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Is n/18 an integer? 1. 5n/18 is an integer 2. 3n/18 is an [#permalink]
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25 Jan 2009, 15:03
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Is n/18 an integer? 1. 5n/18 is an integer 2. 3n/18 is an integer == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 18:03
giantSwan wrote: Is n/18 an integer?
1. 5n/18 is an integer
2. 3n/18 is an integer For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\) 1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient. 2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient. Therefore, A.
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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 18:34
That was what I got, but it is incorrect.



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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 18:54
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I. 5n/18 is an integer 5n/18 = i (integer) n/18 = i/5 (1) when i=1 its not integer when i=5 its integer not sufficient II. 3n/18 is an integer II. 3n/18 is an integer 5n/18 = j (integer) n/18 = j/6 > (2) when j=1 its not integer when j=6 its integer from (1) and (2) i/5 =j/6 6i= 5 j from the above equation it is clear that i must be multiple of 5 and j must be multiple of 6 so i= 5k (kinteger) n/18 = i/5 = 5k/5= k=integer C
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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 20:57
Kool trap qn there.... This one actually makes think whether to go back to strategically solving every DS in eqns, rather than making assumptions and go wrong easily...



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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 21:34
AdrianPeterson wrote: giantSwan wrote: Is n/18 an integer?
1. 5n/18 is an integer
2. 3n/18 is an integer For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\) 1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient. 2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient. Therefore, A. I also chose A. Can someone explain why A is insufficient?



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Re: DS  tough divisibility [#permalink]
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25 Jan 2009, 21:49
chicagocubsrule wrote: AdrianPeterson wrote: giantSwan wrote: Is n/18 an integer?
1. 5n/18 is an integer
2. 3n/18 is an integer For \(\frac{n}{18}\) to be an integer, n must be a multiple of 18. \(18 = 2*3^2\) 1) As 5 is not one of the factors of 18, n must be a multiple of 18 for \(\frac{5n}{18}\) to be an integer. Therefore, sufficient. 2) Because 3 is one of the factors, you cannot know whether n/18 is an integer. If n=6, then the condition would be satisfied and the answer would be no (Is n/18 an integer?). If n=18, the condition would also be satisfied and the answer would be yes. Insufficient. Therefore, A. I also chose A. Can someone explain why A is insufficient? In 1) n could be any multiple of 3.6 or 18.
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Re: DS  tough divisibility [#permalink]
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26 Jan 2009, 03:45
I think the answer is 'C'.
In statement 1, n can be 3.6.  A is insufficient; In statement 2, n can be 6.  B is insufficient;
When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient.



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Re: DS  tough divisibility [#permalink]
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26 Jan 2009, 12:47
excellent work
OA is C



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Re: DS  tough divisibility [#permalink]
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26 Jan 2009, 15:13
Ibodullo wrote: I think the answer is 'C'.
In statement 1, n can be 3.6.  A is insufficient; In statement 2, n can be 6.  B is insufficient;
When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient. in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer



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Re: DS  tough divisibility [#permalink]
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26 Jan 2009, 16:05
ALD wrote: Ibodullo wrote: I think the answer is 'C'.
In statement 1, n can be 3.6.  A is insufficient; In statement 2, n can be 6.  B is insufficient;
When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient. in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer I guess what he meant was not that n could be 3 or 6, but that n could be the number 3.6 or its multiples. By the way, nice catch x2suresh. My factoring method blinded me from the possibility of n as a non integer.
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Re: DS  tough divisibility [#permalink]
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26 Jan 2009, 22:01
ALD wrote: Ibodullo wrote: I think the answer is 'C'.
In statement 1, n can be 3.6.  A is insufficient; In statement 2, n can be 6.  B is insufficient;
When we consider both statements together, n could be any number divisible by 18. (18, 36, 54, 72 ... etc). I couldn't find a number n that could could meet both statements together, but is not divisible by 18. So 'C' is sufficient. in statement 1 i think 'n' can only be a multiple of 18 and not 3 or 6 can you pls explain how in statement 1 'n' can be a multiple of 3 or 6 and satisfy the condition that 5n/18 is an integer Yes, I meant 3,6. Sorry for confusion.



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Re: DS  tough divisibility [#permalink]
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28 Jan 2009, 09:52
x2suresh wrote: I. 5n/18 is an integer
5n/18 = i (integer) n/18 = i/5 (1) when i=1 its not integer when i=5 its integer not sufficient
II. 3n/18 is an integer II. 3n/18 is an integer 5n/18 = j (integer) n/18 = j/6 > (2) when j=1 its not integer when j=6 its integer
from (1) and (2)
i/5 =j/6
6i= 5 j from the above equation it is clear that i must be multiple of 5 and j must be multiple of 6
so i= 5k (kinteger)
n/18 = i/5 = 5k/5= k=integer
C Nice proof. Most of the time, when the question looks simple, it makes us to think in a wring direction easily. In this case, every where in the question and stmts, it is referring to Integers, we forget the fact that n could be factors as well. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.




Re: DS  tough divisibility
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