deepakb wrote: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
Even*odd*even = multiple of 8:
Given 3 consecutive integers {even, odd, even}, the product will always be a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.
n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8 (making an odd integer that is 1 less than a multiple of 8).
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying 12 more favorable choices for n.
Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.
The correct answer is
D.
Last edited by
GMATGuruNY on Sun Jul 31, 2011 4:26 am, edited 1 time in total.
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