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

$\displaystyle 26x+14y=(26,14) \Rightarrow $ min x =?

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- Dec 19th 2008, 01:56 PMSeaDivisibility (gcd) 10
$\displaystyle x\in \mathbf{Z^{+}}$ and $\displaystyle y\in\mathbf{Z}$

$\displaystyle 26x+14y=(26,14) \Rightarrow $ min x =? - Dec 19th 2008, 04:44 PMo_O
$\displaystyle 26x + 14y = 2 \ \Leftrightarrow \ 13x + 7y = 1$

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

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

Since we want positive x's, we want:

$\displaystyle \begin{aligned} x = -1 + \frac{14}{(26,14)}t & > 0 \\ -1 + 7t & > 0 \\ t & > \frac{1}{7} \approx 0.14286\end{aligned}$

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