I need help in understanding to solve these kind of question.
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1) will be divisible by 3?
Any help will be much appreciated. sorry I, dont have right answer for this question.
n(n+1) will be divisible by 3 if either n or (n+1) is divisible by 3. Note that both cannot be divisible by 3 at the same time.
Every number can be written as 3a or 3a+1 or 3a+2.
1 is of the form 3a+1
2 is of the form 3a + 1
3 is of the form 3a
4 is of the form 3a + 1
and so on...
If n is either 3a or 3a+2 (which means that (n+1) will be of the form 3a + 3 which is same as 3a), n(n+1) will be divisible by 3. This means out of 3 consecutive integers, 2 values of n will make n(n+1) divisible by 3. We have 32 complete groups of 3 consecutive integers each. (1,2 ,3), (4, 5, 6,) etc.
So out of every group, 2 values will make n(n+1) divisible by 3.
Hence probability that n(n+1) is divisible by 3 is 2/3.
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