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
Answer : option (A)
Solution :
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
with
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