elementary number theory - Solving linear congruences $12x equiv 1pmod {77}$

I am having some repeated trouble getting the correct answer on linear congruences. Consider the following

$$12x \equiv 1 \pmod {77}$$

$12$ and $77$ are relatively prime so this congruence has a solution. We search for a linear combination of $12$ and $77$ using the extended Euclidean algorithm.

$$77=12(6)+5\\ 12=5(2)+2\\ 5=2(2)+1$$

We now solve for the remainders

$$1=5-2(2)\\ 2=12-5(2)\\ 5=77-12(6)$$

Back substituting we find

$1=77(5)+12(-32)$

The solution to this congruence is $45$ and $-32 \equiv 45$ (mod 77). What am I failing to do properly as to get the first positive solution?

Answer

Your answer is completely correct. More generally, the solution is $45+77k: k\in \mathbb{Z}$ as stated above, because

$$45+77k\equiv 45\mod 77.$$
If you're concerned about getting the negative answer first, it is simple to just add $77$ as you did to find the first positive value.