Toolkit 115

Solving Linear Congruences

9x5(mod7)x?(mod7)9x\equiv5\pmod7\Longrightarrow x\equiv ?\pmod7

Solution 1

9x5(mod7)2x5(mod7)9x\equiv5\pmod7\Longrightarrow2x\equiv5\pmod7
8x20(mod7)x6(mod7)\Longrightarrow8x\equiv20\pmod7\Longrightarrow x\equiv6\pmod7

Solution 2

9x5(mod7)2x2(mod7)9x\equiv5\pmod7\Longrightarrow2x\equiv-2\pmod7
x1(mod7gcd(7,2))x6(mod7)\Longrightarrow x\equiv-1\pmod{\frac{7}{\gcd(7,2)}}\Longrightarrow x\equiv6\pmod7

There are infinitely many ways, but all of them try to achieve the common goal of reducing the coefficient of x.

11x3(mod17)x?(mod17)11x\equiv3\pmod{17}\Longrightarrow x\equiv ?\pmod{17}

Solution 1

×222x6(mod17)5x6(mod17)\times2\quad 22x\equiv6\pmod{17}\Longrightarrow5x\equiv6\pmod{17}
×315x18(mod17)2x18(mod17)\times3\quad15x\equiv18\pmod{17}\Longrightarrow-2x\equiv18\pmod{17}
÷(2)x98(mod17)\div(-2)\quad x\equiv-9\equiv8\pmod{17}

Solution 2

11x6x3(mod17)11x\equiv-6x\equiv3\pmod{17}
÷(3)2x116(mod17)\div(-3)\quad2x\equiv-1\equiv16\pmod{17}
x8(mod17)\Longrightarrow x\equiv8\pmod{17}