Use the Euclidean Algorithm to find a member $x$ of the canonical complete residue system modulo $213$ that satisfies $24x \equiv 123 (\bmod 213)$.
My work so far:
$ 24x-123=213y $
$24x-213y=123 $
Using the Euclidean Algorithm:
$-213=-8(24)-21$
$\ \quad 24=-1(-21)+3$
$\ -21=-7(3)+0$
So
$3=24+1(-21)$
$3=24+1(-213+8(-24))$
$3=9(24)+1(-213)$
Then I multiply by 41 to get 123
$123=369(24)+41(-213)$
But $369$ is not in the canonical complete residue system for $213$. What am I doing wrong??
Then I need to find all the solutions to $24x\equiv 123 (\bmod 213)$. Help!
369 seems fine; $$24(369) = 8856 \equiv 123 \bmod 213$$ (If you care, $8856 = 213(41) + 123$)
The corresponding number in the "canonical complete residue system" is probably $156$, as $$369 \equiv 156 \bmod 213$$
As for actually getting a member of the "canonical complete residue system modulo 213", there's at least one thing that I'm noticing in your proof. In that final step, you multiplied by 41 to get your solution.
If you think that step actually gets you what you're looking for, you're basically assuming that the answer is going to be a multiple of 41. Nowhere in your proof indicates that this has to be true.
The beginning of my answer is pointing out that 369 is a solution. It might not be the canonical solution, but it's a solution. You need one more step somewhere to get your "canonical answer" (hence my suggestion to try 369's residue mod 213)
As for finding all of the solutions, suppose you have your canonical answer $x$, and you also found the smallest positive integer $y$ such that $24y \equiv 0 \bmod 213$. What would
$$24(ky + x)$$
be congruent to mod 213 for all $k \in \mathbb{Z}$?