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Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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23 Jun 2006, 15:06

i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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23 Jun 2006, 15:10

You are right..

If I pick 7 then 7 x 8 x 9 will also be divisible by 8....

GMATCUBS21 wrote:

i dont get it, if one of the ways of an odd # is even why cant it be divisible by odd, i mean a number like 7 would be divisible by 8. N+1=8 hence thats at least one odd # tat can be divisible by 8, so i dont understand why its 48.....and cant be any odd #'s

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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23 Jun 2006, 15:16

well u said n is even, and i just told u that n can also be odd, so how do u get 48, please explain......im not as good at math as u...so how do u do this one....its not that u do all even's because i just showed u 7 works....so how do u do it

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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22 Jul 2006, 10:32

If I pick any no from the set 7, 8, 9, 14, 15, 16 ..............94,95,96. then I can satisfy the condition n(n+1)(n+2) divisibel by 8. Such 36 no I can pick so probability is 36/96 So the answer is 3/8.

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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22 Jul 2006, 11:07

I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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22 Jul 2006, 11:16

pravinkawadkar wrote:

I think I am right now. One need to pick up even integer. If 'n' is even and not divisible by 4 then n+2 will be divisible by 4 so n*(n+2) will be divisible by 8. There are total 48 even integers and ans should 1/2. I think I am right now. I need to be careful about trap.

Pravin

When n is
n n+1
7 8
15 16
....
95 96.. total 12.

48+12 = 60 possible outcome. Thus 5/8 => D
_________________

Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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24 Apr 2015, 05:51

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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. \(\frac{1}{4}\) B. \(\frac{3}{8}\) C. \(\frac{1}{2}\) D. \(\frac{5}{8}\) E. \(\frac{3}{4}\)

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

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8; B. \(n+1\) is itself divisible by 8;

(Notice 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.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8.

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