$2 \cdot 49$ Find the smallest natural number $n$ that satisfies the following properties:
(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 $n$.


［Solution］Let the required natural number be
$$
n=10 x+6 \text {, }
$$

where $x$ is a natural number.
Let the number of digits in $n$ be $m$. According to the problem, we have
$$
6 \cdot 10^{m-1}+x=4(10 x+6) \text {, }
$$

which simplifies to
$$
\begin{array}{l}
13 x=2 \cdot 10^{m-1}-8, \\
2 \cdot 10^{m-1}=13 x+8,
\end{array}
$$

Thus, we can divide the number $200 \cdots 0$ by 13 until the first remainder is 8. This way, we obtain the smallest natural number
$$
x=15384 .
$$

Therefore, the smallest required natural number is 153846.