Prove that the product of three …

Let us three consecutive  integers be, x, x + 1 and x + 2 respectively.

[• Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2. ( it's proved )]

Let x = 3p ...................................(i)
where p belongs to an integer and does not equal to zero ( 0 ).

=> x is divisible by 3.

If x = 3p + 1 
where p belongs to an integer and does not equal to zero ( 0 ).

then x + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) 
which is divisible by 3.

If x = 3p + 2
where p belongs to an integer and does not equal to zero ( 0 ).

then x + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) 
which is divisible by 3.

So that x, x + 1 and x + 2 is always divisible by 3.
=>  x (x + 1) (x + 2) is divisible by 3.

Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.

Let x = 2q
where q belongs to an integer and does not equal to zero ( 0 ).

=> x is divisible by 2

If x = 2q + 1 
where q belongs to an integer and does not equal to zero ( 0 ).

then x + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) 
which is divisible by 2.
 
So that x, x + 1 and x + 2 is always divisible by 2.
⇒ x (x + 1) (x + 2) is divisible by 2.

Since x (x + 1) (x + 2) is divisible by 2 and 3.
 
∴ x (x + 1) (x + 2) is divisible by 6.