16.2.13 * Find the smallest natural number such that when the unit digit of this number is moved to the first position, the resulting number is 5 times the original number.
Let the number we are looking for be $10a + b$, where $b \in \{1, 2, \cdots, 9\}$, and $a$ is an $n$-digit number, then $5(10a + b) = 10^n \cdot b + a$. Therefore,
$$
49 \cdot a = \underbrace{99 \cdots 95}_{n-1 \uparrow 9} \cdot 6,
$$
$49 \nmid b$, so $7 \mid 99 \cdots 95$. Among numbers of the form $99 \cdots 95$, the smallest one divisible by 7 is 99995. At this time, $b = 7, a = 14285$. The number we are looking for is 142857.