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## Probability

alex.gellatly GMAT Destroyer!
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Probability Mon May 21, 2012 7:59 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
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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Anurag@Gurome GMAT Instructor
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Mon May 21, 2012 8:09 pm
alex.gellatly 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

Case 1: n is even
Let us assume that n = 2k, for any integer k.
Then n(n + 1)(n + 2) = 2k(2k + 1)(2k + 2) = 4k(2k + 1)(k + 1)
Now either k or k + 1 will be even, so 8 will be a multiple of n(n + 1)(n + 2).

Number of even integers between 1 ans 96, inclusive = {(96 - 2)/2} + 1 = 48

Case 2: If n + 1 is divisible by 8
n + 1 = 8a, where a ≥ 1
n = 8a - 1
8a - 1 ≤ 96
8a ≤ 97
a ≤ 12.1 implies 12 integers.

Also, when n and n + 2 are even, n + 1 will be odd, and when n + 1 is divisible by 8, then n and n + 2 will be odd. This means, there is no repetition.

Total integers = 48 + 12 = 60

Therefore, probability that n(n + 1)(n + 2) will be divisible by 8 = 60/96 = 5/8

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Mon May 21, 2012 8:15 pm
n(n+1)(n+2) should be divisible by 8.

If n is even then it is divisible by 8, so we have 48 even numbers.

If n is 7, then n+1 is divisible by 8. Similar manner all odd numbers having n+1 as multiple of 8 will be divisible by 8. 96/8 = 12 multiples of 8 will be available.

So 48+12/96 = 60/96 = 5/8

IMO D.

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