I don't understand highlighted part, which means I don't know how to simply congruences.
how can 20y1 = 1 (mod3) can be simply 2y1 = 1 (mod 3)
how 2 y1 = 1 (mod3) change to 2y1 = 2 (mod3)
I don't understand every steps....
I only know what a = b (mod 3)
means that b is remainder of a/3 ....
Is there any rules? for it ?
On
\begin{array}{r|l} 20y_1 \equiv 1 \pmod 3 & \text{Given} \\ 2y_1 \equiv 1 \pmod 3 & \text{Because $20 \equiv 2 \pmod 3$.} \\ 4y_1 \equiv 2 \pmod 3 & \text{Multiply both sides by $2$.} \\ y_1 \equiv 2 \pmod 3 & \text{Because $4 \equiv 1 \pmod 3$.} \\ \end{array}
Why is $20 \equiv 2 \pmod 3$?
$\quad$ Because $3$ goes into $20$ six times with a remainder of $2$.
$\quad$ Or because $3$ divides $(20-2)$.
Why multiply both sides by $2$?
$\quad$ Because $\dfrac 12 \equiv 2 \pmod 3$. (See the next question.)
Why is $4 \equiv 1 \pmod 3$?
$\quad$ Because $3$ goes into $4$ one times with a remainder of $1$.
$\quad$ Or because $3$ divides $(4-1)$.
First note \begin{align} 2^{-1} \equiv 2\mod 3 \end{align} then it follows \begin{align} y_1 \equiv2^{-1}*2y_1 \equiv 2^{-1}\ast1 \equiv 2\mod 3. \end{align}
Edit: Let me explain a bit more.
You start out with the equation \begin{align} 20y_1 \equiv 1 \mod 3 \end{align} and you basically want to solve for $y_1$. Of course, if this is just regular old middle school algebra then we would immediately multiply both sides by $20^{-1}$ to get \begin{align} y_1 \equiv 20^{-1} \mod 3. \end{align} But then we need to figure out what is $20^{-1}$.
However, the problem could be further simplified if we first reduced $20$ to $2$ since we are in the mod $3$ world to get \begin{align} 2y_1 \equiv 1 \mod 3 \end{align} which means \begin{align} y_1 \equiv 2^{-1} \mod 3. \end{align} Finally, observe \begin{align} 2*2 \equiv 4 \equiv 1 \mod 3 \end{align} which means $2^{-1} = 2$. Thus, we have the desired result.