Question
Sun April 17, 2011 By:

# prove that the product of 3 consicutive positive interger is divisible by 6 ?

Mon April 18, 2011

Any positive integer is of the form 3q, 3q + 1 or 3q + 2 for some integer q.

Let a, a + 1, a + 2 be any three consecutive integers.

Case 1: a = 3q

a (a + 1) (a + 2) = 3q (3q + 1) (3q + 2) = 3q (even number, say 2t) = 6qt

[Since, product of 3q + 1 and 3q + 2 being the product of consecutive integer is even]

Case 2: a = 3q + 1

a (a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3) = (even number, 2t) (3) (q + 1) = 6(qt + t)

Case 3: a = 3q + 2

a (a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4) = multiple of 6, for every q

Hence, the product of three consecutive positive integers is divisible by 6.

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