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Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]
21 Jun 2007, 20:05

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

Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]
21 Jun 2007, 22:48

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

Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]
22 Jun 2007, 02:17

[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

Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]
22 Jun 2007, 09:23

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?

Re: probability that n(n + 1)(n + 2) will be divisible by 8? [#permalink]
22 Jun 2007, 09:48

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, 09:53, edited 2 times in total.