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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 19 Sep 2008, 09:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)(n + 2) will be divisible by 8?

A.1/4
B.3/8
C.1/2
D.5/8
E.3/4

Last edited by Nihit on 19 Sep 2008, 09:53, edited 1 time in total.
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Re: integer n [#permalink] New post 19 Sep 2008, 09:50
Buddy,

I think you forget to write answers in options A, B, C, D and E.

By the way, I come up with anser 1/2.
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Re: integer n [#permalink] New post 19 Sep 2008, 09:56
Sorry..

It is 5/8. I miss some number.. :oops:
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Re: integer n [#permalink] New post 19 Sep 2008, 09:57
thanks mate, yeah E is the answer , am somehow getting 1/2 as well. Can u explain ?
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Re: integer n [#permalink] New post 19 Sep 2008, 10:35
I'm sure I'd get it wrong in real q though I'd think it is a bit suspicious to have a simple 1/2.

It's 5/8, or 60/96. From the net and simple calculation, we could get easy 48. However, we must notice that there are more numbers that are divisible by 8 under the term of (n)(n+1)(n+2). We have 12 more. For instance, 14x15x16, 15x16x17, 16x17x18...all three can be divided by 8.
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Re: integer n [#permalink] New post 19 Sep 2008, 13:15
probably would get this wrong in exam but..
all even numbers : n(n+1)(n+2) seem to be divisible by 8.(48 in all)
all odd numbers, every odd number that is followed by a multiple of 8 will be divisble by 8, so we have 7,15,23,31,39,47,55,63,71,79,87,95. (12 in all)
so probability = (48 + 12) / 96 = 60/96 = 5/8
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Re: integer n [#permalink] New post 19 Sep 2008, 14:32
i get 5/8..

its like this..we know that every 1 number in 8 numbers will be divisible by 8...right.

so if n=8 then the whole equation is divisible by 8..if (n+1) is 8 then the same thing and so on..

so basically in 8 numbers we know 3 numbers will make the whole stem divisible by 8..

but now lets see there are 2 such numbers that are not in the 8--16 range that would make the stem divisible they are when n=2 or n=4

so total numbers is 5/8
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Re: integer n [#permalink] New post 19 Sep 2008, 14:56
Nihit wrote:
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)(n + 2) will be divisible by 8?

A.1/4
B.3/8
C.1/2
D.5/8
E.3/4


3 consecutive integers are devesible by 8 , if one is a multiple of 4 and anothoer is a multiple of 2 , or one of them is a multiple of 8 . i think this is the key ..
Re: integer n   [#permalink] 19 Sep 2008, 14:56
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If an integer n is to be chosen at random from the integers

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