Question

14. Using Euclid's Division Lemma, prove that the product of three consecutive natural numbers is always divisible by 3.

Answer 1

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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.

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


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# Nikky