Question

prove that the product of three consecutive positive integers is divisible by 6

Answer 1

let the numbers be n, n+1, n+2
product =n(n+1)(n+2)

lets prove this by induction
let n=1
1*2*3=6
 is divisible by 6

assume for n=k, k*(k+1)(k+2) is divisible 6
so consider n=k+1
(k+1)(k+2)(k+3)=k(k+1)(k+2) + 3(k+1)(k+2)
we already know k(k+1)(k+2) is divisible  by 6
lets consider 3(k+1)(k+2)
it is clearly divisible by 3
(k+1)(k+2) is even so divisible 2
 so 3(k+1)(k+2) is divisible by 6 too

 hence proved by induction

Answer 2

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

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

∴ n = 3p or 3p + 1 or 3p + 2, where p is some 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, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

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

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

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

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

So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.

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

∴ n (n + 1) (n + 2) is divisible by 6.