A number theory problem by Christopher Boo

To prove that $n(n + 1)(n + 2)(n + 3)$ is divisible by $1 \times 2 \times 3 \times 4 = 24,$ we need to check for divisibility by 8 and by 3.

Among any four consecutive positive integers, there is exactly one multiple of 4 and two multiples of 2 (including the multiple of 4). Thus, the product of any four consecutive positive integers must be divisible by $2 \times 4 = 8.$

Among any four consecutive positive integers, at least one must be a multiple of 3. Thus, the product of any four positive consecutive must be divisible by 3.

Since the product of four consecutive positive integers is divisible by both 8 and 3, we conclude that it is also divisible by 24. $\blacksquare$

I will prove the general case:

$k! \mid \displaystyle \prod_{m=0}^{k-1} (n+m)!$

Note that the RHS is just $\dfrac{(n+k-1)!}{(n-1)!}$ .

Furthermore, since all binomial coefficients are integers, it follows that $\dbinom{n+k-1}{n-1}$ is an integer.

Thus, $\dfrac{(n+k-1)!}{k!(n-1)!}$ is an integer, so

$k! \mid \dfrac{(n+k-1)!}{(n-1)!}$

Thus, proven.

Apply the pigeonhole principle. Since this is the product of 4 consecutive numbers, at least one of them must be divisible by 4. A similar argument states that it is divisible by 2 and 3, and hence we are done.