Recurrence and modular arithmetic incorrect solution 
Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$. (Source: 2017 AIME I)

This is my solution:
We wish to find the least $n$ such that $a_n\equiv 0\pmod{99},$ with the recurrence relation $a_n \equiv a_{n-1} + n \pmod{99}.$ Also, for every $n > 10,$ $a_n = \sum_{k=10}^n k = \frac{10 + n}{2} \cdot (n-9),$ so we wish to find the least $n$ such that $$(10 + n)(n - 9)2^{-1} \equiv 50(10+n)(n - 9) \equiv 0 \pmod{99}.$$ $$50(n^2+n-90) \equiv 50(n^2+n+9) \equiv 50n(n+1)+450 \equiv 0 \\ \Longleftrightarrow 50n(n+1) \equiv 45  \Longleftrightarrow n(n+1) \equiv 45\cdot 50^{-1} \equiv 90 \Longleftrightarrow n(n+1) \equiv 90.$$ Then $n\equiv 9 \pmod{99}$, so the least $n>10$ is $108$.
It seems that $n=108$ actually does work in the sense that $99 \mid a_{108}$, but the actual answer is

 45

How should I edit my solution to give the minimum value? I suspect that somewhere along the second line my solution became a little suspect, I'm not sure why it gives the wrong answer.

You did well through the conclusion $n(n+1)\equiv90\bmod99$, but your solution of that was lacking.
I would solve that congruence using the Chinese remainder theorem;
$99$ is the product of $11$ and $9$, which are relatively prime.
$n(n+1)\equiv0\bmod9$ and $n(n+1)\equiv2\bmod11$.
$n\equiv0 $ or $8\bmod9$ and $n\equiv1$ or $9\bmod11$.
Putting these together, we get $n\equiv45, 9, 89$, or $53\bmod99$.
Do you get it now?
