If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?
A)25%
B)50%
C)62.5%
D)72.5%
E)75%
Friday, December 28, 2007
QUESTION#12
Posted by Gaurav & Kunal at 10:41 PM
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4 Comments:
According to me, option E) 75%.
No real method for this one, I thought if we do it for the first 8 numbers, we can generalize it for all sets of 8.
I found that for any n*(n+1)*(n+2) in 1 to 8, there are 2 combinations for which number is not divisible by 8.
Moreover 96 is also divisible by 8, hence we have complete 12 sets, in each of which, 6 out of 8 combinations are divisible by 8.
Hence, answer is 3/4 or 75%.
If n is even n(n+1)(n+2) is always divisible, they are = 48
If n is odd n(n+1)(n+2) is divisible by 8 they are = 12.
So total no.s are 48+12 = 60.
The probability is = 60/96 = 62.5.
Hence the answere is (C).
Correct me if I am wrong.
I think C should be the answer
same approach as explained above by Soloman
Soloman if n is even n(n+1)(n+2) is always divisible, they are = 48 ???? How did u find this 48 ??
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