Is there a positive integer which, when its first digit is removed, a) becomes 57th of its original value, b) becomes 58th of its original value?
Let the first digit of the $n$-digit number $A$ be $a$, and the number obtained by deleting the first digit be $B$. Then,

$$
A=a \cdot 10^{n-1}+B
$$

In case a), $A=57 B$, so from (1) we have

$$
a \cdot 10^{n-1}=56 B=8 \cdot 7 \cdot B
$$

Since every positive integer can be uniquely factored into primes, the prime number 7 on the right side must appear in the prime factorization of $a$. However, we know that $a$ is a digit, so $a=7$. From (2), we get

$$
10^{n-1}=8 B, \text{ or } B=5^{3} \cdot 10^{n-4}
$$

The above steps are reversible, so in case a), precisely the numbers of the form $7125 \cdot 10^{k}$ $(k \geqq 0)$ have the stated property.

In case b), $A=58 B$, so from (1) we have

$$
a \cdot 10^{n-1}=57 B=3 \cdot 19 \cdot B
$$

The number on the left side is not divisible by 19, because $a$ is a digit, so (3) cannot hold. Therefore, there is no positive integer that reduces to its 58th when its first digit is deleted.