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 :

with
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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 :