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Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

\(n(n+1)(n+2)\) is divisible by 8 in two cases:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.

This was difficult. I was thinking either N can be multiple of 8 or n+1 can be multiple of 8 pr n+2 a multiple 8. With this I just got 36 choices. I missed sequences like 2*3*4
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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

\(n(n+1)(n+2)\) is divisible by 8 in two cases:

1. When \(n\) is even: \(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}\)

Answer: D.

Thanks for such a good explanation !!!

I was stuck in this for more then 3 minutes and couldn't find the solution.
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Re: If an integer n is to be chosen at random from the integers [#permalink]

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01 Nov 2013, 04:45

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

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17 Oct 2014, 07:23

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

What are the cases when \(n(n+1)(n+2)\) is divisible by 8?

This means that \(n(n+1)(n+2)\) must have 2*2*2 in its prime factorization.

If n is even, we'll have 2 even numbers in \(n(n+1)(n+2)\), one of which is divisible by 4. This means that \(n(n+1)(n+2)\) is divisible by 8. Therefore, our probability is at least 1/2. We can eliminate choices A,B, and C immediately.

The other possibility is if we have a multiple of 8 in the middle. This means that n will have to be odd, so we're not considering overlapping cases. This happens exactly 1/8 of the time, or (96/8) = 12 times.

1/2 + 1/8 = 5/8.

Answer: D.

gmatclubot

Re: If an integer n is to be chosen at random from the integers
[#permalink]
17 Oct 2014, 07:23

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