Considering a generalized version of the puzzle, "If m! x n! = X! then m x n = ?"

(X, 1) and (X, 0) will always be valid "degenerate" solutions (producing answers "X" and "0", respectively). Let's ignore them onward.

Interestingly, for X=10 we have the pair (7, 6) which works only due to the fact that 6! = 8 x 9 x 10. In other words, 6! can be restructured to form the "tail" of a different factorial.

My intuition is that it happens because of 6! containing relatively many prime factors, allowing them to be rearranged to form the "tail".

This made me wonder how often that happens in general (for different X), with the intuition being, it's extremely rare.

So I had ChatGPT write me a program to check this as far as possible (read: until I think I understood the pattern and got fed up waiting for 8! to pop up):

6! = 5! x 3!
10! = 7! x 6!
24! = 23! x 4!
120! = 119! x 5!
720! = 719! x 6!
5040! = 5039! x 7!

It seems that "n" always has to be small, which makes sense because we need to tightly control the prime factors involved in the completion of the tail.

But that also means that, from a certain point, the only solutions seem to be of form m = X - 1, n! = X. (In fact, X=10 is the only example that breaks this rule, making it, in my eyes, the only "interesting" solution.)

It's not a rigorous proof, of course, that this is the only asymptotic solution form, but at least a strong heuristic. Also it helped build a wandwavy proof that there are infintely many X for which a non-degenerate solution exist.