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Homework Help: Sum equals product problem

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data

    If the product of the numbers R and 11/S is the same as their sum, find the value of S.

    2. Relevant equations


    3. The attempt at a solution

    I am suspecting that the only set of 2 numbers that have the same sum and product is 2 and 2.

    So I guess R is 2, 11/S is also 2.

    That gives S = 5.5

    Please will anyone verify if this is correct?
  2. jcsd
  3. Sep 5, 2013 #2
    Is this the complete question?
  4. Sep 5, 2013 #3
    It probably isn't. Graph by wolfram alpha attached.
    Also integer solutions if thats what the question is asking: (-11,12);(-1,22);(11,10).
    EDIT: P.S. to solve a equation in two variables you always need two equations.

    Attached Files:

    Last edited: Sep 5, 2013
  5. Sep 5, 2013 #4


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    Assuming, as others have suggested, that R and S are integers, what equation in which all terms are integers expresses the given condition? Can you deduce anything by thinking about factors?
  6. Sep 5, 2013 #5
    That was the complete question, extract from Olympiad.
  7. Sep 5, 2013 #6
    That's cool, haven't thought of the solutions you create.

    I didn't understand your graph though.

    Is there an algebra way of solving this though? I don't think graphical calculator is allow in exam.
  8. Sep 5, 2013 #7
    I am not sure what you mean by 'thinking about factor', I don't know the actual value of R and S so I cannot determine their factor. Is there any specific theorem that I should try?

    the equation I made was:

    R + 11/S = R x 11/S

    which got me nowhere.
  9. Sep 5, 2013 #8


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    That's fine, but that equation has many solutions. R=2, S=11/2 works, and so does R=3, S=22/3. Etc, etc. Can you post a reference to the exact Olympiad question?
  10. Sep 6, 2013 #9

    Ray Vickson

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    That gives you RS + 11 = R*11, or 11*(R-1) = R*S. Thus, R*S must be divisible by 11, and that suggests looking at some simple values like R = 11 or S = 11, etc, assuming you want positive integer values of R and S. If all you want are real values, you can just put S = 11*(R-1)/R and let R be anything you want (but not zero).
  11. Sep 6, 2013 #10
    A reference would be nice...Were there options in the question.
    The graph just shows that there are infinite solutions possible. ie. Each point is a solution
    Ignore the zig-zag lines at x axis, its just Mathematica having a fit.
  12. Sep 7, 2013 #11
    Here's the paper.

    Well I also failed to do question 2, please help if possible^^. thanks!

    Attached Files:

  13. Sep 7, 2013 #12
    O yes this is cool!

    I have actually got to the equation you wrote, but didn't realise that it means infinite sets of solution!

  14. Sep 7, 2013 #13
    There weren't options in the question.

    I have attached the original document in previous post.

    Some of the questions are somewhat confusing I think.

    Like question 3, I think the answer can only be an expression, not a value, but it doesn't say. I think the answer is 11/6 log2 (Q) ], but there's no solution come with the document, quite annoying.
  15. Sep 7, 2013 #14
    Well, as you have to find four values: P Q R S
    You will have to solve third question for R first before attempting the fourth one which requires the value of R to get S. To get R you will have to get Q and for Q you will need P.
    So start from the start and solve question 1 for P first plug it into q.2 get Q plug into q.3 get R and use it finally for S.
  16. Sep 7, 2013 #15


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    Very clever! So it is not four problems but a single one:biggrin:. And everything fits. R and S are really positive integers.

  17. Sep 7, 2013 #16
    And cruel, imagine the plight of the guy who does everything correctly except the first one...no marks for steps just answers.
    Also it has a time multiplication factor, never seen that before.

    EDIT: HEY! ehild, you just told OP the answer!
  18. Sep 7, 2013 #17


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    I do not think so... You gave the principle of solution. :approve: It was very clever of you!

    Last edited: Sep 7, 2013
  19. Sep 7, 2013 #18
    OMG i feel so unintelligent, i thought these questions are all separate!!! No wonder I don't understand half of the questions in other events...

    well thanks for pointing it out anyway
  20. Sep 7, 2013 #19


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    Rather nasty of them to chain 4 questions into 1, but thankfully, all of them are fairly elementary. What level is this for, just out of curiosity?
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