4. Find all natural numbers whose representation does not contain zero and has the following property: if we omit any digit in it, we get a number that is a divisor of the original number.
SOLUTION. The sought number $n$ contains at least two digits. Let's write it in the form $n=10a + b$, where $a$ is the number that results from erasing the last digit $b$ of the number $n$. According to the problem, $a \mid 10a + b$. From this, $a \mid b$. Considering also that $b \neq 0$, $a$ must be a single-digit number, so $n$ is a two-digit number with non-zero digits $a, b$, where $b = ka, k \in \mathbb{N}$.

If we erase the digit $a$ in the number $n$, the remaining number $b$ must divide the original number $n = 10a + b$. From this, we sequentially get $b \mid 10a, ka \mid 10a, k \mid 10$, and thus $k \in \{1, 2, 5\}$. Substituting into $b = ka$ gives us three possible cases: $b = a, b = 2a$, and $b = 5a$, and in each case, we can easily determine the suitable pairs of digits $a, b$. This way, we find out what the sought numbers $n = 10a + b$ must look like.

Conclusion: The solutions to the problem are the numbers: $11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88$, and $99$. By checking, we can verify that all of them satisfy the conditions of the problem.

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