Let n be any positive integer.
Thus, the three consecutive positive integers are n, n+1 and n+2.
We know that any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)
So,
For n= 6q,
⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2)
⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]
For n= 6q+1,
⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)
⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)]
For n= 6q+2,
⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)
⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)]
For n= 6q+3,
⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)
⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]
For n= 6q+4,
⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)
⇒ n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+2)(3q+1)(2q+2)]
For n= 6q+5,
⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)
⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)]
⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+5)(q+1)(6q+7)]
Hence, the product of three consecutive positive integers is divisible by 6.
