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Re: If an integer n is to be chosen at random from the integers [#permalink]

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25 Sep 2013, 04:50

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chunjuwu 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

please explain, thank you.

For n(n+1)(n+2) to be divisible by 8, either n has to be even ( because if, n and n+2 are even, then n is divisible by 8) or n+1 as a whole should be divisible by 8 There are 96 values for n. The possibility of n to be even is (96/2) = 48 and the possibility of n to be divisible by 8 is (96/8)= 12 Therefore the total prob = (48+12)/96 = 60/96 = 5/8

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