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Maths puzzle

  1. May 30, 2004 #1

    jcsd

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    I've nicked this puzzle (and slightly modified it) from another forum:

    If the sum of a set of numbers is 100, what is their highest possible product if:

    a) they are non-negative reals

    b) they are non-negative integers
     
  2. jcsd
  3. May 30, 2004 #2

    AKG

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    Would it be from this thread?
     
  4. May 30, 2004 #3

    jcsd

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    No, it's not from that forum. The original puzzle only had part b) in it, part a) is my own addition, but I see that someone's already posted a very simlair puzzle.
     
  5. May 30, 2004 #4

    mathman

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    Since the maximum is always achieved when the numbers are all equal, the problem reduces to finding the maximum of x100/x, and then calculating the products for the integers bracketing 100/x. The answer is x=e. I'll let you work out the final answer.
     
  6. May 31, 2004 #5

    jcsd

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    I've solved it already, it was meant to be a quiz.

    The answer to part a) as the fact that the maximum value of x^1/x is e^1/e, is quite well known, though it's helpful for part b) it's not necessrily needed to be known.
     
    Last edited: May 31, 2004
  7. May 31, 2004 #6

    uart

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    Yes but e^(100/e) is not a valid solution in this case as the numbers must add to 100, and no integer multiple if e is equal to 100. Note that the number of numbers is an integer regardless of whether or not the actual numbers themselves are constrained to be integers.

    Anyway I think the solution to "part a" is (100/37)^37 and the solution to "part b" is 4 * 3^32, though I haven't totally confirmed these values.
     
    Last edited: May 31, 2004
  8. May 31, 2004 #7

    jcsd

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    Of course yes.

    well your defintely right on part b), and you certainly look right for part a)



    the question I really wnated to ask for part a) I suppose was then which value of xy = 100 gives the highest vlaue for x^y.
     
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