How to solve linear congruences as a newbie? Please, before referring to another problem on the site or giving a link, I would like to say I read most of them such as: This, This, and many others but due to the (subtle) difference in my question, I'm having a really hard time to apply the methods used in other questions. 
We yesterday had our first lecture about Number Theory. I have been trying to work on this problem since yesterday evening. 
$$16a + 17b + 18c \equiv 19\pmod{100}$$
It's given that $ 1 \leq a,b,c \leq 99$ and that $a = 95\ and\ c=11$. We want to know what $b$ is.
I know that the answer is $53$ ( I used the naive way to get the answer) but I fail to do it according to the official methods. Can someone please explain this?
I tried this: ( i filled both a and c in)
$1718 +17b \equiv 19\ (mod\ 100)$
$17b \equiv 19-1718\ (mod\ 100)$
$17b \equiv -1699\ (mod\ 100)$
$$ x \equiv \frac{-1699}{17} \equiv \frac{-1699}{17} \equiv \frac{1}{17}\ i\ got\ stuck\ here$$
I also tried to work with it simplified:
$$17b \equiv 1 (mod\ 100)$$ I thought i was almost done here since i was able to write it like this: 
$$b \equiv \dfrac{1}{17} (mod\ 100)$$
So i thought there is some $b$ number if i divide it by a $1/17$ i get $n$ as an answer and $100$ as rest but even doing so yielded in a wrong answer. Can someone please help?

HINT.-As a newbie you could act as follows:
First at all, note that $17$ is invertible modulo $100$ because it is not divisible by $2$ nor by $5$ so $17b \equiv 1 (mod\ 100)$ has solution.
Since $17n=17+17\cdots+17$ (n times), you could add successively $17$  plus $17$ until you get a number of the form $100x+1$ (this requires to be some patient having into account you know the answer is $53$).
Another shorter way is to add successively $100$ adding $1$ and dividing by $17$ until you have an exact quotient which is precisely your answer.This way requires just $9$ times instead of $53$.
