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What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1

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What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 05:18
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What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Source: http://www.GMATinsight.com

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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 05:48
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Source: http://www.GMATinsight.com

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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 05:54
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Source: http://www.GMATinsight.com

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:06
vasuca10 wrote:
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8

Thanks, but I was not able to understand what evenly divisible meant, thats why I chose to ignore that part.
I think that I get it now, I guess that means that it will leave and even quotient .
_________________

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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:08
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8

Thanks, but I was not able to understand what evenly divisible meant, thats why I chose to ignore that part.
I think that I get it now, I guess that means that it will leave and even quotient .

Evenly divisible means that you have no remainder
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:15
GMATinsight wrote:
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8

Thanks, but I was not able to understand what evenly divisible meant, thats why I chose to ignore that part.
I think that I get it now, I guess that means that it will leave and even quotient .

Evenly divisible means that you have no remainder

That means my hunch was correct. anyways thanks GMATinsight for telling that.
so going by my original method. will the probability equal to 5/8 ?
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:25
I m getting 5/8 GmatDaddy

Posted from my mobile device
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:30
My very first post said the same !
Anyways thanks for conforming !!
vasuca10 wrote:
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Source: http://www.GMATinsight.com

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8

_________________

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If my post was of any help to you, You can thank me in the form of Kudos!!

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Posts: 2371
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GMAT: INSIGHT
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:41
Just out of curiosity is the probability $$\frac{5}{8}$$ ?
I solved it in a not at all efficient manner.

GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Its written as evenly divisible by 8 and not divisible by 8
If it was the later part then answer would have been 5/8
Every 5 out of 8 products are divisible by 8
Hence 5/8

Thanks, but I was not able to understand what evenly divisible meant, thats why I chose to ignore that part.
I think that I get it now, I guess that means that it will leave and even quotient .

Evenly divisible means that you have no remainder

That means my hunch was correct. anyways thanks GMATinsight for telling that.
so going by my original method. will the probability equal to 5/8 ?

Yes probability is 5/8
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 06:46
How is the ans a? That doesn’t seem right
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 07:49
rahulkashyap wrote:
How is the ans a? That doesn’t seem right

Take any 8 consecutive integer values of 8 and check how many of them satisfy the given constraint... you will fine 5 favourable cases and 3 unfavourble cases hence probability will be 5/8 if we know that n is an integer
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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08 Sep 2018, 08:15
1
Well you'd have to assume the range of n in that case. The probability would be different if the range were something like 1 to 20

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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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09 Sep 2018, 06:25
shouldn't the answer be d. in any case out of 10 consecutive integer values of n, for 5 values the product is divisible by 8.
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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09 Sep 2018, 12:35
1
GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Source: http://www.GMATinsight.com

$$? = P\left( {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{8} = \operatorname{int} } \right)$$

(1) Although the ratio we are asked to obtain

$$\frac{{\,\,\,\# \,\,\left\{ {\,\,n\left( {n + 1} \right)\left( {n + 2} \right){\mkern 1mu} {\mkern 1mu} \,\,{\text{divisible}}{\mkern 1mu} {\mkern 1mu} {\text{by}}{\mkern 1mu} {\mkern 1mu} {\text{8}}\,\,\,:\,\,\,n{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \operatorname{int} \,\,\,} \right\}\,\,\,}}{{\# \,\,\left\{ {\,\,\,n\left( {n + 1} \right)\left( {n + 2} \right)\,\,\,:\,\,\,n\,\,{\mkern 1mu} \operatorname{int} \,\,\,} \right\}}}$$

DOES have a precise mathematical meaning and a unique value... the reasoning to obtain it rigorously is out-of-GMAT`s scope.

Anyway, the rationale the examiner´s (probably) had in mind is important:

---------------------------------------------------------------------------------------------------
n=1 implies n(n+1)(n+2) is not divisible by 8
n=2 implies n(n+1)(n+2) IS divisible by 8
n=3 implies n(n+1)(n+2) is not divisible by 8
n=4 implies n(n+1)(n+2) IS divisible by 8
n=5 implies n(n+1)(n+2) is not divisible by 8
n=6 implies n(n+1)(n+2) IS divisible by 8
n=7 implies n(n+1)(n+2) IS divisible by 8
n=8 implies n(n+1)(n+2) IS divisible by 8

n=9 is the beginning of a new cycle, identical to the n=1 situation
---
n=16 is the end of this new cycle, identical to the n=8 situation

Hence in every cycle we have 5 favorable cases among 8 equiprobable possibilities (*) , hence the answer is 5/8.
----------------------------------------------------------------------------------------------------------
(*) The real issue is to formalize this reasoning... we have an infinite number of blocks to evaluate, and
although things "behave" in the same way... the justification is far out-of-GMAT´s universe!

(2) We cannot use the fact that n is an integer here... that´s where we are able to BIFURCATE:

> If n is an integer, the argument shown above (between the parallel lines) is PERFECT and the answer would be 5/8
(The fact that we are dealing with integers in the interval [1,96] avoids the "infinite blocks" higher-level analysis!)

> If n is not necessarily an integer, we could choose n in the interval [1,96] such that n=1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, ... , 95, 95.5, 96 (for instance).
In this case, it is not 5 favorable cases for every 8 (most n´s will give n(n+1)(n+2) not an integer, thefore the "no divisibility by 8" is more frequent)!

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 02:17
any specific reason of selecting 8 integers and then counting divisble by 8 in them ? y cant i choose 10,15 or say 16 for that mattter???

GMATinsight wrote:
rahulkashyap wrote:
How is the ans a? That doesn’t seem right

Take any 8 consecutive integer values of 8 and check how many of them satisfy the given constraint... you will fine 5 favourable cases and 3 unfavourble cases hence probability will be 5/8 if we know that n is an integer
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 02:44
Pallavim28

Yeah, This is actually a pattern that repeats itself after every 8 integers. Give it a try yourself/
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 03:06
Hey Thanks for a prompt reply .. I considered that i might need more info about whats the sample space or the number of integers abnd ehnce did not marked say. but whatsoever the integer si(guess would get the same ans every time) as their is a pattern

Pallavim28

Yeah, This is actually a pattern that repeats itself after every 8 integers. Give it a try yourself/
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 03:18
[quote="GMATinsight"]What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Why option B is wrong because I am getting answer as 1/4 from option B.
And pls someone explain how we can get the probability using option A.

Anurag Jain
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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 03:28
jainanurag470

To define a probability you need a sample space or the number of all possible outcomes.
Statemetn 2 does not define if n is integer or not, hence we can not count the total no of possible cases.

jainanurag470 wrote:
GMATinsight wrote:
What is the probability of n(n+1)(n+2) being evenly divisible by 8?

1) n is an integer
2) 1≤ n ≤ 96

Why option B is wrong because I am getting answer as 1/4 from option B.
And pls someone explain how we can get the probability using option A.

Anurag Jain

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Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1  [#permalink]

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10 Sep 2018, 03:46
1

To define a probability you need a sample space or the number of all possible outcomes.
Statemetn 2 does not define if n is integer or not, hence we can not count the total no of possible cases.

Thank you.
But I still doubt option A because sample space is not clear in option A too.
for eg. If I take first 8 positive intgers as my sample space then probability will be = 3/8
And if I take first 9 positive intgers as my sample space then probability will be = 3/9=1/3

Anurag Jain
Re: What is the probability of n(n+1)(n+2) being evenly divisible by 8? 1 &nbs [#permalink] 10 Sep 2018, 03:46

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