How to solve a linear congruence $8n+9\equiv 0\pmod{\!1163}$ Problem : 
Solve in $\mathbb N$ : 
$8n+9\equiv 0\pmod{1163}$
The mathematics give me : $n=435+1163k$ , $k\in\mathbb N$ 
$1163$ is a prime number 
$8n\equiv -9\pmod{1163}$ 
$8n\equiv 1154\pmod{1163}$ 
I don't know any ideas to complete my work ? 
I have to see your hints
 A: $$8n\equiv 1154\;(mod \;1163)$$ $$\Rightarrow 4n\equiv577\; (mod \;1163) $$ $$\Rightarrow 4n\equiv-586\; (mod \;1163) $$ $$\Rightarrow 2n\equiv-293\; (mod \;1163) $$ $$\Rightarrow 2n\equiv870\; (mod \;1163) $$ $$\Rightarrow n\equiv435\; (mod \;1163) $$
A: A different method: 
Let, $$\frac{8n+9}{1163}=m \in\mathbb{Z^{+}}$$
$$\begin{align}n=\frac{1163m-9}{8} \in\mathbb{Z^{+}} \Rightarrow\frac{(145\times8+3)m-8-1}{8} \in\mathbb{Z^{+}}  \Rightarrow \frac{3m-1}{8} \in\mathbb{Z^{+}} \Rightarrow \frac{8k+1}{3} \in\mathbb{Z^{+}} \Rightarrow \frac{9k+1-k}{3} \in\mathbb{Z^{+}} \Rightarrow \frac{k-1}{3} \in\mathbb{Z^{+}} \Rightarrow k=3z+1 \end {align}$$
$$m=\frac {8(3z+1)+1}{3}=8z+3$$
where,
$$\frac{3m-1}{8}=k \in\mathbb{Z^{+}}$$
$$\frac{k-1}{3}=z \in\mathbb{N_0}$$
Finally, we get

$$n=\frac{1163m-9}{8}=\frac{1163(8z+3)-9}{8}=1163z+435$$ where $$z \in\mathbb{N_0}$$

A: Here's an idea:
$8n\equiv1154\equiv2317\equiv\color{red}{3480}\bmod1163$.
Can you solve it now?
A: If you are a one trick pony you can get the answer
using the (not always useful) shortcut found here:
$\; 8n \equiv -9\pmod{1163} \; \text{ iff }$
$\; 8n \equiv 1154\pmod{1163} \; \text{ iff }$
$\; 4n \equiv 577\pmod{1163}$
And since $4 \mid 1164$, the solution is
$\quad x = \Large(\frac{1164}{4}) \normalsize (577) = (291)(577) = 167907 \equiv 435 \pmod{1163}$
Of course a better trick is the one given by Laassila souhayl ($+1$).
