Question

3. Prove that the product of three consecutive positive integer is divisible by 6.​

Answer 1

SoluTion :-

Let the three consecutive positive integers be n, n+1 and n+2.

As a number is divided by 3, the remainder obtained is either 0,1 or 2.

Thus, n = 3p or 3p + 1 or 3p + 2, where p is an integer.

⇒ If n = 3p, then n is divisible by 3.

⇒  If n = 3p+1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3 (p+1) is divisible by 3.  

⇒ If n = 3p+2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3 (p+1) is divisible by 3.

 

So, one of the numbers is always divisible by 3 :-

n(n+1)(n+2) is divisible by 3

Whenever a number is divided by 2, the remainder is either 0 or 1.

Therefore, n = 2q or 2q + 1 (where q is an integer.)

⇒ If n = 2q, then n and n + 2 = 2q + 2 = 2(q+1) is divisible by 2.

⇒ If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2(q+1) is divisible by 2.

 

So, one of the numbers is always divisible by 2.

∵ n(n+1)(n+2) is divisible by 2 and 3.

∴ n(n+1)(n+2) is divisible by 6.