n(n+1)(n+2) is a fancy way of saying the product of three consecutive integers

if n is odd, n+1 is even, and n+2 is odd

n(n+1)(n+2) = odd x even x odd = even x odd = even

if n is even, n+1 is odd, and n+2 is even

n(n+1)(n+2) = even x odd x even = even

(if you don't understand this, you should review number properties of odd's and evens)

from above, n is even regardless if n is even or odd; thus A and B are false we can eliminate them.

when n is odd, n(n+1)(n+2) is even so C is false.

Now we have choices D and E left.

choice D:

any three consecutive positive integers will have one of the numbers divisible by 3.

proof:

when a positive integer is divided by 3, it has remainders of 0 (i.e. divisible by 3), 1, or 2

if n has a remainder of 0, it is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3

if n has a remainder of 1, n+1 has a remainder of 2, and n+2 has a remainder of 0. n+2 is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3

if n has a remainder of 2, n+1 has a remainder of 0, and n(n+1)(n+2) is divisible by 3

Eliminate D.

We could pick by process of elimination but for the sake of completion:

If n is even, n can be expressed as 2k, for k an integer

if n = 2k

then n + 2 is also even and n + 2= 2m for m an integer

n(n+1)(n+2) = 2k(n)(2m) = 4knm

here knm is an integer, so 4knm and consequently n(n+1)(n+2) is divisible by 4.

enjoy!

p.s. for more OG alternate explanations, click here