CONSECUTIVE NUMBERS AND DIVISIBILITY (BY 2, 4 AND 8)

Few facts to note:

1. On the number line, odd and even numbers alternate. For example: 1 is odd, 2 is even. So, for any two consecutive number one will be odd and the other even.
 2. Between two consecutive even numbers, one will be divisible by 4 and the other will only be divisible by 2 and not by 4.

For example:
 2 and 4 -> 2 is not divisible by 4 but 4 is.
 4 and 6 -> 4 is divisible by 4 but 6 is not.

Let’s consider a product of any three consecutive numbers n(n+1)(n+2)

Case 1: n is odd:

n(n+1)(n+2) -> odd * even * odd

Takeaway: The product of three consecutive numbers n(n+1)(n+2) is definitely divisible by 2 if n is odd.
 
 Case 2: n is even:
 
 n(n+1)(n+2) -> even * odd * even
 n and n+2 are two consecutive even numbers. As we know between two consecutive even numbers, one is divisible by 4 and other is only divisible by 2 and not 4. Hence, n(n+2) will be divisible by 2, 4 and 2*4 =8

Takeaway: The product of three consecutive numbers n(n+1)(n+2) is definitely divisible by 2, 4 and 8 if n is even.
 
 In this case we can also say the last two digits of the product are divisible by 4 (a number is only divisible by 4 if its last two digits are) and the last three digits of the product are divisible by 8 (a number is only divisible by 8 if its last three digits are).