champagne
Message-ID: <166@luke.UUCP>
Date: Sat, 1-Jun-85 12:33:33 EDT
Article-I.D.: luke.166
Posted: Sat Jun  1 12:33:33 1985
Date-Received: Tue, 4-Jun-85 00:24:22 EDT
References: <179@ubvax.UUCP> <262@zaphod.UUCP>
Organization: Benetics Corp, Mt.View, CA
Lines: 45

> 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?

In reading the above explanation and solution, I agree with the answer.
FOUR.  However, I didn't bother with all the advanced algebra.  If you
step through the first few squares, you will find that 4, 6, and 16 all
produce squares that (1) are odd multiples of 10 and (2) end in 6.
Since the total sale price must be a square (each sheep sold for a price
equal to the total number of sheep), and at least three answers end in 6
(producing a remainder of four), why go on and on and on...

Can I have a glass? (:->