$10 \cdot 27$ Find the smallest natural number $n$ that satisfies the following conditions:
(1) The last digit of $n$ is 6;
(2) If the last digit 6 of $n$ is moved to the front of the other digits, the resulting new number is 4 times $n$.
(4th International Mathematical Olympiad, 1962)
[Solution 1] Let the required natural number be $n=10 x+6$,
where $x$ is a natural number.
Let the number of digits of $n$ be $m$. By the problem, we have
$$
\begin{array}{c}
6 \cdot 10^{m-1}+x=4(10 x+6), \\
13 x=2 \cdot 10^{m-1}-8, \\
2 \cdot 10^{m-1}=13 x+8 .
\end{array}
$$

Thus, divide the number $200 \cdots 0$ by 13 until the first remainder is 8, which will give us the smallest natural number $x$.
At this point, we get
$$
x=15384 \text {. }
$$

Therefore, the smallest natural number sought is 153846.
[Solution 2] Let the smallest natural number sought be

By the problem, we have
$$
\begin{array}{l}
n=\overline{a_{m} a_{m-1} \cdots a_{2} 6}, \\
4 n=\overline{6 a_{m} a_{m-1} \cdots a_{2}} .
\end{array}
$$

Since the unit digit of $n$ is 6, then the unit digit of $4 n$ is 4, i.e., $a_{2}=4$. At this point,
$$
n=\overline{a_{m} a_{m-1} \cdots a_{3} 46} \text {, }
$$

Since $46 \cdot 4=184$, the tens digit of $4 n$ is 8, i.e., $a_{3}=8$. At this point,
$$
n=\overline{a_{m 1} a_{m-1} \cdots a_{4} 846} \text {, }
$$

Since $846 \cdot 4=3384$, the hundreds digit of $4 n$ is 3, i.e., $a_{4}=3$. At this point,
$$
n=\overline{a_{m} a_{m-1} \cdots a_{5} 3846} \text {, }
$$

Since $3846 \cdot 4=15384$, the thousands digit of $4 n$ is 5, i.e., $a_{5}=5$. At this point,
$$
n=\overline{a_{m} a_{m-1} \cdots a_{6} 53846} \text {, }
$$

Since $53846 \cdot 4=215384$, the ten-thousands digit of $4 n$ is 1, i.e., $a_{6}=1$, at this point,
$$
n=\overline{a_{m} a_{m-1} \cdots a_{7} 153846} \text {, }
$$

Since $153846 \cdot 4=615384$, the leading digit of $4 n$ is now 6, thus
$$
\begin{array}{l}
4 n=615384 \\
n=153846
\end{array}
$$

This is the smallest natural number $n$ sought.