Cool property of the number $24$

In$\;4\;$ consecutive integers $\;(n-1)\,,\,n\,,\,(n+1)\,,\,(n+2)\;$ there are exactly two even and two odd ones.

Of the even ones, exactly one is divisible by $\; 4\;$ so the whole product is divisible by $\;8\;$ , and since at least one of the four numbers is a multiple of three the whole thing is divisible by $\;2^3\cdot 3=24\;$ .

We have $$n(n-1)(n-2)(n-3)=\frac{n!}{(n-4)!}=4!\times\frac{n!}{4!(n-4)!}=24\times\binom{n}{4},$$ where the binomial coefficient $\binom{n}{4}$ is known to be an integer.

For more, see these previous questions: 1, 2, 3