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Junior | Next Rank: 30 Posts
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by apuso » Mon Jan 18, 2010 8:30 pm
help required for the below 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)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
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Junior | Next Rank: 30 Posts
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by diegow77 » Mon Jan 18, 2010 9:30 pm
this is definitely a hard one, I'd have to guess

what I did first was did
123= not divisible
234= 24
345= not divisible
456=120
567= not divisible
678=336
789= not divisible
8910=720

I got a ratio of 1/2. I only suspect that as the numbers get bigger it would only make it less divisible to 8 meaning that any probability higher than 1/2 is false and can be eliminated.

I'm not sure if this pattern continues throughout integers past 10.

What is the answer by the way?
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by ace_gre » Mon Jan 18, 2010 10:31 pm
Hi, Here is my approach..

Consider the case when n is even,
n * n+1 * n+2 ==>n and n+2 is even and is divisible by 8.
From 1-96, there are 48 even integers for n.

Also n *n+1 * n+2 is divisible by 8 when n+1 is divisible by 8. ( hence n is odd)
There are 12 such cases.

So total probability = (48 + 12) / 96 = 5/8

IMO D
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by apuso » Tue Jan 19, 2010 7:21 am
the correct ans is option D. 5/8

thanks ace_gre for the explanation!!!
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