N3 (4-1, Poland) Find the smallest positive integer $n$ with the following properties:
(1) The unit digit of $n$ is 6;
(2) If the unit digit 6 of $n$ is moved to the front of the other digits, the resulting new number is 4 times $n$.


Solution 1 Let $n=10a+6$, where $a$ is an $m$-digit number. Then
$$
\begin{array}{c}
6 \times 10^{m}+a=4(10a+6). \\
a=\frac{2\left(10^{m}-4\right)}{13}.
\end{array}
$$

Since $a$ is a positive integer, $(2,13)=1$, hence $13 \mid 10^{m}-4. m$ is at least 5, at this point $n=$ 153846.
Therefore, the minimum value of $n$ is 153846.
Solution 2 Let $n=\overline{a_{k} a_{k \cdots 1} \cdots a_{2} 6}$, then according to the problem, we have
$$
m=\overline{6 a_{k} a_{k-1} \cdots a_{2}}=4 n.
$$

Since the last digit of $n$ is 6, $a_{2}=4$. Substituting $a_{2}=4$ into $n$ and multiplying by 4, we can find $a_{3}=8$, then substituting $a_{3}=8$ into $n$ and multiplying by 4, we get $a_{4}=3$, and so on, until the first time a digit in $m=4n$ is 6, we find $a_{5}=5, a_{6}$ $=1$.
Therefore, the minimum value of $n$ is 153846.