flock of bulls?
herd of geese?
Hi,
A many many years ago, this transaction had taken place between 2 animal traders. They sold one flock of bulls at their disposal. They received sale proceeds for each bull equivalent to number of bulls in that flock.
From that sale proceeds, they purchased goats and kids at the rate of Dollar 10/- each for one goat and one kid. Post equal distribution of goats and kids, one trader got 1 goat more and another got one kid and compensation in terms of Dollars.Now, how much Dollars, he had taken to have equal distribution.
Since the price of each bull was equal to the number of bulls, the total take is a perfect square.
Since one trader got one more goat, the ten's digit is odd.
A glance at the table of squares will tell you how many kids the other trader got and therefore how many dollars he was compensated.
Hi PaddyMac,
You are on right track. Your answer about sales proceeds is correct. it is perfect square having ten's digit odd. Now you have to find out unit's digit in that perfect square.Post that, you will know the compensation he had taken to have equal distribution.
The puzzle here is to figure out what the puzzle is. Let me restate it as I understand it, following Paddy's clarification:
The number of bulls sold is the same as the price received per bull (hence the amount of money they receive is a perfecrt square). With the proceeds they bought goats at \$10 each and kids at \$1 each, and the two ended up with the first trader having one more goat than the second trader, and the second trader having one more kid than the first (in other words, there are an equal odd number of goats and kids). The trader who received the extra kid instead of a goat gets \$9 compensation. Question is: what were the proceeds they received from sales of the bulls?
The problem is that there are multiple answers, none of them having to do with odd tens digits, so it seems my understanding of the problem is incorrect. For example proceeds could be \$64, in which case they buy 5 goats for 5x10=\$50 and 5 kids for 5 x \$1 = \$5, and have \$9 left over as compensation for the trader who gets 3 kids and 2 bulls. Other possibilities are \$196 and \$300. So it seems I still don't understand the question - please clarify as best you can. Thanks.
PS - if you place a backslash in front of the dollar sign then the sign will display correctly - otherwise thr forum software thinks teh dollar sign is a delimiter for LaTeX.
OK, I think I see now - the goat & kid pair together cost \$10. What threw me is that Paddy's post #5 and Vinod's confirmation in post #6 are incorrect - he had siggested that the goat costs \$10 and the kid costs \$1. Since there is an odd number of both animals, and the cost per set is \$10, the tens digit is indeed odd. And the question as posed by Vinod: "how much Dollars, he had taken to have equal distribution" should actually be "how many dollars were left over?" And yes, for a perfect square with odd tens digit the ones digit is always the same.
As noted in post #9 I had figured that out. It wasn't my assumption that the kid cost $1 - that's what Paddy had said, and you had agreed - hence my confusion about this. But it's all clear now. The answer is: there was $6 left over for the trader who got the last kid instead of the last goat.
This is my assumed scenario (arbitrarily setting the herd size).
If they had 16 bulls sold at $16 each, the traders have $256. Buying goats at $10 each, they buy 25. They buy kids with the left over $6.
Splitting the goats, A gets 13 and B gets 12. Trying to make the split as even as possible, B takes the kids.
To make the split even, A pays B $2.
Where did the \$2 come from? Didn't they use up their last \$6 buying kids?
My solution is this: with 16 bulls they get proceeds of \$256, with which they buy 25 goats & 25 kids (at \$10 per pair). A takes 13 goats and 12 kids, B takes 12 goats and 13 kids, and B also takes the remaining \$6.