$11 \cdot 42$ For an integer $x \geqslant 1$, let $p(x)$ be the smallest prime that does not divide $x$, and $q(x)$ be the product of all primes less than $p(x)$. In particular, $p(1)=2$, and if there is some $x$ such that $p(x)=2$, then $q(x)=1$.
The sequence $x_{0}, x_{1}, x_{2}, \cdots$ is defined by:
$$
\begin{array}{c}
x_{0}=1, \\
x_{n+1}=\frac{x_{n} p\left(x_{n}\right)}{q\left(x_{n}\right)}
\end{array}
$$

where $n \geqslant 0$, find all integers $n$ such that $x_{n}=1995$.
(36th International Mathematical Olympiad Preliminary Question, 1995)
[Proof] Clearly, from the definitions of $p(x)$ and $q(x)$, for any $x$, $q(x)$ divides $x$, thus
$$
x_{n+1}=\frac{x_{n}}{q\left(x_{n}\right)} \cdot p\left(x_{n}\right)
$$

which means that for any $n$, $x_{n+1}$ is a positive integer.
Additionally, it is easy to prove by mathematical induction that for all $n$, $x_{n}$ has no square factors.
Therefore, a unique code can be assigned to $x$ based on whether it is divisible by a certain prime number. The specific method is as follows:
Let $p_{0}=2, p_{1}=3, p_{2}=5, \cdots$ be the sequence of all prime numbers in ascending order.
Let $x>1$ be any number without square factors. Suppose $p_{m}$ is the largest prime that divides $x$.

Divide $x$ by $p_{i}$. If $p_{i}$ divides $x$, set the code $S i=1$; otherwise, set $S i=0$. Then the code for $x$ is $\left(1, S_{m-1}, S_{m-2}, \cdots, S_{1}, S_{0}\right)$.
Define $f(x)=\frac{x p(x)}{q(x)}$.
If the last bit of $x$'s code is 0, then $x$ is odd, $p(x)=2, q(x)=1, f(x)=2 x$, and the code for $f(x)$ is the same as $x$'s code except that the last 0 is replaced by 1. If the last several bits of $x$'s code are $011 \cdots 1$, then the last several bits of $f(x)$'s code are $100 \cdots 0$. If the code traverses all binary numbers, then the code for $f(x)$ can be obtained by adding 1 to $x$'s code.

Given $x_{1}=2$ and for $n \geqslant 2$, $x_{n+1}=f\left(x_{n}\right)$, the code for $x_{n}$ can directly equal the binary representation of $n$.
Therefore, when $x_{n}=1995=3 \cdot 5 \cdot 7 \cdot 19$, there exists a unique $n$ such that the code for $x_{n}$ is
10001110 (the numbering from right to left is based on $2 \nmid 1995, 3 \mid 1995, 5 \mid 1995, 7 \mid 1995, 11 \nmid$
By $(10001110)_{2}=(142)_{10}$.
Thus, $n=142$ is the integer $n$ such that $x_{n}=1995$.