от Добромир Глухаров » 27 Ное 2013, 17:06
[tex]7x\equiv 1(mod\ 5)\ (1)[/tex]
[tex]7\equiv 7-5\equiv 2(mod\ 5)[/tex]
[tex]2x\equiv 1(mod\ 5)\ |.3[/tex]
[tex]6x\equiv 3(mod\ 5)\ (2)[/tex]
[tex](1)-(2):\ 7x-6x\equiv 1-3(mod\ 5)\Rightarrow x\equiv -2\equiv 5-2\equiv 3(mod\ 5)[/tex]
Окончателно: [tex]x\equiv 3(mod\ 5)[/tex] или [tex]x=5k+3,k\in\mathbb{Z}[/tex]