Two brothers inherited a certain number of sheep. They sold them and got the same amount of dollars for each sheep as there were sheep in the herd/flock. They collected the money only in 10 dollar notes plus a certain extra amount below 10 which was given to them as coins. Now they split the money among themselves by alternately taking a 10 dollar not from the pile until all were gone.
In the end the younger brother complained: "This wasn't fair, you took the first not and also the last note. So you got 10 dollars more than me."
So the elder brother gave him all of the coins.
The brother complained again, that those were less than 10 dollars, so he should still get more money.
"That's true", said the elder brother, and he signed a cheque for the remaining sum.
What was the amount on the cheque?
The problem has more than one solution as far as I can see:
the sum is a square, so any square number is a possible solution for the amount of money. With the condition that the elder brother took one more 10$ Bill we can conclude that any square number can be excluded which statisfies the condition int(number/10) even.
That leaves the solutions: 16 (one bill for the elder,none for the younger, check over 3$) I shorten this to (1,0,3)
36 (2,1,3)
196 (10,9,2)
256 (13,12,2)
576 (29,28,2)
676 (34,33,2)
1156 (58,57,2)
since there is an infinite number of squares fulfilling the condition the number of solutions is infinite.
However there is probably a value of a sheep in a certain range, picking the right square. Or there is a minimum amount of money a check can be written out on....
Originally posted by crazyblueThere had to be an odd number of notes. Obviously the loose change was less than $10. To be fair, the first brother would have to add together the final $10 with the loose change and divide that value equally. This would give the second brother less than $10 but more than $5.
Two brothers inherited a certain number of sheep. They sold them and got the same amount of dollars for each sheep as there were sheep in the herd/flock. They collected the money only in 10 dollar notes plus a certain extra amount below 10 which was given to them as coins. Now they split the money among themselves by alternately taking a 10 dollar not from ...[text shortened]... brother, and he signed a cheque for the remaining sum.
What was the amount on the cheque?
There are two unknowns: the loose change and the cheque. Their total is also unknown - it is not $10 or $5 but some value in between. We might eliminate one unknown by subtraction (Total - Change = Cheque for instance) but is that helping much if we have no values?
Another unknown is the number of sheep, Y. The price received in dollars was Y squared. Ysquared / 10 yields an odd number plus loose change. As that is unknown I can't make a neat expression linking these things together.
Look forward to the solution - I don't have one.
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Represent the number of sheep as 10x + Y
Y is the units in the number of sheep.
Num sheep squared must have an odd number of 10s.
(10x + Y)^2 = 100x^2 + 20xy + y^2
clearly 100x^2 and 20xy have an even number of 10s, so y^2 must have an odd number of 10s.
Remembering y was units, and checking the 10 possible values it is clear that y must be 4 or 6 because everything else has an even number of 10s when squared.
as 4^2 = 16 and 6^2 = 36, we now now that (10x + Y)^2 ends in a 6.
i.e. the number of dollars got for the sheep must end in a 6
So the younger brother gets the odd 6 dollars and is still 4 dollars short. Assuming the cheque evens things up, it must be for 2 dollars.