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

Write the recursion $\bmod 99$
$$a_n = a_{n-1} + 35\bmod 99;\;a_1=24$$
we get the general solution:
$$a_n=35 n-11$$
and $35n-11\equiv 0\bmod 99$ if $$35n\equiv 11\bmod 99\tag{1}$$
Consider the inverse of $35\bmod 99$, that is $17$.
$17\cdot 35\equiv 1 \bmod 99$
Multiply both sides of $(1)$ by $17$
$$n\equiv 187\bmod 99\to n\equiv 88\bmod 99$$
