9.5. Find all numbers $a$ such that for any natural $n$ the number $\operatorname{an}(n+2)(n+4)$ is an integer.

(O. Podlipsky)
Answer. $a=\frac{k}{3}$, where $k$ is any integer.

Solution. Substituting $n=1$ and $n=2$, we get that the numbers $15a$ and $48a$ are integers. Therefore, the number $48a - 3 \cdot 15a = 3a$ is also an integer. Thus, $a = \frac{k}{3}$ for some integer $k$.

It remains to show that all numbers of this form work. Indeed, one of the three consecutive even (or odd) numbers $n, n+2, n+4$ is divisible by 3; hence, $n(n+2)(n+4)$ is divisible by 3, and therefore $a n(n+2)(n+4) = k \frac{n(n+2)(n+4)}{3}$ is an integer.

Comment. It is proven that any number of the specified form works - 2 points.

It is proven that the number $a$ must have the specified form - 4 points.

Only the correct answer - 1 point (this point is not added to the others).