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Thread: Find the number of pairs (x,y)

  1. #1
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    Find the number of pairs (x,y)

    Good night

    Let n be an odd integer >=5.


    1) Find the number of pairs (x,y) of positive integers which satisfy the equation x+2y=n.

    2) Find the number of triples (x,y,z) of positive integers which satisfy the equation x+y+2z=n.


    1) the answer is (n-1)/2

    I put
    x+2y=n

    y = (n-x)/2

    the n is odd then n= 2k + 1 and
    x+2y=n then x = 1

    y = (n-1)/2

    ???





    2) the answer is [(n-1)(n-3)]/4

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  2. #2
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    Re: Find the number of pairs (x,y)

    Quote Originally Posted by webster View Post
    Let n be an odd integer >=5.
    1) Find the number of pairs (x,y) of positive integers which satisfy the equation x+2y=n
    2) Find the number of triples (x,y,z) of positive integers which satisfy the equation x+y+2z=n.
    1) If $n$ is even, say $8$, then $x$ must be even and less than $8$. WHY?
    Are there $\left\lfloor {\dfrac{{n - 1}}{2}} \right\rfloor$ (floor function) even integers less the $n$.
    How many of those even integers will actually work?

    If $n$ is odd then $x$ must be odd less than $n$.

    So what is the complete answer to #1 ?.



    Now for #2. If $n$ odd then $x+y$ must be even. Now you work it out.
    Thanks from topsquark
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    Re: Find the number of pairs (x,y)

    1) if n = 8 for example then x = {0,2,4,6}
    if n = 9 then x = {1,3,5,7}

    floor function?
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  4. #4
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    Re: Find the number of pairs (x,y)

    Quote Originally Posted by webster View Post
    1) if n = 8 for example then x = {0,2,4,6}
    if n = 9 then x = {1,3,5,7}
    floor function?
    What I did was to give you was give in the hope that you would explore this for yourself.
    In $x+2y$ that sum is odd only if x is odd. Working with the set of positive integers
    if $x+2y=n$ then if $n$ is odd, $\left(\forall y\le\left\lfloor {\frac{n}{2}} \right\rfloor\right))(\exists x)[x+2y=n]$. Can you prove that?
    So how many integers, $y$, can paired with some $x$ to solve $x+2y=n~?$

    Now is case #2, $x+y+2z=n$ if $n$ is odd then $x+y$ must be odd.
    Thus $x~\&~y$ must have different parity, i.e. one is even and one is odd.
    Example: Suppose that $n=17$. Because $1\le x,~1\le y,~\&~~1\le z$ .
    Here are a few combinations:
    $\begin{array}{*{11}{c}} n&z&x&y \\ {17}&7&1&2 \\ {17}&7&2&1 \\ {17}&6&1&2 \\ {17}&6&4&2 \\ {17}&5&5&2 \\ {17}&5&4&3 \\ {17}&5&3&4 \\ {17}&5&6&1 \\ {17}&4&4&5 \end{array}$

    How many more can you find?
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  5. #5
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    Re: Find the number of pairs (x,y)

    2)

    The problem is equivalent to solving a system of 2 equations

    t+2z=n

    and

    x+y=t

    The first equation has (n-1)/2 solutions namely

    t=n-2z for each z, 1\leq  z\leq  \frac{n-1}{2}

    The second equation has t-1 solutions {(1,t-1),(2,t-2),...,(t-1,1)} for a given t

    Therefore the total number of solutions is

    \sum _{z=1}^{(n-1)/2} (n-2z-1)
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  6. #6
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    Re: Find the number of pairs (x,y)

    Can I use this

    \binom{n-1}{2-1}=\left ( n-1 \right )!\\\\\\\frac{\left ( n-1 \right) }{1.2}




    or

    x + 2y = n
    one solution is (-n, n)
    (x - (-n)) + 2(y-n) = 0

    all the solutions are given by (x,y)= (-n+2t, n-t)

    -n+2t>0
    t>n/2

    n-t>0
    t<n

    n/2<t<n

    ?? and now?
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