elementary number theory - How do I solve $32x equiv 12 pmod {82}$?

I am able to solve simpler linear congruences, for example $3x \equiv 2 \pmod 5$. What I would do in this case is use that $0 \equiv 10 \pmod 5$ and then utilising a theorem: $3x \equiv 12 \pmod 5$. Then I can divide by $3$ leaving me $x \equiv 4 \: \left( \mathrm{mod} \: {\frac{5}{\mathrm{GCD}(5,3)}} \right) \quad \Longleftrightarrow x \equiv 4 \pmod{5}$ which means the solution is $x = 4k + 5$ where $k \in \mathbb{Z}$.

But I cannot apply the same method to this congruence:
$$32x \equiv 12 \pmod {82}$$
This is how far I got:
$$8x \equiv 3 \: \left( \mathrm{mod} \: \frac{82}{\mathrm{GCD}(82, 4)} \right)$$
$$\Updownarrow$$
$$8x \equiv 3 \pmod {41}$$

What could I do next? Please provide solutions without the Euclidean algorithm.

EDIT:

What I found later is that I can say that
$$0 \equiv 205 \pmod {41}$$
And then I can add it to the congruence in question and divide by $8$.
So I guess my question is essentially 'How can I find a number that is a multiple of $41$ (the modulus) and which, if added to $3$ gives a number that is divisible by $8$?'

I reckon the Euclidean algorithm is something which gives an answer to these kinds of questions?!

Answer

Hint :you can do like this $$\quad{8x \equiv 3 \pmod {41}\\ 8x \equiv 3+41 \pmod {41}\\8x \equiv 44 \pmod {41} \div4 \\ 2x \equiv 11 \pmod {41}\\2x \equiv 11+41 \pmod {41}\\2x \equiv 52 \pmod {41}\div 2\\x \equiv 26 \pmod {41}\\x=41q+26}$$