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If an integer n is to be chosen at random from the integers

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If an integer n is to be chosen at random from the integers [#permalink] New post 30 Sep 2013, 19:59
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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

Last edited by Bunuel on 01 Oct 2013, 00:07, edited 1 time in total.
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Re: what is the probability that n(n + 1)) will be divisible by [#permalink] New post 30 Sep 2013, 20:40
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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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Re: what is the probability that n(n + 1)) will be divisible by [#permalink] New post 01 Oct 2013, 00:07
Expert's post
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


Similar question to practice: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html

Hope this helps.
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Kudos [?]: 2 [0], given: 15

Re: If an integer n is to be chosen at random from the integers [#permalink] New post 01 Oct 2013, 10:11
Hi Karishma/Bunuel,

I was approaching this question in this ways:

since in 1 to 96 ---- 32 number will be divisible by 3.

in n(n+1) there will be another 16 number which will be divisible by 3

Hence 32+16 = 48

so the probability should be 48/96 = 1/2.

Please help me in understanding what i am doing wrong

Thanks,
Ankur
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Re: If an integer n is to be chosen at random from the integers [#permalink] New post 01 Oct 2013, 18:40
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ankurjohar wrote:
in n(n+1) there will be another 16 number which will be divisible by 3



I don't understand how you got this.

You are right that in 1 to 96, there are 32 numbers divisible by 3. When you pick 2 consecutive numbers out of 1 to 96, 32 times n will be a multiple of 3 and 32 times (n+1) will be a multiple of 3. Hence n*(n+1) will be a multiple of three 64 times out of the 96 times you can pick a different value of n.

1* 2
2* 3
3* 4
4* 5
5* 6
6* 7

Note that for 6 different values of n, n(n+1) is divisible by 3 four times.
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Re: If an integer n is to be chosen at random from the integers   [#permalink] 01 Oct 2013, 18:40
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