11.5. Find all numbers $a$ such that for any natural $n$ the number $a n(n+2)(n+3)(n+4)$ is an integer.

(O. Podlipsky)
Answer. $a=\frac{k}{6}$, where $k$ is any integer.

Solution. Substituting $n=1, n=3$ and $n=4$, we get that the numbers $60a, 630a$, and $24 \cdot 56a$ are integers. Therefore, $a$ is a rational number, and the denominator $q$ in its irreducible form is a divisor of the numbers 60, 630, and $24 \cdot 56$. Consequently, their greatest common divisor (GCD) is also divisible by $q$. Since GCD $(60,630)=$ $=30$, and $24 \cdot 56$ is not divisible by 5, then $q$ is a divisor of the number 6, and $a=k / 6$ for some integer $k$.

It remains to show that all numbers of this form are suitable. Indeed, one of the three consecutive numbers $n+2, n+3$, $n+4$ is divisible by 3, and one of the consecutive numbers $n+2$, $n+3$ is divisible by 2; therefore, $n(n+2)(n+3)(n+4)$ is divisible by 2 and by 3, and hence by 6. Therefore, $\operatorname{an}(n+2)(n+3)(n+4)=$ $=k \frac{n(n+2)(n+3)(n+4)}{6}-$ is an integer.

Comment. It is proven that any number of the specified form is suitable - 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 combined with others).