Problem 3. For natural numbers $x$ and $y$ and an odd integer $z$, it is known that $x! + y! = 48z + 2017$. Find all possible such triples of numbers $(x, y, z)$. (Recall that $1! = 1, 2! = 1 \cdot 2, n! = 1 \cdot 2 \cdot \ldots \cdot n$) Answer: $(1; 6; -27), (6; 1; -27), (1; 7; 63), (7; 1; 63)$.
Solution. The expression $x!+y!$ must be an odd number, so $x=1$ or $y=1$.

Let $x=1$, then $1+y!=48z+2017$ or $y!=48z+2016$. Since $2016=48 \cdot 42$, $y!$ is divisible by 48, hence $y \geqslant 6$. If $y \geqslant 8$, then $y!$ is divisible by 96 and $2016=96 \cdot 21$. In this case, $z$ would be even, which contradicts the problem's condition. Therefore, $y=6$ or 7.