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# is n/18 an integer? I. 5n/18 is an integer II. 3n/18 is an

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Manager
Joined: 30 Dec 2008
Posts: 121
is n/18 an integer? I. 5n/18 is an integer II. 3n/18 is an [#permalink]

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18 Jan 2009, 21:51
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is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer
Director
Joined: 29 Aug 2005
Posts: 836

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19 Jan 2009, 05:02
cul3s wrote:
is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer

Stmt 1. try n= 18 and n=18/5; Insuff

Stmt2. Insuff.

Stmt 1 & 2: Only multiples of 18 satisfy both, in which case n/18 is an integer. So, C
Manager
Joined: 27 May 2008
Posts: 198

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19 Jan 2009, 08:17
A

18 ==> $$3^2$$*2

So for n/18 a integer, n should be atleast $$3^2$$*2

Stmt1 says 5n/18 -- integer ==> since 5 is not a factor of 18, one can clearly conclude tht n/18 also as a integer.

Stmt2 - 3n/18 is a integer -- here n can be 3*2, to make 3n/18 ==> 1 which doesn't mean n/18 is an integer.

So A
Manager
Joined: 04 Jan 2009
Posts: 235

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19 Jan 2009, 10:11
cul3s wrote:
is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer

yes.
5n/18 is integer=>n =18 or its multiple and hence, n/18 is integer.
3n/18 is integer=>n=6 or its multiple=>sometimes integer and sometimes not. Hence insufficient.

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tusharvk

Director
Joined: 29 Aug 2005
Posts: 836

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19 Jan 2009, 10:28
tusharvk wrote:
cul3s wrote:
is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer

yes.
5n/18 is integer=>n =18 or its multiple and hence, n/18 is integer.
3n/18 is integer=>n=6 or its multiple=>sometimes integer and sometimes not. Hence insufficient.

Stmt 1 would be suff if n were integer. n can be, say, 18/5.
Manager
Joined: 30 Dec 2008
Posts: 121

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19 Jan 2009, 12:44
ok...the answer is C. see below for explanation. I get part 1 and 2 counterexamples, but Im confused with combining 1 and 2 and LCM stuff (bold part). Any effort in putting those in simple words would be much appreciated.

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(1) INSUFFICIENT: We are told that 5n/18 is an integer. This does not allow us to determine whether n/18 is an integer. We can come up with one example where 5n/18 is an integer and where n/18 is NOT an integer. We can come up with another example where 5n/18 is an integer and where n/18 IS an integer.

Let's first look at an example where 5n/18 is equal to the integer 1.

If 5n/18 = 1, then n/18 =1/5. In this case n/18 is NOT an integer.

Let's next look at an example where 5n/18 is equal to the integer 15.

If 5n/18 = 15, then n/18= 3. In this case n/18 IS an integer.

Thus, Statement (1) is NOT sufficient.
(2) INSUFFICIENT: We can use the same reasoning for Statement (2) that we did for statement (1). If 3n/18 is equal to the integer 1, then n/18 is NOT an integer. If 3n/18 is equal to the integer 9, then n/18 IS an integer.

(1) AND (2) SUFFICIENT: If 5n/18 and 3n/18 are both integers, n/18 must itself be an integer. Let's test some examples to see why this is the case.

The first possible value of n is 18, since this is the first value of n that ensures that both 5n/18 and 3n/18 are integers. If n = 18, then n/18 is an integer. Another possible value of n is 36. (This value also ensures that both 5n/18 and 3n/18 are integers). If n = 36, then n/18 is an integer.

A pattern begins to emerge: the fact that 5n/18 AND 3n/18 are both integers limits the possible values of n to multiples of 18. Since n must be a multiple of 18, we know that n/18 must be an integer. The correct answer is C.

Another way to understand this solution is to note that according to (1), n = (18/5)*integer, and according to (2), n = 6*integer. In other words, n is a multiple of both 18/5 and 6. The least common multiple of these two numbers is 18. In order to see this, write 6 = 30/5. The LCM of the numerators 18 and 30 is 90. Therefore, the LCM of the fractions is 90/5 = 18.

Again, the correct answer is C.
Manager
Joined: 02 Sep 2008
Posts: 103

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20 Jan 2009, 17:07
is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer

Statement 1: 5n/18 is integer now we can also write this as 5 * n/18 as integer. Now we have not doubt that '5' is integer so for any thing 5x to be integer x has to be integer. here x is n/18 so since 5n/18 is integer n/18 is also an integer. - Sufficient.

Statement 2: 3n/18. Now in this case too we can apply the same analogy but here 18 is divisible by 3 (which is not possible in statement 1 as 18 is not divisible by 5) which means 3n/18 reduces to n/6. Now we can not certainly tell that if n/6 is integer then 3n/18 will also be integer. Hence this statement is not enough.

Hence IMO A.
SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York

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21 Jan 2009, 03:41
cul3s wrote:
is n/18 an integer?

I. 5n/18 is an integer
II. 3n/18 is an integer

good question.

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: multiples DS   [#permalink] 21 Jan 2009, 03:41
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