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If an integer n is to be chosen at random from the integers 1 to 96,

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

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New post 23 Jun 2006, 15:39
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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. \(\frac{3}{4}\)

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Apr 2015, 08:25, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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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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New post 23 Jun 2006, 15:57
Messed up this ... deleting.

Last edited by haas_mba07 on 23 Jun 2006, 16:29, edited 2 times in total.
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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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New post 23 Jun 2006, 16:06
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
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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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New post 23 Jun 2006, 16:10
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
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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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New post 23 Jun 2006, 16:16
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
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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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New post 23 Jun 2006, 16:41
I think I finally got the right one.. although my first answer 1/2 still remains the same.

Between 1-10 : you have 5 numbers you can pick to get nx (n+1) x (n+2) divisible by 8 : {2, 4, 6, 7, 8}

Continuing this from 10-20 : 5 more and so on.

In total from 1-90 you have 9x5 = 45 numbers.

Between 91-96 you can pick 3 more to get the satisfying condition: 92, 94, 96

Therefore in total you can pick : 48
Probability = 48/96 = 1/2
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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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New post 22 Jul 2006, 10:52
Is N(N+1)(N+2) multiple of 8?

1 2 3
2 3 4
3 4 5
.....
7 8 9

When N is even number, "N(N+1)(N+2)" is always multiple of 8.
Thus probability = 48/96 = 1/2

When N is odd number and N+1 is multiple of 8, N(N+1)(N+2) is multiple of 8
Thus, probability = 12/96 = 1/8

1/2+1/8 = 5/8
or
(48+12)/96 = 5/8

Ans is D.
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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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New post 22 Jul 2006, 11:32
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.

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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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New post 22 Jul 2006, 11:35
The set should be 6,7,8, 14,15,16.....94,95,96. Ans is same. I commit typing mistake.

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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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New post 22 Jul 2006, 12:07
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.

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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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New post 22 Jul 2006, 12:16
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
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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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New post 22 Jul 2006, 17:09
GMATCUBS21 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



D IMO. If n>=2 then for any even n, n(n+1)(n+2) is divisible by 2 and 4, and therefore by 8.

Further, every 8th number will be divisible by 8
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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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New post 23 Jul 2006, 01:52
5/8

For all N = even N(N+1)(N+2) will be divisible by 8: Total = 48
For all N+1 = multiple of 8, N(N+1)(N+2) will be divisible by 8: Total = 12

Prob = 60/96 = 5/8
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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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New post 26 Jul 2006, 03:34
For all even numbers N . n(n+1)(n+2) is divisible by 8. Even numbers = 48

For all odd numbers
From 1-10 - 1 odd number
2- 10 - 1 odd number

90-96 - 1

hence total odd numbers = 12

Total numbers = 60

Probabiltiy = 60/96 = 5/8
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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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Re: If an integer n is to be chosen at random from the integers 1 to 96, [#permalink]

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New post 24 Apr 2015, 08:26
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.

Answer: D.

Similar question: divisible-by-12-probability-121561.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
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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, 08:26
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