${\displaystyle 17x+13y=405}$ ${\displaystyle 17x+13y=405(mod13)}$ ${\displaystyle 17x=405(mod13)}$ ${\displaystyle 4x=2(mod13)}$ ${\displaystyle x=7(mod13)}$

${\displaystyle x=13z+7(z\in \mathbb{Z})}$

${\displaystyle 17(13z+7)+13y=405}$ ${\displaystyle 17\cdot 13z+119+13y=405}$ (${\displaystyle -119}$) ${\displaystyle 17\cdot 13z+13y=286}$ (${\displaystyle /13}$) ${\displaystyle 17z+y=22}$ ${\displaystyle y=22-17z}$

Lösung: Alle Paare ${\displaystyle (13z+7,22-17z)}$ mit ${\displaystyle z\in \mathbb{Z}}$

${\displaystyle x\ge 0}$, ${\displaystyle y\ge 0}$

${\displaystyle 13z+7\ge 0\iff z\ge -\frac{7}{15}}$ ${\displaystyle 22-17z\ge 0\iff z\le \frac{22}{17}}$

${\displaystyle \Rightarrow -\frac{7}{13}\le z\le \frac{22}{17}}$ mit ${\displaystyle z\in \mathbb{Z}}$ ${\displaystyle \Rightarrow z=0,z=1}$

Lösungen: ${\displaystyle (7,22)}$, ${\displaystyle (20,5)}$