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  1. #1
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    Find the number

    Find the total number of positive integer solutions of:

    $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$
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  2. #2
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    Quote Originally Posted by perash View Post
    Find the total number of positive integer solutions of:

    $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$
    $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{2m}{mn} + \frac{3n}{mn} = \frac{2m + 3n}{mn}$

    So you have $\displaystyle \frac{2m + 3n}{mn} = \frac{1}{2006}$.


    Therefore $\displaystyle 2m + 3n = 1$ and $\displaystyle mn = 2006$.


    Try to solve these equations simultaneously.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{2m}{mn} + \frac{3n}{mn} = \frac{2m + 3n}{mn}$

    So you have $\displaystyle \frac{2m + 3n}{mn} = \frac{1}{2006}$.


    Therefore $\displaystyle 2m + 3n = 1$ and $\displaystyle mn = 2006$.


    Try to solve these equations simultaneously.
    HEllo : remark

    is not :

    conclusion :

    THANKS
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  4. #4
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    $\displaystyle \frac ab=\frac cd \Leftrightarrow \exists k \in \mathbb{Q} ~:~ \begin{cases} a=ck \\ b=dk\end{cases}.$

    So here, we have $\displaystyle \exists k \in\mathbb{Q} ~:~ \begin{cases} 2m+3n=k \\ mn=2006k\end{cases}.$

    And since m and n are positive integers, we can choose $\displaystyle k\in\mathbb{N}$

    We get $\displaystyle 2m+3n-\frac{mn}{2006}=0$

    $\displaystyle \Rightarrow m=\frac{3n}{2-\frac{n}{2006}}=\frac{3\cdot 2006 \cdot n}{4012-n}$

    For m to be an integer, we need 4012-n to divide the numerator.

    I conjecture that the number of possibilities is the number of divisors of $\displaystyle 3\cdot 4012$ (more or less, because we have to remember that n<4012)
    But I really don't know yet how to prove it

    $\displaystyle 2m + 3n = 1$
    I bet you can't find 2 positive integers such that it works
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  5. #5
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    A little mistake

    Quote Originally Posted by Moo View Post
    We get $\displaystyle 2m+3n-\frac{mn}{2006}=0$

    $\displaystyle \Rightarrow m=\frac{-3n}{2-\frac{n}{2006}}=\frac{3\cdot 2006 \cdot n}{n-4012}$
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  6. #6
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    $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$
    To begin with, since $\displaystyle m$ and $\displaystyle n$ are positive integers, neither fraction on the LHS can be greater than $\displaystyle \frac{1}{2006}$, (because then, the other would have to be negative).
    So, if $\displaystyle \frac{2}{n} < \frac{1}{2006}$ then $\displaystyle n>4012$ and similarly $\displaystyle m>6018$.
    Multiplying throughout by $\displaystyle mn$, $\displaystyle 2m+3n=\frac{mn}{2006}$, so $\displaystyle mn$ must be an integer multiple of 2006, (since the LHS is an integer).
    Let $\displaystyle mn = 2006k$ where $\displaystyle k$ is an integer, then $\displaystyle n=\frac{2006k}{m}$.
    Substituting, $\displaystyle 2m + \frac{3.2006k}{m} = k$, so $\displaystyle 2m^2 - km + 6018k = 0$.
    Solving this for $\displaystyle m,\\\\\ m = \frac{k \pm \sqrt{k^2 - 48144k}}{4}$. What we need now is that $\displaystyle \sqrt{k^2 - 48144k}$ should be an integer and in such a way that the resulting values of $\displaystyle m$ and $\displaystyle n$ are both integers.
    One obvious possibility is $\displaystyle k = 0$, leading to $\displaystyle m = 12036$ and $\displaystyle n = 8024$, and this produces the solution $\displaystyle \frac{2}{8024}+\frac{3}{12036}=\frac{1}{2006}$.
    Sadly, at this point I grind to a halt , I don't know of a systematic way of finding further values of $\displaystyle k$ such that $\displaystyle k^2-48144$ is a perfect square. (Help anyone ?)
    However, notice that the calculated solution cancels down to $\displaystyle \frac{1}{4012}+\frac{1}{4012}=\frac{1}{2006}$, which is kind of obvious and does suggest a second solution.
    $\displaystyle \frac{1}{2006} = \frac{5}{10030} = \frac{2}{10030} + \frac{3}{10030}$.
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  7. #7
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    Ooops !! .
    I should have said that one obvious value of $\displaystyle k$ was $\displaystyle k = 48144$ leading to $\displaystyle m = 12036$.
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  8. #8
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    Quote Originally Posted by BobP View Post
    Sadly, at this point I grind to a halt , I don't know of a systematic way of finding further values of $\displaystyle k$ such that $\displaystyle k^2-48144k$ is a perfect square. (Help anyone ?)
    Let $\displaystyle k^2-48144k$ be $\displaystyle r^2$.

    Then $\displaystyle k^2-48144k+24072^2=r^2+24072^2$

    now, if $\displaystyle 24072=2uv$
    or $\displaystyle 12036=uv$

    then, $\displaystyle (k-24072)^2=(u^2+v^2)^2$

    hence, it comes down down to factorise 12036 into two factors(u,v), and for each pair we will get two values of k(one positive and one negative).
    Since we are interested only in positive values,
    $\displaystyle k=(u-v)^2+48144$

    Prime factorisation of 12036 will be helpful for finding factors:

    $\displaystyle 12036=2^2.3.17.59$

    Example:
    u=102
    v=118
    gives k=48400,r=3520
    Last edited by malaygoel; Jul 2nd 2009 at 05:42 AM. Reason: making it better
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  9. #9
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    Thanks for that, a useful idea that I don't recall seeing before.
    It does produce further solutions, m=12980 n=7480, m=2018036 n=4024, m=8054090 n =4015. (I don't know whether I got all of them !)
    I've still not got my head round it totally though, since this method doesn't seem to pick up my original solution of m=12036 n=8024, and I'm not sure whether there might not be others.
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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by BobP View Post
    I've still not got my head round it totally though, since this method doesn't seem to pick up my original solution of m=12036 n=8024, and I'm not sure whether there might not be others.
    Your solutions were trivial...i.e. the discriminant is zero for your solutions. I am sure that solutions from my method(it gives 12 solutions) and your solutions make the complete set...or at least I can hope so.
    Last edited by malaygoel; Jul 4th 2009 at 09:17 PM.
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  11. #11
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    Hello, perash!

    I had a good start on it, but I haven't finished it yet . . .


    Find the number of positive integer solutions of: . $\displaystyle \frac{2}{n} + \frac{3}{m} = \frac{1}{2006}$

    Multiply by $\displaystyle 2006mn\!:\quad 4012m + 6018 n \:=\:mn$

    Solve for $\displaystyle n\!:\quad n \;=\;\frac{4012m}{m - 6018}$


    Since $\displaystyle n$ is a positive integer, then there are two cases:

    . . [1] $\displaystyle m-6018$ is a factor of 4012

    . . [2] $\displaystyle m-6018$ is a factor of $\displaystyle m$



    Case [1]: $\displaystyle 4012 \:=\:2^2\!\cdot\!17\!\cdot\!59$ . . . which has 12 divisors.
    . . So we have 12 solutions:

    . . $\displaystyle \begin{array}{|c||c|c|} \hline
    \text{divisor} & m & n \\ \hline \hline
    1 & 6019 & 24,\!148,\!228 \\
    2 & 6020 & 12,\!088,\!156 \\
    4 & 6022 & 6,\!040,\!066 \\
    17 & 6035 & 1,\!424,\!260 \\
    34 & 6052 & 714,\!136 \\
    59 & 6077 & 413,\!236 \\ \hline\end{array}$ . . . $\displaystyle \begin{array}{|c||c|c|} \hline
    \text{divisor} & m & n \\ \hline \hline
    68 & 6086 & 359,\!074 \\
    118 & 6136 & 208,\!624 \\
    236 & 6254 & 106,\!318 \\
    1003 & 7021 & 28,\!084 \\
    2006 & 8024 & 16,\!048 \\
    4012 & 10,\!030 & 10,\!030 \\ \hline \end{array}$



    Case [2]: I found three more solutions . . . so far.

    . . $\displaystyle \begin{array}{|c|c|} \hline m & n \\ \hline \hline
    6021 & 8,\!052,\!080 \\ 6024 & 4,\!028,\!048 \\ 6069 & 477,\!428 \\ \vdots & \vdots \\\hline \end{array}$


    At this point, I desperately needed a nap . . .

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