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

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

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21 Jun 2007, 20:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

OA: D

Last edited by humtum0 on 21 Jun 2007, 21:12, edited 1 time in total.
VP
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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21 Jun 2007, 21:05
humtum0 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. 4/1
B. 8/3
C. 2/1
D. 8/5
E. 4/3

OA: D

I think you have the ratio reverse....
I get 5/8...
Here is how...
Case1: (odd)(even)(odd) => instances when the even number is divisible by 8 = 96/8 = 12
Case2: (even)(odd)(even) => Because any even number you pick, this will always be divisible by 8, instances when picking even number = 96/2 = 48

P = (12+48) / 96 = 5/8
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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21 Jun 2007, 22:53
humtum0 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

OA: D

D. it sahould be > 1/2. D is only choice >1/2.
VP
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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21 Jun 2007, 23:24
Himalayan wrote:
humtum0 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

OA: D

D. it sahould be > 1/2. D is only choice >1/2.

Also, E is greater than 1/2
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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21 Jun 2007, 23:48
bkk145 wrote:

I think you have the ratio reverse....
I get 5/8...
Here is how...
Case1: (odd)(even)(odd) => instances when the even number is divisible by 8 = 96/8 = 12
Case2: (even)(odd)(even) => Because any even number you pick, this will always be divisible by 8, instances when picking even number = 96/2 = 48

P = (12+48) / 96 = 5/8

Perfect !

Director
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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22 Jun 2007, 03:17
[quote="humtum0"]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

Probability = Fav. Outcomes/Total Possible Outcomes

Total outcomes for n = 96-1+1 = 96

Favorable outcomes:

Case1:
n(n+1)(n+2) is divisible by 8 whenever n is even.
# of even integers bet. 1 & 96 = 96/2= 48
=> n can take 48 possible values

Case2:
n(n+1)(n+2) is divisible by 8 whenever (n+1) is divisible by 8 & (n+1) is divisible by 8 whenever (n-1) is 1 less than a multiple of 8 bet. 1 & 96.

=> (n+1) is divisible by 8 whenever n = 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, => n can take 12 additional possible values

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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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22 Jun 2007, 10:23
bkk145 wrote:
Case1: (odd)(even)(odd) => instances when the even number is divisible by 8 = 96/8 = 12

Please help me understand this a little better. How did you derive the first case? How did you know there would be 8 instances where odd*even*odd = 12?
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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22 Jun 2007, 10:48
hd54321 wrote:
bkk145 wrote:
Case1: (odd)(even)(odd) => instances when the even number is divisible by 8 = 96/8 = 12

Please help me understand this a little better. How did you derive the first case? How did you know there would be 8 instances where odd*even*odd = 12?

since the three numbers are consecutive then:

odd*even*odd

n=8*1 then 7*8*9
n=8*2 then 15*16*17
n=8*3 then 23*24*25

n=8*5 then 39*40*41

and so on...

even*odd*even

n=3 then 2*3*4
n=5 then 4*5*6

and so on...

Last edited by KillerSquirrel on 22 Jun 2007, 10:53, edited 2 times in total.
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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22 Jun 2007, 10:51
KillerSquirrel wrote:
since the three number are consecutive then:

odd*even*odd

n=8*1 then 7*8*9
n=8*2 then 15*16*17
n=8*3 then 23*24*25
n=8*4 then 39*40*41

Brain fart. I think a collective "duh" is in order. Thanks!
Director
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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]

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22 Jun 2007, 12:47
bkk145 wrote:
Himalayan wrote:
humtum0 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

OA: D

D. it sahould be > 1/2. D is only choice >1/2.

Also, E is greater than 1/2

from 1 to 10, 6 are divisible by 8 and the pattren repeats. so it should be 6/10 which is >1/2. so it is 5/8.

Now I realize that your approach is cool.
Re: probability that n(n + 1)(n + 2) will be divisible by 8?   [#permalink] 22 Jun 2007, 12:47
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