Example 12. Given $x=19^{94}-1, y=2^{m} \cdot 3^{n} \cdot 5^{l}$ $(m, n, l$ are non-negative integers, and $m+n+l \neq 0)$. Find the sum $S$ of all divisors of $x$ that are of the form $y$.
$$
\begin{array}{l}
\text { Solve } x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots-C_{94}^{93} \cdot 20 \\
=2^{3}\left(2 n_{1}-235\right) \\
=2^{3}\left[2\left(n_{1}-118\right)+1\right] .\left(n_{1} \in \mathbb{N}\right) \\
x=(18+1)^{94}-1 \\
=18^{94}+C_{94}^{1} \cdot 18^{93}+\cdots+C_{94}^{93} \cdot 18 \\
=3^{2}\left(3 n_{2}+188\right) \\
=3^{2}\left[3\left(n_{2}+62\right)+2\right] .\left\{n_{2} \in \mathbb{N}\right) \\
x=(20-1)^{94}-1 \\
=20^{94}-C_{94}^{1} \cdot 20^{93}+\cdots+C_{94}^{92} \cdot 20^{2} \\
-C_{94}^{93} \cdot 20 \\
=5\left(5 n_{3}+376\right) \\
=5\left[5\left(n_{3}+75\right)+1\right] .\left(n_{3} \in \mathbb{N}\right) \\
\end{array}
$$
$$
\therefore y=2^{m} \cdot 3^{n} \cdot 5^{l} \text { satisfies } 0 \leqslant m \leqslant 3,0 \leqslant n \leqslant
$$
$2,0 \leqslant l \leqslant 1$, and $m+n+l \neq 0$. Therefore,
$$
\begin{aligned}
S= & \left(2^{0}+2^{1}+2^{2}+2^{3}\right)\left(3^{0}+3^{1}+3^{2}\right) \\
& \cdot\left(5^{0}+5^{1}\right)-1 \\
= & 1169
\end{aligned}
$$