$10 \cdot 24$ Find all natural numbers $n$ such that the fractions $\frac{1}{n}$ and $\frac{1}{n+1}$ can both be expressed as finite decimal fractions.
(Moscow Mathematical Olympiad, 45th, 1982)
[Solution] If the fractions $\frac{1}{n}$ and $\frac{1}{n+1}$ can be expressed as finite decimal numbers, then it must be that
$$
\left\{\begin{array}{l}
n=2^{a} \cdot 5^{b}, \\
n+1=2^{c} \cdot 5^{d} .
\end{array}\right.
$$

where $a, b, c, d$ are non-negative integers.
Thus, we have $\quad 2^{c} \cdot 5^{d}-2^{a} \cdot 5^{b}=1$.
When $a \leqslant c, \quad b \leqslant d$, we have
$$
2^{a} \cdot 5^{b}\left(2^{c-a} \cdot 5^{d-b}-1\right)=1 .
$$

That is
$$
\left\{\begin{array}{l}
2^{a} \cdot 5^{b}=1, \\
2^{c-a} \cdot 5^{d-b}-1=1 .
\end{array}\right.
$$

From this, we solve $a=0, b=0, c=1, d=0$.
Thus, $n=1, n+1=2$.
When $c \leqslant a, \quad b \leqslant d$, we have
$$
2^{c} \cdot 5^{b}\left(5^{d-b}-2^{a-c}\right)=1 \text {. }
$$

That is
$$
\left\{\begin{array}{l}
2^{c} \cdot 5^{b}=1, \\
5^{d-b}-2^{a-c}=1 .
\end{array}\right.
$$

From this, we solve $c=0, \quad b=0$, thus
$$
5^{d}-2^{a}=1 .
$$

Next, we find all non-negative integer solutions to (1).
Considering $\bmod 3$, we have
$$
\begin{array}{l}
2^{2 k} \equiv 1 \quad(\bmod 3) \\
2^{2 k+1} \equiv 2 \quad(\bmod 3)
\end{array}
$$

where $k$ is a non-negative integer.
Thus, from
$$
1=5^{d}-2^{a} \equiv 2^{d}-2^{a}(\bmod 3),
$$

we know that $d$ is odd, and $a$ is even, so
$$
\begin{aligned}
2^{a} & =5^{d}-1 \\
& =(5-1)\left(5^{d-1}+5^{d-2}+\cdots+1\right) . \\
& 5^{d-1}+5^{d-2}+\cdots+1
\end{aligned}
$$

Notice that
is the sum of $d$ odd numbers, and is a factor of $2^{a}$, so it can only be 1, thus $d=1$, and hence $a=2$.
That is, the non-negative integer solutions to (1) are only $a=2, d=1$.
At this time
$$
n=2^{a} \cdot 5^{b}=4 \text {. }
$$

Therefore, only
$$
n=1 \text { and } n=4 \text {. }
$$