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Math Help - Divisibility (gcd) 10

  1. #1
    Sea
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    Divisibility (gcd) 10

    x\in \mathbf{Z^{+}} and y\in\mathbf{Z}

    26x+14y=(26,14) \Rightarrow min x =?
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    26x + 14y = 2 \ \Leftrightarrow \ 13x + 7y = 1

    By inspection, we can see that a solution is given by: x = -1, \ y = 2

    Thus, all solutions are given by: x = -1 + \frac{14}{(26,14)}t, \ \ \ y = 2 - \frac{26}{(26,14)}t where t \in \mathbb{Z}

    Since we want positive x's, we want:
    \begin{aligned} x  = -1 + \frac{14}{(26,14)}t & > 0 \\ -1 + 7t & > 0 \\ t & > \frac{1}{7} \approx 0.14286\end{aligned}

    Thus, for all t > \tfrac{1}{7}, we have positive x's as our solution.
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