Okay so I'm having difficulty figuring out something specifically, I'm currently working on congruence classes, now the teacher gave us an example of :
$[6]\cdot x=[14]$ in $\mathbb{Z}/16\mathbb{Z}$
and it eventually equates to
$[6]\cdot[5]=[30]$
I'm having difficulty understanding how to solve for the $x$ in the following equation:
$[42]\cdot x=[27]$ in $\mathbb{Z}/15\mathbb{Z} $
The problem is I don't understand how they seem to solve for the $x$ in most equations, I can't seem to solve the pattern here, if someone could help me out that would be great. I understand everything else, except this speed bump.
Hint: the class of $42$ in $\mathbb{Z}/15\mathbb{Z}$ is the same of $12$ or of $-3$, in fact you can write $42$ as $12+2\cdot 15$ or as $-3 + 3\cdot 15 $ and in $\mathbb{Z}/15\mathbb{Z}$ two integers are in the same class if and only if their difference is a multiple of $15$.
The same holds for the class of $27$ in $\mathbb{Z}/15\mathbb{Z}$.
Now, $-3$ is not invertible: how to solve the congruence?
Well, is the same of solve $[-3](X-1)=0$ in in $\mathbb{Z}/15\mathbb{Z}$...