Two numbers x and y are chosen at random (without replacement) from amongst the numbers 1, 2, 3,……..2004. the probability that x3+y3 is divisible by 3 is

We divide the number in three groups 3k+1   type {1,4,7,...........,2005} 3k+2   type {2,5,8,...........,2006} 3k+3   type {3,6,9,...........,2007} x3+y3  is divisible by 3. If x and y both belong to 3rd group or one of them belongs to the first group and the other to the second group. So favorable number of cases =669C2+669×669 Total number of cases =2007C2 ∴ desired probability 669×6682+669×6692007×20062=669×20062007×2006=13