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sudhagar
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 09:39
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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
25%
50%
62.5%
72.5%
75%



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sudhagar
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 09:40
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I found the answer, but it took me more time. Looking for any shortcuts?
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laxieqv
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 09:54
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This was discussed here: https://www.gmatclub.com/phpbb/viewtopic ... ht=#136703
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automan
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 09:56
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All the products starting by an even integer are divisible by 8. And all the products starting by an odd number are not divisible by 8. And all products starting by 7, 15, 23, etc are also divisble by 8.

1*2*3 is not divisible by 8
2*3*4 is divisible by 8
3*4*5 is not divisible by 8
4*5*6 is divisible by 8
5*6*7 is not divisible by 8
6*7*8 is divisible by 8
7*8*9 is divisible by 8
8*9*10 is divisble by 8

so the total number of divisible numbers is 96/2 + 96/8.

The probablity is 60/97, is C
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olorin
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 12:24
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I am not good with series questions.

This proves it.
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sudhagar
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If n is an integer from 1 to 96, what is the probability for 11 Oct 2005, 13:25
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automan: Thats the same method I employed. Looks like I need to be quick at the same time being careful too.



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