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simplythebest
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 Posted: Thu Jun 14, 2007 1:34 am Post subject: Set Question : Probality |
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| A random number N is chosen from 1 to 96 inclusive, what is the probability that the number N(N+1)(N+2) is divisible by 8? |
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jayhawk2001
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| Posted: Thu Jun 14, 2007 1:51 pm Post subject: Re: Set Question : Probality |
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| simplythebest wrote: | | A random number N is chosen from 1 to 96 inclusive, what is the probability that the number N(N+1)(N+2) is divisible by 8? |
5/8 ?
If n is even, then n*(n+2) will yield a product which is a multiple of 8. So,
1/2 the numbers will satisfy this.
If n is odd, the condition will be satisfied when n+1 is a multiple of 8.
There is 1 such number for every 4 consecutive odd numbers (7,
15, 23 etc.).
So, total = 1/2 + 1/2*1/4 (where 1/2 is probability of odd and
1/4 is probability of n+1 being multiple of 8 in the list of odd integers).
So, 1/2 + 1/8 = 5/8 |
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800GMAT
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| Posted: Thu Jun 14, 2007 8:31 pm Post subject: |
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agree with 5/8
all even numbers will satisfy the given condition
i.e. 96/2 = 48
when n is odd, multiples of 8 will satisfy the given condition i.e. 12 multiples
(48+12)/96 = 5/8 |
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800GMAT
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| Posted: Sun Jun 17, 2007 10:23 pm Post subject: |
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| what is the OA........... |
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AMalik
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| Posted: Mon Jun 18, 2007 1:46 am Post subject: |
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Hi,
From 1 to 10, there are 5 sets, which are divisible by 8.
(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)
So till 96, there will be 12 * 5 such sets = 60 sets
so probability will be 60/96=5/8 |
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simplythebest
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| Posted: Mon Jun 18, 2007 2:26 am Post subject: |
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| Perfect... |
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