For $n$ positive integer, let $F(n) = 1! × 2! × 3! × 4! × \cdots × n!$, product of factorial(i) for $i$ in $[1\ldots n]$.
Let $G(n) = \{i \in [1\ldots n],\text{ such that }n\mid F(i)\}$.
It is obvious that n belongs to G(n) that makes it a non empty set.
Can we find min(G(n)) without using brute-force? is there any mathematical result based on this or any specific theorem?
for eg: $n=4$ answer is $3$
$F(1) = 1! = 1$ , not divisible by $4$
$F(2) = 1! × 2! = 2$ , not divisible by $4$
$F(3) = 1! × 2! × 3! = 12$ , divisible by $4$
$F(4) = 1! × 2! × 3! × 4! = 288$ , divisible by $4$
So $G(4) = {3, 4}$.
I was thinking if we can use prime factorisation of $n$ here somehow!