Prove that sum of three consecutive number is divisible by 6?

1.	Prove that sum of three consecutive number is divisible by 6?
Answer» Let three consecutive positive integers be, n, n + 1 and n + 2.\'When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.\xa0∴ 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, ⇒\xa0n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.If n = 3p + 2, ⇒\xa0n + 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, when 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\xa0⇒\xa0n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.If n = 2q + 1 ⇒\xa0n + 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.Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6.

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