1. Joined
    29 Apr '05
    Moves
    827
    07 Jul '11 11:45
    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?
  2. SubscriberPonderableonline
    chemist
    Linkenheim
    Joined
    22 Apr '05
    Moves
    653643
    07 Jul '11 12:33
    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....
  3. Joined
    29 Apr '05
    Moves
    827
    07 Jul '11 12:40
    so this was too easy, your solution looks more than good. just look at the question again, it's already answered 😉
  4. Standard memberfinnegan
    GENS UNA SUMUS
    Joined
    25 Jun '06
    Moves
    64930
    07 Jul '11 13:06
    Originally posted by crazyblue
    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 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.

    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.
  5. SubscriberPonderableonline
    chemist
    Linkenheim
    Joined
    22 Apr '05
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    653643
    07 Jul '11 15:07
    Originally posted by crazyblue
    so this was too easy, your solution looks more than good. just look at the question again, it's already answered 😉
    actually not since we have 2 or 3 as the number on the check (only if we expect a sheep to cost more then 6$ we do have the solution).
  6. Joined
    26 Apr '03
    Moves
    26771
    07 Jul '11 23:205 edits
    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.
  7. SubscriberPonderableonline
    chemist
    Linkenheim
    Joined
    22 Apr '05
    Moves
    653643
    13 Jul '11 15:35
    Actually my problem was that I calculated the difference between 36 and 40 wrong 🙁 All the the squares have 6 as last digit and thus the sum on the check is 2. (and that is all that had been asked)
  8. Joined
    29 Apr '05
    Moves
    827
    15 Jul '11 07:39
    yep, that looks all correct now. well done!
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