## Task 1 - 240831

Determine all those natural numbers whose sixth power, in its decimal digit representation, contains exactly once the digits 2, 4, 5, exactly twice the digits 8, 9, and no other digits!
I. If a natural number $a$ has the required properties, then:

$a^{6}$ is a seven-digit number, so $1000000 \leq a^{6} \leq 9999999$.

Since $15^{6}=(225 \cdot 15)^{>} 3300^{2}>9999999$, it follows that $10 \leq a \leq 14$.

Because $10^{6}$ ends in the digit 0, $11^{6}$ in the digit 1, and $14^{2}$, and thus also $14^{6}=\left(14^{2}\right)^{3}$, ends in the digit 6, $a$ can only be one of the numbers 12, 13.

Since $13^{3}=169 \cdot 13$ ends in the digit sequence ...97 and thus $13^{6}=\left(13^{3}\right)^{2}$ ends in the digit sequence ...09, only the possibility $a=12$ remains.

II. Indeed, $12^{6}=2985984$, from which it is clear that the number 12 has all the required properties.