11.1. Find all solutions in natural numbers for the equation: $x!+9=y^{3}$.
Answer. $x=6, y=9$

Solution. If $x \geq 9$ then $x$ ! is divisible by 27, $x$ ! + 9 is divisible by 9 and not divisible by 27. Then $y^{3}$ is divisible by 3, which means $y$ is divisible by 3 and $y^{3}$ is divisible by 27 - a contradiction. Therefore, $x \leq 8$. Checking the values $x=1,2,3, \ldots, 8$ gives the only solution $x=6, y=9$.

Grading criteria. Only the answer with verification: 1 point.

Using divisibility by 3 and estimating $x \leq 8: 5$ points.

Checking the values $x=1,2,3, \ldots, 8: 2$ points.