prove that product of 3 consecutive positive integers is always divisible by 6

Let three consecutive positive 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.

∴

*n*= 3*p*or 3*p*+ 1 or 3*p*+ 2, where*p*is some integer.If

*n*= 3*p*, then*n*is divisible by 3.If

*n*= 3*p*+ 1, then*n*+ 2 = 3*p*+ 1 + 2 = 3*p*+ 3 = 3(*p*+ 1) is divisible by 3.If

*n*= 3*p*+ 2, then*n*+ 1 = 3*p*+ 2 + 1 = 3*p*+ 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, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴

*n*= 2*q*or 2*q*+ 1, where*q*is some integer.If

*n*= 2*q*, then*n*and*n*+ 2 = 2*q*+ 2 = 2(*q*+ 1) are divisible by 2.If

*n*= 2*q*+ 1, then*n*+ 1 = 2*q*+ 1 + 1 = 2*q*+ 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.Since, *n* (*n* + 1) (*n* + 2) is divisible by 2 and 3.

∴

*n*(*n*+ 1) (*n*+ 2) is divisible by 6.
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