A Four Product

We use the fact that if $a|b$ and $b|c$ , then $a|c$ .

The product of any four consecutive integers is divisible by $4!=24$

So,let $P$ be the product of the four consecutive integers,then: $8|24$ and $24|P$ .therefore, $\boxed{8|P}$

Out of the four numbers, two of the numbers are even. Of those two numbers exactly one is a multiple of 4. So 8 will divide their product.

We know product of 3 consecutive numbers is divisible by $8$ , therefore

$(x-1)(x)(x+1) \equiv 0 \mod 8 \\ (x-1)(x)(x+1)(x+2) \equiv 0 \times (x+2) \mod 8 \\ \boxed{\therefore (x-1)(x)(x+1)(x+2) \equiv 0 \mod 8}$