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 $n$ for which $a_{n}$ is divisible by $99 ?$

We have to find. $a_n \equiv 0\pmod{99}$
$$ a_n=100a_{n-1}+134 \\ \implies a_n-a_{n-1}\equiv 35 \pmod{99}$$
Now how do I proceed from here?
I Concluded some results, verified for smaller values, which proved to be wrong. How does doing this get me the $n$? Or am I even correct to proceed like this?
The (unofficial) solution isn't very good either:

$a_{3}=253534 \\
a_{4}=25353534 \\
\therefore a_{n}=2 \underbrace{535353 \ldots 53}_{(n-1) \text { Times } 53}4$
Now, $a_n \rightarrow$ divisible by $99 \Rightarrow$ by $\ 9 \ \& \ 11$ both.
Sum of digits $=6+8(n-1)$
To be divisible by 9
$\mathrm{n}=7,16,25,34,43,52,61,70,79,88, \ldots$
$a_{7}=2\underbrace{535353535353 }_{6 \text { Times } 53} 4$
But $a_{7} \rightarrow$ Not divisible by 11 .
$a_{16}=2\underbrace{5353535353 \ldots \ldots 53 }_{15 \text { Times } 53}4$
Similarly, $a_{16} \rightarrow$ Not divisible by 11 .
Now, $n=88$
$a_{88}=2 \underbrace{5353 \ldots \ldots 53}_{87 \text { Times } 53}4$
Divisibility by $11 \rightarrow|(2+3+3 \ldots \ldots)-(5+5+\ldots+ 4)|$
$$
\begin{array}{l}
=|263-439| \\
=176
\end{array}
$$
$\therefore$ Least $n=88$

Hints are more appreciated than the solution.
 A: Your method can work quite well. For all $1 \lt i \le n$, note $a_i - a_{i-1}\equiv 35 \pmod{99}$ means each $a_i$ is congruent to $35$ more than the previous one of $a_{i-1}$. Thus, starting from $a_1$ and repeating this $n - 1$ times, you get
$$a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$$
Since $99 = 9(11)$, you can split \eqref{eq1A} into
$$35n - 11 \equiv 0 \pmod{9} \implies 8n - 2 \equiv 0 \pmod{9} \implies 4n \equiv 1 \pmod{9} \tag{2}\label{eq2A}$$
$$35n - 11 \equiv 0 \pmod{11} \implies 2n \equiv 0 \pmod{11} \implies n \equiv 0 \pmod{11} \tag{3}\label{eq3A}$$
Note \eqref{eq3A} means $n = 11k, \; k \in \mathbb{Z}$. You can thus use this to determine $k$ from \eqref{eq2A}.
A: Claim 1: a number is divisible by $9$ if and only if the sum of its digits (written in base $10$) is $0\pmod 9$.
Proof: if $n=\sum_{k=0}^md_k10^k$, then since $10^k\equiv 1\pmod 9$, we have $n\equiv \sum_{k=0}^m d_k\pmod 9$. Thus $9$ divides $n$ if and only if $9$ divides $\sum_{k=0}^md_k$.
Claim 2: a number is divisible by $11$ if and only if the alternating sum of its digits is $0\pmod {11}$.
Proof: the same thing, but now $10^k\equiv (-1)^k\pmod{11}$.
Now $a_n$ is divisible by $99$ if and only if it is divisible by $9$ and $11$, and the sequence is designed so we can easily tell what the digits are. Writing $b_n$ for the sum of the digits of $a_n$, we have
$$b_2=7+7, b_3=7+8+7,b_4=7+8+8+7,$$
etc. Mod $9$, each successive term subtracts one. Inductively, one sees that $a_n$ is divisible by nine if and only if $n=9k+7$ for some $k$.
Now let $c_n$ be the alternating sum of the digits. We have
$$c_2=2-5+3-4,c_3=2-5+3-5+3-4,$$
etc. In other words,
$$c_2=-4,c_3=-6,c_4=-8,$$
etc. Inductively, $c_n=-2n$, and $a_n$ is divisible by $11$ if and only if $n=11k$.
All together, $a_n$ is divisible by $99$ if and only if $n\equiv 7\pmod 9$ and $n\equiv 0\pmod {11}$. It is then straightforward to use the Euclidean algorithm to find the set of solutions $\{88+99k:k\in\Bbb Z\}$.
