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probability ques

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by MarkMeier20 » Sun Jul 31, 2011 3:37 am
When is n*(n+1)*(n+2) divisible by 8?

1. When n is divisible by 8 (12 cases)
2. When n+1 is divisible by 8 (12 cases)
3. When n+2 is divisible by 8 (12 cases)
4. When n is divisible by 4 (24 cases)

Therefore n*(n+1)*(n+2) is divisble by 8 in 60 of 96 cases.
60/97=5/8
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by ntamhane » Sun Jul 31, 2011 3:52 am
@MarkMeier20 Did you actually write down the cases? if so how much time did you spend?

Or is there a shorter way?
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by Frankenstein » Sun Jul 31, 2011 4:04 am
Hi,
n is either even or odd
n is even -> n=2k. So, n(n+1)(n+2) = 2k(2k+1)(2k+2) = 4k(k+1)(2k+1).
k and k+1 are consecutive integers. So, one of them will always be even . Hence k(k+1) is even.
So, whenever n is even, n(n+1)(n+2) is multiple of 8.
Even numbers between 1 and 96 are 96/2 = 48

n is odd -> n is odd, (n+2) is odd. So, for n(n+1)(n+2) to be a multiple of 8, (n+1) should be a multiple of 8.
So, n+1 can be 8,12,..96 -> 12

So, probability is (48+12)/96 = 5/8.
Cheers!

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by GMATGuruNY » Sun Jul 31, 2011 4:28 am
I posted a solution here:

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