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

## Probability

##### This topic has expert replies

- neerajkumar1_1
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- GMATGuruNY
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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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- neerajkumar1_1
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n(n+1)(n+2) represent 3 consec numbers... and we can divide all the 3 consec number series in even ones and odd ones...

if n is even, then (n+2) will be even, and the product will be divisible by 8

therefore half the series will be multiple of 8... i.e 96/2 = 48

also if n is odd, then (n+2) will be odd, so (n+1) all by itself will have to be a multiple of 8

therefore 96/8 = 12 multiples of 8, which will be exactly the middle number...

hence we have 48 + 12 = 60 series which will be a multiple of 8 out of a total of 96

=60/96 = 5/8