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
 A: It's an easy one-liner using a fractional form of mDL = mod Distributive Law, viz.
Notice  $\,\ \dfrac{\color{#c00}{11}}{35}\bmod \color{#c00}{11}(9)\,=\, \color{#c00}{11}(\color{#0a0}8)\,$ by $\color{#0a0}{\bmod 9\!:\ \dfrac{1}{35}\equiv \dfrac{1}{-1}\equiv 8},\ $ via
Theorem $\ \ \dfrac{\color{#c00}ab}d\bmod \color{#c00}ac\, =\, \color{#c00}a\left(\color{#0a0}{\dfrac{b}d\bmod c}\right)\ \ $ if $\ \ (d,ac) = 1$
Proof $\, $ Bezout $\Rightarrow$ exists $\, d' \equiv d^{-1}\pmod{\!ac}.\,$ Factoring out $\,\color{#c00}a\,$ by mDL
$$\color{#c00}abd'\bmod \color{#c00}ac\, =\ \color{#c00}a(bd'\bmod c)\qquad\qquad\qquad$$
and $\,dd' \equiv 1\pmod{\!ac}\Rightarrow dd' \equiv 1\pmod{\!c},\,$ so $\,d'\bmod c = d^{-1}\bmod c$
A: From $a_n= 24+(n - 1)35 \equiv 0 \pmod{99} $ you can see that $n-1$ is a multiple of $3$. Let it be $3m$, then dividing throughout by 3, $$0\equiv8 +35k \equiv 8+2k \pmod{33} $$ and therefore $$k \equiv -4 \pmod{33}$$
Then take $k=29$ and $n=88$.
N.B.
Another way to look at your 'multiplying by 9' idea is that you can, of course, do this but you must multiply the base by the factor as well! In general this is unlikely to be a useful procedure. In your case you would change an equation mod $99$ into one mod $891$.
A: 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 $n\equiv 88 \pmod{99}$.
A: 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 $17$
$$n\equiv 187\bmod 99\to n\equiv 88\bmod 99$$
A: $35n\equiv11\bmod99$ could also be solved with the Chinese remainder theorem.
It's equivalent to $-n\equiv2\bmod9$ and $n\equiv0\bmod11$.
Can you find a multiple of $11$ between $0$ and $99$ that is $2$ less than a multiple of $9$?
