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Math Help - Number of solutions

  1. #1
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    Number of solutions

    Number of integral solutions of x+y+z=0 with x>=5 , Y>=-5 and z >=-5 ?

    How to solve this by combination ?
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  2. #2
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    I assume that you mean integer solutions.

    Off the top of my head I don't see how to do this with a single combination because different choices for x lead to different possibilities for y and z, and order seems to matter.

    For example, if x = 5, then there are 6 possibilities for the pair (y,z): (-5, 0), (-4, -1), ..., (0,-5)
    Similarly, if x = 6, then there are 5 possibilities for the pair (y,z).
    Continuing in this fashion, we see that there are 6+5+4+3+2+1=21 possibilities for the triple (x,y,z).
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  3. #3
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    Also, it's the same number as in y + z >= 5 where y <= 5 and z <= 5.
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  4. #4
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    But i have answer as 136 !.

    I don't think it is possible to have these many combination in the given range.

    To have the equation, we need the sum of y+z to be negative.
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  5. #5
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    Quote Originally Posted by kumaran5555 View Post
    But i have answer as 136 !.
    Between this thread and that, I am starting to think that your textbook (or whichever source of answers you have) is faulty.
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  6. #6
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    I am really sorry to have that kind of material.

    Apologies.
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  7. #7
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    Maybe this will work better -

    Find coefficient of  x^0   in (x^5 + x^6 + .....)(x^{-5} + x^{-4} + .....)(x^{-5} + x^{-4} + .....)
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  8. #8
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    Quote Originally Posted by kumaran5555 View Post
    But i have answer as 136 !.

    I don't think it is possible to have these many combination in the given range.

    To have the equation, we need the sum of y+z to be negative.
    Have you copied the question correctly? If the problem is to find the number of integral solutions of x+y+z=0 with x>=5 , y>=–5 and z >=–5, then the answer is indeed 136, as you can check by using aman_cc's useful hint.
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  9. #9
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    Thanks.

    The problem statement says integral solutions only.

    But can anyone tell me till how long i have to expand the series.

    As per the question, i think there should be of some range.
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  10. #10
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    This link expands the sum for x\ge 5 which does not give 136.

    This link expands the sum for x\ge -5 which does give 136.
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  11. #11
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    Hello, kumaran5555!

    \text{Number of integral solutions of: }\: x+y+z\:=\:0
    \text{with }x \ge 5,\;y \ge -5,\;z \ge -5

    After approaching it five ways and getting five different answers,
    . . I made a brute-force List.


    The maximum value for \,x is x = 10.

    . . Then: . (x,y,z)\:=\:(10,\text{-}5,\text{-}5)


    Then we have:
    . . \begin{array}{c} (10,\text{-}5,\text{-}5) \\<br />
(9,\text{-}5,\text{-}4)\quad(9,\text{-}4,\text{-}5) \\<br />
(8,\text{-}5,\text{-}3)\quad(8,\text{-}4,\text{-}4)\quad(8,\text{-}3,\text{-}5) \\ <br />
(7,\text{-}5,\text{-}2)\quad(7,\text{-}4,\text{-}3)\quad(7,\text{-}3,\text{-}4)\quad(7,\text{-}2,\text{-}5) \\<br />
(6,\text{-}5,\text{-}1)\quad(6,\text{-}4,\text{-}2)\quad(6,\text{-}3,\text{-}3)\quad(6,\text{-}2,\text{-}4)\quad(6,\text{-}1,\text{-}5) \\<br />
(5,\text{-}5,0)\quad(5,\text{-}4,\text{-}1)\quad(5,\text{-}3,\text{-}2)\quad(5,\text{-}2,\text{-}3)\quad(5,\text{-}1,\text{-}4)\quad(5,0,\text{-}5) <br />
\end{array}


    I found 21 solutions.

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  12. #12
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    Thanks for your answer.
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