$\{k \in \mathbb Z / 8722 \quad | \quad k \cdot 6076 = 5292 \quad in \quad\mathbb Z /8722\}$
The expression is equivalent to:
$k \cdot 6076 \equiv_{8722}5292$
cancelling by $gcd(6076,5292,8722)=98$:
$k \cdot 62 \equiv_{89}54$
$62^{-1} \quad mod \quad 89 = 56 \quad mod \quad 89$
Now I have to get the possible Elements for $k$:
$87, 87 + 89 = 176, 87 + 89 \cdot 2 = 265, 87 + 89 \cdot 3 = 354, ....$
Question: How can I get the cardinality of all the possible numbers without counting them by hand? It would take too much time for $k \le 8722$.
You divided through by $ \gcd(6076, 5292, 8722) = 98$ - and that's how many answers there will be.
So when you went from:
$k \cdot 6076 \equiv 5292 \bmod 8722$
to
$k \cdot 62 \equiv 54 \bmod 89$
correctly dividing through by $98$, this "folded" the space down to intervals of size $89$ each, of which there were $98$ copies, with a valid answer in each copy.
You found that $87\cdot 62 \equiv 54\bmod 89 $ and so your answers are $\{87{+}89t\}$ for $0\le t\le97$