10.29 Try to find the smallest natural number that satisfies the following property: its first digit is 4, but when the first digit is moved to the end, its value becomes $\frac{1}{4}$ of the original.
(46th Moscow Mathematical Olympiad, 1983)
[Solution] Let the number we are looking for be
$$
x=\overline{4 a b \cdots c}=4 \cdot 10^{n}+A .
$$

where $A=\overline{a b \cdots c}$ is an $n$-digit number.
Moving the first digit to the end, we get the number
$$
y=\overline{a b \cdots c 4}=10 A+4 .
$$

According to the problem, we have
that is,
$$
4 \cdot 10^{n}+A=4(10 A+4) \text {, }
$$

or
$$
\begin{array}{l}
39 A=4 \cdot 99 \cdots 96, \\
13 A=4 \cdot 33 \cdots 32 .
\end{array}
$$

Since $(13,4)=1$, then $33 \cdots 32$ must be divisible by 13.
The smallest value of the form $33 \cdots 32$ that is divisible by 13 is 33332.
Therefore,
$$
A=4 \cdot \frac{33332}{13}=10256 \text {. }
$$

Thus, the smallest value of the number $x$ we are looking for is 410256.