(2) Let $x$ be a natural number. If a sequence of natural numbers $x_{0}=1, x_{1}, x_{2}, \cdots$, $x_{l-1}, x_{l}=x$, satisfies $x_{i-1}<x_{i}, x_{i-1} \mid x_{i}, i=1,2, \cdots, l$. Then $\left\{x_{0}\right.$, $\left.x_{1}, x_{2}, \cdots, x_{l}\right\}$ is called a divisor chain of $x$, and $l$ is the length of this divisor chain. $L(x)$ and $R(x)$ represent the length of the longest divisor chain of $x$ and the number of the longest divisor chains of $x$, respectively.

For $x=5^{k} \times 31^{m} \times 1990^{n}(k, m, n$ are natural numbers), try to find $L(x)$ and $R(x)$.
For $x=5^{k} \times 31^{m} \times 1990^{n}$, for any chain of its factors, define $y_{i}=$ $\frac{x_{i}}{x_{i}-1}(i=1,2, \cdots, l)$. Clearly, $y_{i}$ is a natural number greater than 1, and
$$
y_{1} y_{2} \cdots y_{l}=x=2^{n} \times 5^{k+n} \times 31^{m} \times 1990^{n}
$$

Thus, $L(x)=\max l=n+(k+n)+m+n=k+m+3 n$ (the maximum value can be reached when $y_{i}$ is a prime factor of $x$).

Notice that $y_{1}, y_{2}, \cdots, y_{l}$, when $l$ takes the maximum value $k+m+3 n$, are exactly a permutation of $n$ 2's, $k+n$ 5's, $m$ 31's, and $n$ 199's. Conversely, from each such permutation, we can obtain a set $\left\{y_{1}, y_{2}, \cdots, y_{l}\right\}$, and thus $\left\{x_{0}, x_{1}, \cdots, x_{l}\right\}$ (since $\left.x=y_{i} y_{i-1} \cdots y_{1}\right)$.

Therefore, $R(x)$ is equal to the number of permutations of $n$ 2's, $k+n$ 5's, $m$ 31's, and $n$ 199's. That is,
$$
R(x)=\frac{(k+m+3 n)!}{n!n!m!(k+n)!}
$$