9.4. For four consecutive natural numbers, it is known that the largest of them is a divisor of the product of the other three. Find all values that the largest of these numbers can take.
Answer: 6.

First solution. Let our numbers be $n-3, n-2, n-1, n$, where $n-3 \geqslant 1, n \geqslant 4$. Since the numbers $n-1$ and $n$ are coprime, $(n-3)(n-2)$ is divisible by $n$. Note that $(n-3)(n-2) = n^2 - 5n + 6 = n(n-5) + 6$, and since both numbers $(n-3)(n-2)$ and $n(n-5)$ are divisible by $n$, 6 is also divisible by $n$. Therefore, $n \leqslant 6$, and since the largest number is not greater than 6, we need to check the cases $n=4, n=5$, and $n=6$. Clearly, only $n=6$ works.

Second solution. Let our numbers be $n, n+1, n+2, n+3$. Note that if the largest of these numbers is divisible by a prime $p$, then one of the remaining numbers must also be divisible by $p$, but then their difference, which is no more than 3, must also be divisible by $p$, so $p \leqslant 3$. Therefore, $n+3 = 2^m 3^k$.

If $n+3$ is divisible by 9, then the numbers $n+1$ and $n+2$ are not divisible by 3, so $n$ must be divisible by 9. But then $3 = n+3 - n$ would be divisible by 9. This contradiction shows that $k \leqslant 1$. If $n+3$ is divisible by 4, then the numbers $n$ and $n+2$ are not divisible by 2, so $n+1$ must be divisible by 4. But then $2 = n+3 - (n+1)$ would be divisible by 4. This contradiction shows that $m \leqslant 1$. Therefore, the possible values for the largest number are $1, 2, 3, 6$. Clearly, only the last option works.

Comment. Only the answer - 1 point.

Proved that the largest number has the form $2^m 3^k - 2$ points.