Cows,sheep and a lamb!
 

:smile:

Two men sold their herd of x cows at x dollars per head. With the proceeds they bought sheep at $10 each and a single lamb costing less than $10.
Each man received the same number of animals but the one receiving the lamb had to be compensated so as to make the division equitable.
How much money did he receive from the other man ?

Mally :coffee:

[SPOILER]Assuming that they had no money left over after the purchase, $2.
"How" is left as an exercise to the reader[/SPOILER] :wink:

The problem is not well posed. The compensation was it paid from the cow money or the second mans pocket?

I'd say

[SPOILER]10 cows=100$=9 sheep+1 lamb+5$[/SPOILER]

Remark that these are the prices fro the 60's.

H.

I do not think you gave enough data, even supposing the price of a cow is an integer : the price of a lamb could be 0; 1; 4; 5; 6 or 9, giving rise to a compensation of 0; 4,5; 3; 2,5; 2 or 0,5.

S485122,

We must accept that the price of a cow is an integer because the number of cows was surely an integer.

I think that you are incorrect with respect to the price of the lamb. Please give an example where the price of the lamb is $1. What is a cow worth in that case?


HHH,

Even in the 60s, a cow was worth more than a sheep.
But ignoring that, I think that we must reject your answer because your solution
[SPOILER] allows any price for the lamb. (1 lamb + cash = $10 = 1 sheep)
Since the solution is not unique, it does not satisfy the puzzle[/SPOILER]

[spoiler]
Each person receives the same number of animals.
So, s is an odd number of sheep

Our equation is:
x^2 = 10s + l

The only case where s is odd and l < 10 is when s=3 and x=6

So, the man gives the receive of the lamb two dollars 1/2*(20-16).
[/spoiler]

[QUOTE=grandpascorpion;111043][spoiler]
Each person receives the same number of animals.
So, s is an odd number of sheep

Our equation is:
x^2 = 10s + l

The only case where s is odd and l < 10 is when s=3 and x=6

So, the man gives the receive of the lamb two dollars 1/2*(20-16).
[/spoiler][/QUOTE]

Isn't [spoiler]s=1, l=6, and x=4[/spoiler] also possible? Although, granted, that still results in the amount being [spoiler]two dollars[/spoiler].

It says they bought sheep. So, I assumed it was more than one sheep that they bought. Same result though :)

[QUOTE=grandpascorpion;111048]It says they bought sheep. So, I assumed it was more than one sheep that they bought. Same result though :)[/QUOTE]

[SPOILER]x=4,6,14,16,24,26,...[/SPOILER] :yawn: :razz:

Doh. My bad.

[QUOTE=Wacky;111038]S485122, We must accept that the price of a cow is an integer because the number of cows was surely an integer. I think that you are incorrect with respect to the price of the lamb. Please give an example where the price of the lamb is $1. What is a cow worth in that case?[/QUOTE]Indeed I goofed, the only possible price for a lamb is [spoiler]6[/spoiler], giving rise to a [spoiler]2[/spoiler] compensation. Forgot the 20 constraint on the squares...