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probability that n(n + 1)(n + 2) will be divisible by 8?

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by mridula » Fri Feb 20, 2009 12:40 pm
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


How do we solve this? OA is D
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by dendude » Fri Feb 20, 2009 12:50 pm
I got B. 3/8.
From 1 to 96, there are 12 numbers divisible by 8
i.e 8, 16, 24...etc

n * (n+1) * (n+2) will be divisible by 8 if, n OR n+1 OR n+2 is divisible by 8
There are 12*3 = 36 possibilities for that.

Hence, 36/96 = 3/8
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by mals24 » Fri Feb 20, 2009 12:55 pm
Check this out:

https://www.beatthegmat.com/probability- ... 20261.html
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by ontopofit » Fri Feb 20, 2009 10:53 pm
as we see for a repitition with 8 we shud first see how many such n are thea b/w 1 and 8. here u get this to be 5 so total number of n = 96/8 *(5) = 60.

pbability = 60/96 = 5/8
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