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

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Senior Manager
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If an integer n is to be chosen at random from integers 1 to [#permalink] New post 05 Jul 2005, 14:30
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B
C
D
E

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If an integer n is to be chosen at random from integers 1 to 96 inclusive, what is the probability that n(n+1)(n-1) will be divisible by 8?

a. 1/4
b. 3/8
c. 1/2
d. 5/8
e. 3/4
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Re: PS - Divisible by 8 [#permalink] New post 05 Jul 2005, 14:48
AJB77 wrote:
If an integer n is to be chosen at random from integers 1 to 96 inclusive, what is the probability that n(n+1)(n-1) will be divisible by 8?

a. 1/4
b. 3/8
c. 1/2
d. 5/8
e. 3/4


Ans is 'D'
When n = even, n(n+1)(n-1) will be divisible by 8
Eg when n = 2,
2*3*4 is div by 8
No of possible even numbers = 48

When n = 1, 7, 15, 23 etc n(n+1)(n-1)is div by 8
Eg when n =1, n+1 = 8 (divisible by 8)
n=7, n+1= 16 (divisible by 8)
No of such odd numbers = 12

Total = 48 + 12 = 60 /96 = 5/8
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Re: PS - Divisible by 8 [#permalink] New post 06 Jul 2005, 01:34
rthothad wrote:
ssap wrote:
Ans is 'D'
When n = even, n(n+1)(n-1) will be divisible by 8
Eg when n = 2, SSAP, it should be 2*1*3 (n)(n-1)(n+1)
2*3*4 is div by 8
No of possible even numbers = 48

When n = 1, 7, 15, 23 etc n(n+1)(n-1)is div by 8
Eg when n =1, n+1 = 8 (divisible by 8)
n=7, n+1= 16 (divisible by 8)
No of such odd numbers = 12

Total = 48 + 12 = 60 /96 = 5/8


Is 0 divisble by 8?


Yes, 0 is divisible by 8.

i started by realizing that my denominator is 2X2X2, so the numerator should consist of at least an even no. (got a little messy though)

This was getting too long for me, and time consuming, so i eliminated. Whatever, the probability should be higher than 0.5, so, A, B and C, off.

Reason: All first 12 factors of 8 (i.e. 8, 16, 24,...,96) are included. That means that the 12 numbers of (n+1) and another 12 of (n-1) are included also. So, i have 12X3 = 36. Between 1 to 10, i have 0,2,3,7,9,10,11,12,13,.... (9 numbers at least) all satisfying it. So, 36+9 = 45 (reaching the half way mark). So, my guess, beween (D) and (E) is (E).

I'm pretty sure i'll be correct on this guess! :lol:
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 [#permalink] New post 06 Jul 2005, 04:30
Oops, 0 (zero) is not in the list because it says 1 - 96. And i just realized that while all odd nos seem to work, all even nos do not work. I'm now getting 45 (no. of odd nos between 1 and 96) + 12 (multiples of 8 between 1 and 96) = 57/96 which is approxmately 5/8.

My ans is (D). :wink:
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 [#permalink] New post 06 Jul 2005, 06:47
OA is D.

The way I approached this is as follows:

n = 1,2...95,96

If n is odd, (n-1)*n*(n+1) is divisible by 8 because either n-1 is divisible by 4 and n+1 is divisible by 2 or vice versa. We have 48 odd values of n between 1 and 96

If n is even, only multiples of 8 have (n-1)*n*(n+1) are divisible by 8. We have 12 such values 8,16,24..88,96

Required Probability = (# of favorable outcomes)/(Total # of outcomes) = 60/96 = 5/8
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 [#permalink] New post 06 Jul 2005, 07:02
AJB77 wrote:
OA is D.

The way I approached this is as follows:

n = 1,2...95,96

If n is odd, (n-1)*n*(n+1) is divisible by 8 because either n-1 is divisible by 4 and n+1 is divisible by 2 or vice versa. We have 48 odd values of n between 1 and 96

If n is even, only multiples of 8 have (n-1)*n*(n+1) are divisible by 8. We have 12 such values 8,16,24..88,96

Required Probability = (# of favorable outcomes)/(Total # of outcomes) = 60/96 = 5/8

AJB, I used the similar approach, but in your statement "If n is odd," you can use use only odd numbers greater than 1 so that results in 47 numbers.
3 5 7 9 - 4
11 13 15 17 19 upto 89 - 5 * 8 = 40
91 93 95 - 3
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 [#permalink] New post 06 Jul 2005, 07:04
rthotad,

0 is divisible by 8, so n=1 can be used.
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 [#permalink] New post 06 Jul 2005, 17:08
rthothad wrote:
AJB77 wrote:
rthotad,

0 is divisible by 8, so n=1 can be used.


Good to know that. Thanks AJB


0 is divisible by everything except 0 offcourse (0/0 is undefined). So make sure u don't forget abt 0 when doing questions on factors.
  [#permalink] 06 Jul 2005, 17:08
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