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From: weissler@randvax.UUCP (Robert Weissler)
Newsgroups: net.puzzle
Subject: Re: How to win a bottle of Champagne
Message-ID: <2511@randvax.UUCP>
Date: Fri, 31-May-85 13:20:32 EDT
Article-I.D.: randvax.2511
Posted: Fri May 31 13:20:32 1985
Date-Received: Tue, 4-Jun-85 00:41:42 EDT
References: <179@ubvax.UUCP>
Organization: Rand Corp., Santa Monica
Lines: 50

I thought I'd try my hand at the following puzzle.

>    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 ?

He really said: "Hey, but you've got 6 dollars more than I have."

Here's my approach to the solution:

	There are n sheep, where each sheep sells for n dollars.
	According to the 3rd sentence above, that means the total
	amount of money the brothers made was n**2 dollars.  This
	money was in m $10 bills and less than 10 (say, k) $1 bills,
	so:

		10m + k = n**2	or	n**2 - 10m - k = 0.

	Since the elder brother took the first and last $10 bill,
	m must be odd.  Being the math klutz that I am, I had to
	solve it for k by trying several values for n and m.  Since
	we are talking dollars, n, m and k must be integers.  So:

		Let m = 1, 3, 5, 7...(the odd integers)

		Then 10m = 10, 30, 50, 70...

		So n**2 must be greater than 10m by less than 10 (i.e., k).

		Thus n**2 = 16, 36, -, 64	(25 and 49 don't work)

		And finally k = n**2 - 10m = 6 in each case above.

I'm sure the more mathematically adept can find a really elegant
solution, but I'll let somebody else waste their time on that. (:-)

-Robert

ARPA: Weissler@Rand-unix
UUCP: randvax!weissler