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Math Help - Finding ordered pairs

  1. #1
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    Finding ordered pairs

    The number of the ordered pairs of intergers (m,n) for which mn> or = 0

    and m cubed + n cubed +99mn = 33 cubed is equal to ?

    a. 2 b. 3 c. 33 d. 35 e. 99


    Initially my answer was a. 2 ( 0, 33 ) (33, 0 ) but I know there would be more pairs. How do I find the other pairs without randomly plugging in numbers?

    Vicky.
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  2. #2
    Super Member bigwave's Avatar
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    NA
    Last edited by bigwave; January 7th 2010 at 08:10 PM.
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  3. #3
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    Quote Originally Posted by bigwave View Post
    not sure of the background of this question so this is just my observation on this.

    since the equation = 33^3 which is a fixed number and positive
    then m and n will have be both positive. if you plug any number other than 0 you cannot get the 33^3 only m=0 and n=0 will work
    0^3 + n^3 = 33^3 so if m=0\ n=33
    m^3 + 0^3 = 33^3 so if n=0\ m = 33
    so your answer of #2 (0,33)(33,0) appears to be correct

    the answer to this seems to be just looking at the equation and seeing what really is possible given the constraint was a big hint.

    again this is just my observation. there may be something too that I don't see in this, (my disclaimer anyway)

    hope this helps. (some)

    Thanks for your help.
    But I did find one more pair (-33. -33 ). It was just a good guess though.

    I'm still trying to figure out the rest because the answer is 35 pairs.

    Vicky
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  4. #4
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    Hello Vicky
    Quote Originally Posted by Vicky1997 View Post
    The number of the ordered pairs of intergers (m,n) for which mn> or = 0

    and m cubed + n cubed +99mn = 33 cubed is equal to ?

    a. 2 b. 3 c. 33 d. 35 e. 99


    Initially my answer was a. 2 ( 0, 33 ) (33, 0 ) but I know there would be more pairs. How do I find the other pairs without randomly plugging in numbers?

    Vicky.
    This is quite tricky. I have part of the solution for you - I'm not quite sure how to prove the final part. Look at these expressions:
    (m+n)^3=m^3+3m^2n+3mn^2+n^3
    =m^3+n^3+3mn(m+n) (1)
    and
    p^3-q^3=(p-q)(p^2+pq+q^2) (2)
    So
    m^3+n^3+99mn =33^3

    \Rightarrow (m+n)^3 -3mn(m+n)+99mn = 33^3 from (1)

    \Rightarrow (m+n)^3-33^3-3mn(m+n-33)=0
    \Rightarrow (m+n-33)([m+n]^2+33[m+n]+33^2)-3mn(m+n-33)=0 from (2)
    \Rightarrow (m+n-33)([m+n]^2+33[m+n]+33^2-3mn)=0
    \Rightarrow m+n-33=0 or (m+n)^2+33(m+n)+33^2-3mn=0
    The first of these possibilities, m+n=33, gives 34 integer solutions for mn\ge0; namely (0,33), (1,32), (2, 31), ...,(33,0).

    I'm not yet sure how to proceed with the second one. m = n = -33 is a solution:
    (-66)^2+33(-66)+33^2-3(-33)(-33)
    =4.33^2-2.33^2+1.33^2-3.33^2

    =0
    but I'm not sure how to show that this is the only one.

    Anyway, there are your 35 solutions: (0,33), (1,32), (2, 31), ...,(33,0) and (-33,-33).

    Grandad
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