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

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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

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Intern
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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
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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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
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:
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 (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.

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Re: DS - tough divisibility   [#permalink] 28 Jan 2009, 09:52
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