Task B-4.1. Determine the natural numbers $k$ and $n$ such that

$$
\frac{((3!)!)!}{3!}=k \cdot n!
$$

where $n$ is the largest possible number with the given property.
## Solution.

Let's calculate the left side of the given equality.

$$
\frac{((3!)!)!}{3!}=\frac{(6!)!}{6}=\frac{720!}{6}=\frac{720 \cdot 719!}{6}=120 \cdot 719!
$$

Then the equality

$$
120 \cdot 719!=k \cdot n!
$$

is satisfied for all $n=1, \ldots, 719$ and the corresponding $k$. Since we are looking for the largest possible $n$, it follows that

$$
k=120, n=719
$$

Note. If there are no comments that the given equality has multiple solutions, deduct 1 point.