Solving a linear congruence

Question

Use Euclids algorithm to find the multiplicative inverse of 11 modulo 59 and hence solve the linear congruence:

$11x \equiv 8 \mod59$

My working so far....

$ {11v + 51w = 1}$

Using Euclid's algorithm:

$ {59 = 5 \times 11 + 4}$

${11 = 2\times 4 + 3}$

${4 = 1 \times 3 + 1}$

$ {3 = 1 \times 3 + 0}$

$ {11v + 51w = 1}$

rearrange the equations to make the remainder the subject:

$ {4 = 59 - 5 \times 11}$

$ {3 = 11 -2 \times 4 }$

$ {1 = 4 - 1 \times 3} $

Using backwards substitution:

${1 = 4 -1 \times (11 -2 \times 4)}$

${1 = 3 \times 4 - 1 \times 11}$

$ {1 = 3 \times (59 - 5 \times 11) -1 \times 11}$

$ {1 = 3 \times 59 - 16 \times 11}$

$ {1 = 59(3) + 11(-16)}$

${59v + 11w = 1} $ where v = 3 and w = -16

I'm not sure where to go from here to solve the linear congruence.

Simpler: $\ 11x\equiv 8!\overset{\large \times,5}\iff! -4x\equiv 40\iff x\equiv -10,\ $ see Gauss's Algorithm $\ $ — Bill Dubuque, Jun 11 '15 at 22:10
Or $\ {-}48x\equiv 11x\equiv 8!\overset{\large \div 8}\iff! {-}6x\equiv 1\equiv 60\iff x\equiv -10\ \ $ — Bill Dubuque, Jun 11 '15 at 22:16

score 2 · Answer 1 · answered Jun 11 '15 at 22:07

Good, so now you know $\displaystyle{59(3)+11(-16)=1}$. This tells you that

$$59(3)+11(-16)\equiv 1\pmod{\! 59}$$

$$\iff 11(-16)\equiv 1\pmod{\! 59}$$

Multiply both sides by $8$:

$$11(-16\cdot 8)\equiv 8\pmod{\! 59}$$

$$\iff 11(-128)\equiv 8\pmod{\! 59}$$

$$\iff 11(49)\equiv 8\pmod{\! 59}$$

1 Answers1