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Asked by Sparsh

Apr 27, 2015

Prove that the product of three consecutive positive integer 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

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.