An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?	Choose the correct option:	A) 1/9	B) 1/3	C) 1/2	D) 2/3	E) 5/6
n(n+1) to be divisible by 3 either n or n+1 must be a multiples of 3.   In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3.   Answer: 2/3.
<result>D</result>
An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?