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# Is n/18 an integer? 1. 5n/18 is an integer 2. 3n/18 is an

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Manager
Joined: 05 Feb 2007
Posts: 138
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 re-post 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.
Intern
Joined: 24 Jan 2009
Posts: 18
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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All Day

Manager
Joined: 05 Feb 2007
Posts: 138
Re: DS - tough divisibility [#permalink]

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25 Jan 2009, 18:34
That was what I got, but it is incorrect.
SVP
Joined: 07 Nov 2007
Posts: 1759
Location: New York
Re: DS - tough divisibility [#permalink]

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25 Jan 2009, 18:54
2
KUDOS
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 (k-integer)

n/18 = i/5 = 5k/5= k=integer

C
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Manager
Joined: 17 Dec 2008
Posts: 165
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...
Senior Manager
Joined: 02 Nov 2008
Posts: 255
Re: DS - tough divisibility [#permalink]

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25 Jan 2009, 21:34
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?
SVP
Joined: 29 Aug 2007
Posts: 2452
Re: DS - tough divisibility [#permalink]

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25 Jan 2009, 21:49
chicagocubsrule 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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Manager
Joined: 13 Jan 2009
Posts: 168
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.
Manager
Joined: 05 Feb 2007
Posts: 138
Re: DS - tough divisibility [#permalink]

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26 Jan 2009, 12:47
excellent work

OA is C
Manager
Joined: 15 Apr 2008
Posts: 159
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
Intern
Joined: 24 Jan 2009
Posts: 18
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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Manager
Joined: 13 Jan 2009
Posts: 168
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.
Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua
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 (k-integer)

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 re-post 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   [#permalink] 28 Jan 2009, 09:52
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