518*. Does there exist a natural number that, by erasing its first digit, decreases: a) by $57 ;$ b) by 58 times
$\triangle$ a) Let the desired number be denoted as $10^{k} \cdot a + x$, where $k$ is a natural number, $a$ is the digit to be crossed out, and $x$ is the number formed by the last $k$ digits. We obtain:

$$
10^{k} \cdot a + x = 57 x, \quad 10^{k} \cdot a = 56 x
$$

Since the right-hand side of the last equation is divisible by 7, the left-hand side must also be divisible by 7, i.e., $a$ is divisible by 7, hence $a = 7$. Then,

$$
10^{k} \cdot 7 = 56 x, \quad 10^{k} = 8 x
$$

From this, $10^{k}$ is divisible by 8, i.e., $k \geqslant 3$. For example, if $k = 3$, then

$$
1000 = 8 x, \quad x = 125
$$

and the desired number is 7125. The general form of such numbers is $712500 \ldots 0$, where the number of zeros can take any non-negative integer value.

b) Using the same notation, we obtain:

$$
10^{k} \cdot a + x = 58 x, \quad 10^{k} \cdot a = 57 x
$$

However, the last equation is impossible, since its right-hand side is divisible by 19, while the left-hand side is not.

Answer: a) exists, for example, $7125$; b) does not exist.