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

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VP
Joined: 26 Apr 2004
Posts: 1209
Location: Taiwan
If an integer n is to be chosen at random from the integers [#permalink]

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28 Aug 2004, 10:58
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Difficulty:

95% (hard)

Question Stats:

40% (02:07) correct 60% (01:51) wrong based on 120 sessions

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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

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 25 Sep 2013, 04:53, edited 1 time in total.
Edited the question and added the OA
Manager
Joined: 10 Sep 2013
Posts: 80
Re: If an integer n is to be chosen at random from the integers [#permalink]

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25 Sep 2013, 04:50
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chunjuwu 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

For n(n+1)(n+2) to be divisible by 8, either n has to be even ( because if, n and n+2 are even, then n is divisible by 8) or n+1 as a whole should be divisible by 8
There are 96 values for n. The possibility of n to be even is (96/2) = 48 and the possibility of n to be divisible by 8 is (96/8)= 12
Therefore the total prob = (48+12)/96 = 60/96 = 5/8

The OA is D
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Kudos if I helped

Senior Manager
Joined: 19 May 2004
Posts: 291

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29 Aug 2004, 09:33
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I pick D. 5/8

Every threesome where the middle number is odd works
For example
2,3,4 Works.
4,5,6 Works.

Those account for 1/2 of the cases.

But remember that evert threesome where the middle number is divisible by 8 also works!
Example:
7,8,9 Works.
15,16,17.

Those account for 12/96 of the cases which is: 1/8.

1/2 + 1/8 = 5/8.
Senior Manager
Joined: 05 Feb 2004
Posts: 290
Location: USA

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28 Aug 2004, 14:31
Only even values of "n" satisfy and there are 48 even numbers between 1 to 96 inclusive....hence prob = 48/96 = 1/2....??
Senior Manager
Joined: 25 Jul 2004
Posts: 273

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28 Aug 2004, 18:38
Im gonna guess 3/4's.
If it has a multiple of eight, it works (obviously),
or if it has a multiple of 4 and an even number.

It is definately more than 1/2.
VP
Joined: 26 Apr 2004
Posts: 1209
Location: Taiwan

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29 Aug 2004, 02:16
Hi, SigEpUCI

In my opinion, if we randomly choose an interger which is the multiple of 2, n (n+1) (n+2) must be divided by 8.

Hence, the probability, I supposed, is 48/96 = 1/2.

Anything wrong?
VP
Joined: 26 Apr 2004
Posts: 1209
Location: Taiwan

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30 Aug 2004, 07:20
Dookie wrote:
I pick D. 5/8

Every threesome where the middle number is odd works
For example
2,3,4 Works.
4,5,6 Works.

Those account for 1/2 of the cases.

But remember that evert threesome where the middle number is divisible by 8 also works!
Example:
7,8,9 Works.
15,16,17.

Those account for 12/96 of the cases which is: 1/8.

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

well done, Dookie,

thank you
SVP
Joined: 16 Oct 2003
Posts: 1801

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02 Sep 2004, 09:58
I got 1/2.

Excellent explaination Dookie.
Math Expert
Joined: 02 Sep 2009
Posts: 39744
Re: If an integer n is to be chosen at random from the integers [#permalink]

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25 Sep 2013, 04:53
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   [#permalink] 25 Sep 2013, 04:53
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