For some natural numbers $a,b$, let it be known that $b\leq 8$:
$$b\cdot 99a = x68y - 8$$
where we read $x68y$ as x-thousand sixhundred and eighty-y.
What I have established is:
$$\begin{cases}b\mid (x68y-8)&\Longleftrightarrow x68y\equiv 8\pmod{b}\\
99a\mid (x68y-8)&\Longleftrightarrow x68y\equiv 8\pmod{99a}\end{cases} $$
The objective is to determine $a,b$ and for that I thought I would check all possibilities for $b: 8,7,6,5,4,3,2,1$ and see whether 99 divides any of the numbers produced in each case. This turns out to be quite tedious.
What can I do to optimize my work?
The answer is $b=3$ and $a=9$. Also $x=2, y=1$.
On a hunch I checked whether $b=3$ that means $x68y\equiv 2\pmod{3}$ i.e $3\mid (x68y-2)$ which is the same as $3\mid (x+y+14-2)$ (sum of digits). Since it seems unlikely that $y$ is even I took $y=1$ that leaves $x\in\{2,5,8\}$ and on my first attempt, indeed $99\mid (2681-8)$ as required.
$99\mid (x68y-8)$ is the key. Makes this problem 99 times easier.
Doesn't have to be only this solution, though. $b=1, a=27$ is also a possibility unless I'm missing something.