modular arithmetic - Solve the congruence $59xequiv 3pmod {78}$

The question is: Solve the congruence $59x\equiv 3\pmod {78}$
So I already found the inverse of $59\pmod{78}$ which is $41$.

So $41 \cdot 59\equiv 1\pmod {78}$

The solution is:

$59x\equiv 3\pmod {78}$ multiplied by inverse is

$41 \cdot 59x\equiv 41 \cdot 3\pmod {78}$

$x\equiv 123\pmod {78}$

$x\equiv 45\pmod {78}$

$x = 45$

So I have trouble understanding two parts. One, how did we get $x\equiv 123\pmod {78}$?

Two, in the part where we get $x\equiv 45\pmod {78}$ from $x\equiv 123\pmod {78}$ why is $45\pmod {78}=123\pmod {78}$? I get that $45$ is the remainder when $123$ is divided by $78$, but I don't understand how that makes it so $45\pmod {78}=123\pmod {78}$.

Answer

$(1)$ We get $x\equiv 123$ by multiplying $3 \cdot 41$.

$(2)$ $123 - 78 = 45$: that is, $78\mid (123 - 45)$ which means $x\equiv 123 \equiv 45 \pmod {78}$