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From the integers 1, 2, 3,..., 2004, two distinct numbers x and y are selected at random without replacement. What is the probability that x³ + y³ is divisible by 3?
- 1/3
- 2/3
- 1/6
- 1/4
Correct answer: 1/3
Solution
Since n^3 = n (mod 3), x^3+y^3 is divisible by 3 iff x+y is. Among 1..2004 there are 668 numbers in each residue class 0,1,2. Favourable unordered pairs = C(668,2) (both =0) + 668*668 (one =1, one =2). Dividing by C(2004,2) gives exactly 1/3.
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