If (x,y) is a subset of the first 30 natural numbers, then the probability, that `x^(3)+y^(3)` is divisible by 3, is `(S)/(9)` then `S=`

Let S be the set of the first 21 natural numbers, then the probability of choosing {x, y}inS , such that x^3+y^3 is divisible by 3, is

x^(3^(n))+y^(3^(n)) is divisible by x+y, if

Two number a and b aer chosen at random from the set of first 30 natural numbers.Find the probability that a^(2)-b^(2) is divisible by 3 .

Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that a^(2)-b^(2) is divisible by 3 is:

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of then are divisible by 2 or 3, is

If 4 distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all 4 of them are either divisible by 3 or divisible by 5 is

Two natural numbers x and y are chosen at random. What is the probability that x^(2) + y^(2) is divisible by 5?