1. Find the smallest natural number such that when the last digit of this number is moved to the first position, the resulting number is 5 times the original number.
Let the required number be $a_{1} a_{2} \cdots a_{n-1} a_{n}$ - According to the given conditions, we have
$$
\overline{a_{n} a_{1} a_{2} \cdots a_{n-1}}=5 \cdot \overline{a_{1} a_{2} \cdots a_{n-1} a_{n}} \text {. }
$$

Then
$$
\begin{array}{l}
a_{n} \cdot 10^{n-1}+\overline{a_{1} a_{2} \cdots a_{n-1}} \\
=5\left(\overline{a_{1} a_{2} \cdots a_{n-1}} \cdot \vdots 0+a_{n}\right) \text {, } \\
a_{n} \cdot\left(10^{n-1}-5\right)=49 a_{1} a_{2} \cdots a_{n-1} \text {, } \\
a_{n} \cdot 99 \cdots 95=45 \cdot a_{1} a_{2} \cdots a_{n-1} \text {, (1) } \\
n-21 \\
\end{array}
$$
$\because a_{n}$ is a digit, $\therefore 49 \times a_{n}$, then $7 \mid 99 \cdots 95$. $n-2$ times

In numbers of the form $99 \cdots 95$, the smallest number divisible by 7 is 99995, at this time $n=6$. Thus (1) becomes $a _ { 0 } \cdot 9 9 9 9 5 = 4 9 \cdot \longdiv { a _ { 1 } a _ { 2 } a _ { 3 } a _ { 4 } a _ { 5 } }$, i.e., $a_{0} \cdot 14285$ $=7 \cdot \overline{a_{1} a_{2} a_{3} a_{4} a_{5}}$. From $7 \times 14285$, we get $\alpha_{0}=7$, and $a_{1} a_{2} a_{3} a_{4} a_{5}=14285$. Therefore, 142857 is the required number.