1. Find the bases $z$ of all number systems in which the four-digit number $(1001)_{z}$ is divisible by the two-digit number $(41)_{z}$.
1. Since the digit 4 appears in the notation of a two-digit number, it necessarily follows that $z \geq 5$. From the expanded notations $(1001)_z = z^3 + 1$ and $(41)_z = 4z + 1$, we are looking for exactly those natural numbers $z \geq 5$ for which the number $z^3 + 1$ is a multiple of the number $4z + 1$. Using the Euclidean algorithm, we can find their greatest common divisor. We can proceed by first dividing both expressions as polynomials and then eliminating the fractions:

$$
\begin{aligned}
z^3 + 1 & = \left(\frac{1}{4} z^2 - \frac{1}{4^2} z + \frac{1}{4^3}\right)(4z + 1) + \frac{63}{4^3}, \quad / \cdot 4^3 \\
4^3 (z^3 + 1) & = (16z^2 - 4z + 1)(4z + 1) + 63
\end{aligned}
$$

Since the numbers 4 and $4z + 1$ are coprime, we see from this that the number $4z + 1$ divides the number $z^3 + 1$ if and only if it divides the number 63, i.e., if and only if $4z + 1 \in \{1, 3, 7, 9, 21, 63\}$. From the condition $z \geq 5$, however, it follows that $4z + 1 \geq 21$, so $4z + 1 = 21$ (the equation $4z + 1 = 63$ has no integer solution) and $z = 5$.

Note. The factorization (1) can also be easily revealed by using the known formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: according to this, we can write directly

$$
4^3 (z^3 + 1) = (4^3 z^3 + 1) + 63 = (4z + 1)(16z^2 - 4z + 1) + 63.
$$

For a complete solution, award 6 points. If the solver neglects the condition $z \geq 5$ and considers the number $z = 2$ from the equation $4z + 1 = 9$ as a solution, deduct 2 points. Also deduct 1 point if the check for the solution $z = 5$ is missing and the condition $(4z + 1) \mid 63$ is derived only as necessary (not sufficient as in our solution). Award 4 points for the derivation of this condition. Award no more than 1 point in total for merely setting up the condition $(4z + 1) \mid (z^3 + 1)$ or for merely guessing the solution $z = 5$.