Solve the equation $285x \equiv 177 \pmod{924}$ using continued fraction
My attempt(using Wikipedia notion):
Continued fraction form for $\frac{924}{285}$ is $[3;4,6,1,9]=[q_1;q_2,q_3,q_4,q_5]$
$\frac{924}{285}=\frac{h_n}{k_n}$
We know that $h_nk_{n-1}-h_{n-1}k_n=(-1)^n \Rightarrow$$924k_{5-1}-285h_{5-1}=(-1)^5$.
Thus, when we find $h_{4}$ we'll get the equation:
$-285h_{4} \equiv (-1)^5 \pmod{924} \Rightarrow$
$ 285h_{4} \equiv (-1)^{4}\pmod{924} \Rightarrow$
$ 285h_{4}(-1)^{4} \equiv 1\pmod{924} \Rightarrow$
$x=h_{4}(-1)^{4}177$ is a solution, because $h_{4}(-1)^{4}$ is $285^{-1}$ modulo 924, thus $285x \equiv 285*285^{-1}b \equiv b \pmod{924}$
Let's find $h_{4}$:
$\frac{h_1}{k_1}=\frac31 \Rightarrow h_1=3$
$\frac{h_2}{k_2}=3+\frac14=\frac{13}4 \Rightarrow h_2=13$
Using $h_n=q_nh_{n-1}+h_{n-2}$:
$h_3=6*13+3=81$
$h_4=1*81+13=94$ Bingo.
Thus $x=94*(-1)^4*175$.
This question is from an exam. The correct answer is $x \equiv 153,461,769 \pmod{924}$
Did I do something wrong? how can $x$ have multiple options(edit: how can I find all the options)?
Any help would be highly appreciated!
$285x≡177\pmod{924}$ or $285x+924y=177$ can be solved using the extended euclidean algorithm for the gcd of $285$ and $924$.
The remainder sequence of the euclidean algorithm starts with $r_0=a=924$, $r_1=b=285$ and for the extended variant the Bezout factor sequence for the Bezout identity
$$v_k b≡r_k\;\pmod a\qquad (\text{short version of }u_ka+v_kb=r_k)$$
starts with $v_0=0$, $v_1=1$
Then the algorithm proceeds iterating over $k=1,2,...$
\begin{align} q_k&=r_{k-1}\,\text{div}\,r_k,\\ r_{k+1}&=r_{k-1}-q_kr_k, \\ v_{k+1}&=v_{k+1}-q_kv_k \end{align}
resulting in the table
$$\begin{array}{c|c|c|c|} k& q_k&r_k&v_k\\\hline 0 & & 924 & 0 \\ 1 & 3 & 285 & 1 \\ 2 & 4 & 69 & -3 \\ 3 & 7 & 9 & 13 \\ 4 & 1 & 6 & -94 \\ 5 & 2 & 3 & 107 \\ 6 & & 0 & -308 \\ \end{array}$$
that is, in the end we get the $gcd(285,\,924)=3$ and
$$107\cdot 285 ≡ 3 \pmod{924}.$$
Fortunately, $177=3⋅59$ is divisible by $3$, so $3x≡3⋅59⋅107\pmod{3⋅308}$ has the 3 solutions
\begin{align} x&=59⋅107\mod 308&&=153, \\ x&=153+308&&=461&&\text{ and } \\ x&=461+308&&=769.\end{align}