<s>[INST] Solving a linear congruence 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. 
 [/INST] 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 $8$:   
$$11(-16\cdot 8)\equiv 8\pmod{\! 59}$$  
$$\iff 11(-128)\equiv 8\pmod{\! 59}$$  
$$\iff 11(49)\equiv 8\pmod{\! 59}$$
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