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If an integer n is to be chosen at random from the integers

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Senior Manager
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If an integer n is to be chosen at random from the integers [#permalink] New post 20 Feb 2007, 06:28
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A
B
C
D
E

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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

Correct Answer: D

Thinking naively, I thought the answer was B.

It took me more than 10 min to get to the correct answer. (The answers don't have explanations so I have to figure it out)

Would someone be able to solve this prob in 2~3 min??

Please explain.
Merci beaucoup in advance!

I need help in number theory :?
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 [#permalink] New post 20 Feb 2007, 06:53
Find out how can be n * n+1 * n +2 be divisble by 8.

I think it can be if n is divisble by 4.

With this I get the answer as A which is not OA.
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 [#permalink] New post 20 Feb 2007, 07:03
D

When n is even: n*(n+1)*(n+2) is always divisible by 8

number of even numbers = 96/2 = 48

When n is odd : n*(n+1)*(n+2) is divisible by 8,
only if (n+1) is div. by 8
96/8 = 12

Req. prob. = ( 48+12)/96 = 5/8
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 [#permalink] New post 20 Feb 2007, 07:19
n is a multiple of 2 accounts for 48 numbers. We also need to consider cases where (n+1) is 8. So these results have n being odd of 7,15,23.....95. These gives (96-8)/8 + 1 = 12. So total 60 numbers.

D is the answer.
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 [#permalink] New post 20 Feb 2007, 16:12
Wow...

I hadn't thought of that. That when n is even, n(n+1)(n+2) is divisible by 8. That makes the prob so much easier!

Thanks!

:-D
  [#permalink] 20 Feb 2007, 16:12
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If an integer n is to be chosen at random from the integers

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