Koi bhi problem ho question dalo
Let the 2 consecutive positive integer be x, x+1.Product of these consecutive integers be x(x+1).1. evenLet x=2kProduct = 2k(2k+1) From above equation it is clear that the product is divisible by 2.2. oddLet x=2k+1Product = (2k+1)[(2k+1)+1]From above equation it is clear that the product is divisible by 2

Let the two consecutive positive integers be x and x + 1Product of two consecutive positive integers = x (x + 1)= x2 + xCase (i): x is even numberLet x = 2k⇒x2 + x = (2k)2 + 2k = 4k2 + 2k= 2k ( 2k + 1) Hence the product is divisible by 2Case (ii): x is odd numberLet x = 2k + 1⇒ x2 + x = (2k + 1)2 + (2k + 1) = 4k2 + 4k + 1 + 2k + 1= 4k2 + 6k + 2\xa0= 2 (2k2 + 3k + 1) Clearly the product is divisible by 2From the both the cases we can conclude that the product of two consecutive positive integers is divisible by 2.