## Task 1 - 130811

Determine all possibilities to specify a four-digit odd (natural) number $z$ such that it has the following properties:

(1) The number $z$ has four different digits.

(2) The product of the second and the third digit of $z$ is 21 times as large as the product of the first and the fourth digit.

(3) The smallest of the digits of $z$ is in the first place, and the largest is in the second place.
If $a, b, c, d$ are the digits of a number $z$ in this order and satisfy the given conditions, then due to (3) $ac$. Furthermore, $1 \leq a, b, c, d \leq 9$ and thus due to (1)

$$
b \cdot c \leq 8 \cdot 9=72 \quad \text { and } \quad a \cdot d \geq 1 \cdot 2=2
$$

If $a \cdot d \geq 4$, then according to (2) $b \cdot c=21 \cdot a \cdot d \geq 84$. Therefore, $2 \leq a \cdot d \leq 3$.

Since $ac$ implies $b=9$ and $c=7$.

Thus, the only number $z$ that can satisfy the given conditions is $z=1973$. It indeed satisfies the conditions; it is odd, it consists of four different digits, the product $9 \cdot 7$ is 21 times the product $1 \cdot 3$, and the smallest of its digits, 1, is in the first position, and the largest, 9, is in the second position.