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Question 2 (nov.17th)

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Question 2 (nov.17th)

by bacali » Mon Nov 17, 2008 10:26 am
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


OA: D
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by mals24 » Mon Nov 17, 2008 10:45 am
n(n+1)(n+2) will be divisible by 8 in the following 3 cases:

1: n is a multiple of 8

8, 16, 24, 32......96 - total 12 integers

2: n is an even number from 1-96 inclusive.

Total 48 integers. However since these 48 integers already include the above mentioned 12 multiples of 8 we won't include the previous 12 in the calculation in order to avoid redundancy.

3: n is an odd integer which is 1 less than the multiple of 8

7, 15, 31.....95 = 12 integers.

Total number of integers from 1-96 inclusive = 96

Hence total probability = (12+48 )/96 = 60/96 = 5/8.
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