Solving Linear Congruence: 12x \equiv 1(mod5)

[Ed Aboud]

Hi all.

Can someone please tell me what is going wrong here.

Solve

$$12x \equiv 1(mod5)$$


$$gcd(12,5) = 1$$

By Euclid's Algorithm =>

$$1 = 5.5 - 2.12$$

So r is 5 in this case.
$$x = r ( \frac{b}{d} )$$

Where b is 1 and d = gcd(12,5) = 1
$$x = 5 ( \frac{1}{1} )$$

$$x = 5$$

Ok fair enough but then I solve the congruence using

$$x \equiv b a^\phi^(^m^)^-^1 (mod m)$$

$$x \equiv (1) 12^3 (mod5)$$

$$x \equiv 3 (mod 5 )$$

I know this is the correct solution but what did I do wrong in the other one.

Thanks for the help!

 

[soandos]

don't know if this is valid, but isn't the first expression equivalent to $$2x \equiv 1(mod5)$$
then 2x = "6" mod 5

$$x\equiv 3(mod5)$$
yes i know that one is not supposed to do division, but modulus is prime, and there is a multiplicative inverse that i multiplied by (3)

 

[Ed Aboud]

Ok I'm not very good at this, but why is the first one equivalent to $$2x \equiv 1(mod5)$$.

Did you reduce the 12 (mod 5) ? Are you able to do that?

 

[soandos]

I think so as $$12\equiv 2(mod5)$$