4.47 Let $x$ be a natural number. If a sequence of natural numbers
$$
x_{0}=1, \quad x_{1}, \quad x_{2}, \cdots, \quad x_{l}=x,
$$

satisfies
$$
x_{i-1}<x_{i}, x_{i-1} \mid x_{i}, \quad i=1,2, \cdots, l .
$$

then $\left\{x_{0}, x_{1}, x_{2}, \cdots, x_{l}\right\}$ is called a divisor chain of $x$, and $l$ is the length of this divisor chain. $L(x)$ and $R(x)$ denote the length of the longest divisor chain of $x$ and the number of such longest divisor chains, respectively. For $x=5^{k} \cdot 31^{m} \cdot 1990^{n}(k, m, n$ are natural numbers), try to find $L(x)$ and $R(x)$.
(5th China High School Mathematics Winter Camp, 1990)
[Solution] For any natural number
$$
x=p_{1}^{\alpha} p_{2}^{\alpha} \cdots p_{n}^{\alpha_{n}}
$$

where $p_{1}, p_{2}, \cdots, p_{n}$ are distinct prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are positive integers.
Obviously, the factor chain of $x$ exists and is finite, thus there must be a longest factor chain.
Let $\left\{x_{0}, x_{1}, \cdots, x_{t}\right\}$ be a longest factor chain of $x$. We prove that $\frac{x_{i}}{x_{i-1}}$ must be a prime number $(i=1,2, \cdots, t)$.
In fact, if there exists $i(1 \leqslant i \leqslant t)$ such that $\frac{x_{i}}{x_{i-1}}$ is not a prime, we can assume
$$
\frac{x_{i}}{x_{i-1}}=q_{1} q_{2},
$$

where $q_{1}, q_{2}$ are positive integers greater than 1, then
$$
\left\{x_{0}, x_{1}, \cdots, x_{i-1}, q_{1} x_{i-1}, x_{i}, x_{i+1}, \cdots, x_{t}\right\}
$$

is also a factor chain of $x$.
This contradicts the fact that $\left\{x_{0}, x_{1}, \cdots, x_{i-1}, x_{i}, \cdots, x_{t}\right\}$ is the longest factor chain.
Therefore, for any $1 \leqslant i \leqslant t, \frac{x_{i}}{x_{i-1}}$ must be a prime factor of $x$, thus
$$
t=L(x)=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}.
$$

Conversely, if
$$
\left\{x_{0}, \quad x_{1}, \cdots, x_{m}\right\}
$$

is a factor chain of $x$, and for any $1 \leqslant i \leqslant m, \frac{x_{i}}{x_{i-1}}$ is a prime, by the definition of a factor chain we have
$$
m=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=L(x).
$$

Thus, $\left\{x_{0}, x_{1}, \cdots, x_{m}\right\}$ must be a longest factor chain.
Therefore, starting from 1 and multiplying a prime factor of $x$ step by step until reaching $x$, we get a longest factor chain of $x$, and different orders of multiplying different prime factors result in different longest factor chains, hence
$$
R(x)=\frac{\left(\alpha_{1}+\cdots+\alpha_{n}\right)!}{\alpha_{1}!\cdots \alpha_{n}!}.
$$

For
$$
\begin{aligned}
x & =5^{k} \cdot 31^{m} \cdot 1990^{n} \\
& =2^{n} \cdot 5^{k+n} \cdot 31^{m} \cdot 199^{n}.
\end{aligned}
$$

then
$$
\begin{aligned}
L(x) & =3 n+k+m, \\
R(x) & =\frac{(3 n+k+m)!}{(n!)^{2} m!(k+n)!}.
\end{aligned}
$$