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Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]
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]
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]
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]
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]
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]
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]
24 Apr 2015, 05:51
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Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]
24 Apr 2015, 07:26
Expert's post
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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