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If n is an integer from 1 to 96, what is the probability for

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If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 14 Nov 2011, 04:56
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If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

a. 25%
b. 50%
c. 62.5%
d. 72.5%
e. 75%

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 n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 14 Nov 2011, 05:50
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Total possible ways = 96

Out for all even values of n starting from n=2 to n=96, the expression is divisible by 8.

There are 48 even values and 48 odd values between 1 and 96.

And for n= odd number, n+1 i.e the second term alone is going to be even. And the term must be a multiple of 8 for it to be divisible by 8. So totally we have 12 possibilities of n+1 values for n = odd to be between n=1 and n=95.

So, total number of possible values divisible by 8 = 48(for even n) + 12( for odd n ) = 60
Probability = 60/96 *100% = 62.5%
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 14 Nov 2011, 07:43
For the sake of avoiding redundancy, I second ksp's explanation. I had the same.
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 15 Nov 2011, 20:43
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GMATmission wrote:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

a. 25%
b. 50%
c. 62.5%
d. 72.5%
e. 75%


There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/Pattern Recognition/Intuition - whatever you may want to call it.

We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.

Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:
1*2*3 N
2*3*4 Y
3*4*5 N
4*5*6 Y
5*6*7 N
6*7*8 Y
7*8*9 Y
8*9*10 Y

5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5%
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 16 Nov 2011, 00:16
Wow. That is really helpful karishma. !

Thanks.
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 05 Jan 2012, 00:21
Wonderful explanations ksp and Karishma. Thanks a ton.
Used a logic similar to that used by ksp. Got C 62.5%).
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 29 Aug 2014, 03:27
VeritasPrepKarishma wrote:
GMATmission wrote:
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8?

a. 25%
b. 50%
c. 62.5%
d. 72.5%
e. 75%


There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/Pattern Recognition/Intuition - whatever you may want to call it.

We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.

Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:
1*2*3 N
2*3*4 Y
3*4*5 N
4*5*6 Y
5*6*7 N
6*7*8 Y
7*8*9 Y
8*9*10 Y

5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5%


to KSP:
YOUR startegy is is helpful for those with 1-90/1-80/1-1000 integers but not for 1-96 , that time total items will be 76 and 8 is divided suc items are 47 then 47/76= 0.618=61.8%. which is not the result
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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New post 29 Aug 2014, 03:33
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Re: If n is an integer from 1 to 96, what is the probability for  [#permalink]

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Re: If n is an integer from 1 to 96, what is the probability for &nbs [#permalink] 13 Jan 2019, 13:23
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