$10 \cdot 28$ (1) Find all positive integers whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{25}$ of the original integer.
(2) Prove that there is no such integer whose first digit is 6, and when this 6 is removed, the resulting integer is $\frac{1}{35}$ of the original integer.
(2nd Canadian Mathematical Olympiad, 1970)
[Solution] (1) Let the positive integer with the first digit being 6 be
$$
6 \cdot 10^{n}+m, \quad 0 \leqslant m \leqslant 10^{n}-1 .
$$

From the problem, we have $m=\frac{1}{25}\left(6 \cdot 10^{n}+m\right)$,
$$
\begin{array}{l}
24 m=6 \cdot 10^{n}, \\
m=2^{n-2} \cdot 5^{n} .
\end{array}
$$

Therefore, the number we are looking for has the form
$$
6 \cdot 10^{n}+2^{n-2} \cdot 5^{n}=625 \cdot 10^{n-2}, \quad(n \geqslant 2, \quad n \in \mathbb{N}) .
$$

Thus, the positive integers we are looking for are
$$
625,6250,62500, \cdots \text {. }
$$
(2) The integer we are looking for should satisfy the condition:
$$
\begin{aligned}
m & =\frac{1}{35}\left(6 \cdot 10^{n}+m\right) \\
34 m & =6 \cdot 10^{n} \\
17 m & =3 \cdot 2^{n} \cdot 5^{n} .
\end{aligned}
$$

Since 17 is a prime number, and 17 must be a factor of $3 \cdot 2^{n} \cdot 5^{n}$, which is impossible. Therefore, there is no integer that meets the requirements of the problem.