ankurjohar wrote:

Hi Folks,

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.

Thanks,

Ankur

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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