Show that the product of 3 consecutive numbers are divisible by 6.

1.	Show that the product of 3 consecutive numbers are divisible by 6.
Answer» Thank you!! ✌?? Let three consecutive positive integers be, n, n + 1 and n + 2.\xa0When 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.\xa0If n = 3p + 1,\xa0⇒\xa0n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.\xa0If n = 3p + 2,\xa0⇒\xa0n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.\xa0So, 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.\xa0Similarly, when a number is divided 2, the remainder obtained is 0 or 1.\xa0∴ n = 2q or 2q + 1, where q is some integer.\xa0If n = 2q\xa0⇒\xa0n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.\xa0If n = 2q + 1\xa0⇒\xa0n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.\xa0So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.\xa0⇒ n (n + 1) (n + 2) is divisible by 2.\xa0Hence n (n + 1) (n + 2) is divisible by 2 and 3.∴ n (n + 1) (n + 2) is divisible by 6.\xa0

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