## Task B-4.1.

Determine all natural numbers $n$ for which

$$
\frac{(9!n!)^{2}-(8!(n+1)!)^{2}}{(9!n!)^{2}-18 \cdot(8!)^{2} n!(n+1)!+(8!(n+1)!)^{2}}>0
$$
## Solution.

Let's write the left side of the given inequality in the form

$$
\frac{(9 \cdot 8!)^{2}(n!)^{2}-(8!)^{2}((n+1) \cdot n!)^{2}}{(9 \cdot 8!)^{2}(n!)^{2}-18 \cdot(8!)^{2} n!\cdot(n+1) \cdot n!+(8!)^{2}((n+1) \cdot n!)^{2}}
$$

If we divide both the numerator and the denominator by $(8!n!)^{2}$, we get

$$
\frac{81-(n+1)^{2}}{81-18(n+1)+(n+1)^{2}}
$$

Notice the difference of squares in the numerator and the square of a difference in the denominator.

$$
\frac{(9-(n+1))(9+(n+1))}{(9-(n+1))^{2}}=\frac{(8-n)(10+n)}{(8-n)^{2}}
$$

After canceling out $8-n$ (with the condition $n \neq 8$), we get

$$
\frac{10+n}{8-n}>0
$$

which means $8-n>0$ (the numerator is always positive).

Thus, $n<8$, so the final solution is $n \in\{1,2,3,4,5,6,7\}$.