If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$ 
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$

So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$
But after that what i did is to multiply both sides by $9$ in the congruence
$35n - 11 \equiv 0 \pmod{99}$
$315n \equiv 0 \pmod{99}$
$18n \equiv 0 \pmod{99}$
dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?
thankyou

You need to solve $35n \equiv 11 \pmod{99}$. This can be done via a variation of the (Extended) Euclidean Algorithm.
$$
\begin{array}{ll}
99n \equiv 0 \pmod{99} & (1)\\
35n \equiv 11 \pmod{99} & (2)\\
-6n \equiv -33 \pmod{99} & (3) = (1)-3\times(2)\\
-n \equiv -187 \equiv -88 \pmod{99} & (4) = (2) + 6\times(3) 
\end{array}
$$
Therefore $n\equiv 88 \pmod{99}$.
