$2 \cdot 56$ Find the smallest natural number, such that when its last digit is moved to the first position, the number is multiplied by 5.
[Solution] Let $\overline{a_{1} a_{2} \cdots a_{n-1} a_{n}}$ be the required number. According to the problem, we have $\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n}}$,

Thus,
$$
\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}}=5\left(\overline{a_{1} a_{2} \cdots a_{n-1}} \cdot 10+a_{n}\right), \\
a_{n} \cdot\left(10^{n-1}-5\right)=49 \cdot a_{1} a_{2} \cdots a_{n-1}, \\
a_{n} \cdot \underbrace{99 \cdots 9}_{n-2 \uparrow 9} 5=49 \cdot a_{1} a_{2} \cdots a_{n-1} .
\end{array}
$$

From (1), we know that the left side of the equation is divisible by 49. Since $a_{n}$ is a digit, it cannot be divisible by 49, so $\underset{n-2 \uparrow 9}{99 \cdots 95}$ must be divisible by 7. The smallest such number is 99995, which means $n=6$. Therefore, the required number must be at least a 6-digit number. At this point, (1) becomes
$$
\begin{array}{l}
a_{6} \cdot 99995=49 \cdot \overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, \\
a_{6} \cdot 14285=7 \cdot \overline{a_{1} a_{2} a_{3} a_{4} a_{5}} .
\end{array}
$$
14285 is not divisible by 7, so $a_{6}=7$. Therefore, the required number is 142857.