10.25 Find the smallest natural number, such that when its last digit is moved to the first position, its value increases by 5 times.
(13th All-Russian Mathematical Olympiad, 1987)
[Solution] Let the required number be $10 y + x$, where $0 \leqslant x \leqslant 9$.
By the problem statement,
$$
\begin{array}{c}
5(10 y + x) = 10^{n} x + y, \\
49 y = (10^{n} - 5) x .
\end{array}
$$

Since $0 \leqslant x \leqslant 9$, $x$ cannot be divisible by 49. Therefore, $10^{n} - 5$ must be divisible by 7.
Thus,
$$
\begin{array}{l}
10^{n} - 5 = 99995, \\
49 y = 99995 x, \\
7 y = 14285 x .
\end{array}
$$

Since $7 \nmid 14285$, we have $x = 7, y = 14285$.
The smallest required number is 142857.