一、(满分 20 分) Determine the smallest positive integer $n$, whose last digit is 7, such that if the last digit 7 is moved to the front, the resulting number is 5 times the original number.
Let $n=\overline{a_{k} a_{k-1} \cdots a_{2} a_{1}}, x=\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}}$, where $a_{1}$ $=7, k$ is a natural number greater than 1. Then
$$
\begin{aligned}
n & =\overline{a_{k} a_{k-1} \cdots a_{2}} \times 10+7=10 x+7, \\
5 n & =\overline{7 a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{k-1}+\overline{a_{k} a_{k-1} \cdots a_{3} a_{2}} \\
& =7 \times 10^{k-1}+x .
\end{aligned}
$$

Eliminating $n$ from the above two equations, we get
$$
5(10 x+7)=7 \times 10^{k-1}+x .
$$

Thus, $x=\frac{10^{k-1}-5}{7}$.
When $k$ is the smallest, $x$ is the smallest, and $n$ is also the smallest.
When $k$ takes $2,3,4,5$, $x$ is not a natural number.
When $x=6$, $x=14285$.
Therefore, the smallest positive integer $n$ is 142857: