If two distinct numbers m and n are chosen at random form the set {1, 2, 3, …, 100}, then find the probability that `2^(m) + 2^(n) + 1` is divisible by 3.

Answer

`(49)/(198)`

Explanation

`2^(m) + 2^(n) + 1 = (3 - 1)^(m) + (3 - 1)^(n) + 1`
`=3k + (-1)^(m) + (-1)^(n) + 1`
This is divisible by 3 if both m and n are even.
`therefore` Required probabaility = `(.^(50)C_(2))/(.^(100)C_(2)) = (50 xx 49)/(100 xx 99) = (49)/(198)`