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

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

(A) 1/3

(B) 2/3

(C) 1/6

(D) 1/4