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Asked by Sparsh

Apr 27, 2015

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

Raghunath Reddy

Member since Apr 11, 2014

Sol;

Let us three consecutive 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.

let 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 that n, n + 1 and n + 2 is always divisible by 3.

â‡’ n (n + 1) (n + 2) is divisible by 3.

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

âˆ´ 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 that n, n + 1 and n + 2 is always divisible by 2.

â‡’ n (n + 1) (n + 2) is divisible by 2.

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

âˆ´ n (n + 1) (n + 2) is divisible by 6.

Let us three consecutive 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.

let 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 that n, n + 1 and n + 2 is always divisible by 3.

â‡’ n (n + 1) (n + 2) is divisible by 3.

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

âˆ´ 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 that n, n + 1 and n + 2 is always divisible by 2.

â‡’ n (n + 1) (n + 2) is divisible by 2.

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

âˆ´ n (n + 1) (n + 2) is divisible by 6.

Syeda

Member since Jan 25, 2017

Answer.

Let us three consecutive 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.

let 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 that n, n + 1 and n + 2 is always divisible by 3.

â‡’ n (n + 1) (n + 2) is divisible by 3.

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

âˆ´ 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 that n, n + 1 and n + 2 is always divisible by 2.

â‡’ n (n + 1) (n + 2) is divisible by 2.

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

âˆ´ n (n + 1) (n + 2) is divisible by 6

Let us three consecutive 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.

let 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 that n, n + 1 and n + 2 is always divisible by 3.

â‡’ n (n + 1) (n + 2) is divisible by 3.

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

âˆ´ 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 that n, n + 1 and n + 2 is always divisible by 2.

â‡’ n (n + 1) (n + 2) is divisible by 2.

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

âˆ´ n (n + 1) (n + 2) is divisible by 6