number theory - Linear Congruence using canonical complete residue system

Use the Euclidean Algorithm to find a member $x$ of the canonical complete residue system modulo $213$ that satisfies $24x \equiv 123 (\bmod 213)$.

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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??

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Then I need to find all the solutions to $24x\equiv 123 (\bmod 213)$. Help!