Problem with congruence equation $893x \equiv 266 \pmod{2432}$ I'm trying to solve $893x \equiv 266 \pmod{2432}$.
Firstly, I find the $\operatorname{gcd}(893, 2432)$ using the extended Euclidean Algorithm. When I calculate this, I receive that the gcd is (correctly) $19$, and that $19 = 17(2432) -49(893)$. From this, I know that there are $19$ distinct solutions.
I then divide my above, initial congruence by the gcd, obtaining $47x \equiv 14 \pmod{128}$. 
I know that $\operatorname{gcd}(129, 47) = 1$ and that $1 = 18(128) - 49(47)$.
Therefore $14 = 14(18)(128)  -14(49)(47)$.
This implies that a solution to the congruence $47x \equiv 14 \pmod{128}$ is $x = -14(49) = -686$. 
$-686 ≡ 82 \pmod{128}$, so I substitute $x = -14(49)$ for $x = 82$. 
From this, I gather then that the solution to the congruence is $82 + 128t$, where $t$ is one of $0,1,2,...,18$. However, I believe this is not correct. 
Where did I go wrong, and how might I go about fixing this?
Thank you so much!

$\textrm{After carrying Euclidean Algorithm on 893 and 2432, we get their H.C.F. 19, and then go backwards} \\ \textrm{step by step to get a particular solution.}$
$\therefore 893 ( - 49 ) + 2432 \times 18 = 19$
$\left. {\Rightarrow  893(- 49 \times 14 ) + 2432 ( 18 \times 14) = 19\times 14 } \right.$
$\therefore 893 ( - 49 \times 14 + 128 n ) + 2432 ( 18 \times 4 +47n)=266$
$\textrm{Hence }x= - 686 + 128 n,  \text{where }n\in Z.$
$\textrm{ i.e. }x= 82 + 128 n, \text{where }n\in Z.$
