## Task 1 - 220721

Determine all even natural numbers $z$ with the following properties:

(1) The number $z$ is five-digit, none of its five digits is a 0.

(2) The three-digit number formed by the first three digits of $z$ in this order is a square number.

(3) The three-digit number formed by the last three digits of $z$ in this order is a cube number.

Hint: If $a$ is a natural number, then $a^{2}$ is its square and $a^{3}$ is its cube.
I. If a number $z$ has the required properties, then:

The number formed by the last three digits of $z$ is one of the numbers $5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729$; for due to $4^{3}999$, these are the only three-digit cube numbers.

Since $z$ and thus the last digit of $z$ is even, the only remaining possibilities for the last three digits of $z$ are 216 and 512. Therefore, the number formed by the first three digits of $z$ ends in 2 or 5.

However, there is no square number that ends in 2; for if a natural number $a$ ends in $0, 1, 2, 3, 4, 5, 6, 7, 8$ or $9$, then its square number ends in $0, 1, 4, 9, 6, 5, 6, 9, 4$ or 1, respectively.

Therefore, only the possibility 512 remains for the last three digits of $z$, and the first three digits form a square number ending in 5, i.e., one of the numbers $5^{2}, 15^{2}, 25^{2}, 35^{2}, \ldots$

Of these, due to $5^{2}999$, only $15^{2}=225$ and $25^{2}=625$ are three-digit numbers.

Thus, it is shown that only 22512 and 62512 can have the required properties.