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N is a positive integer. Is 9 a factor of N? 1. 18 is the

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N is a positive integer. Is 9 a factor of N? 1. 18 is the [#permalink]

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New post 25 Aug 2008, 08:42
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N is a positive integer. Is 9 a factor of N?
1. 18 is the factor if n^2
2. 27 is the factor of N^3

How to approach these types of problems?

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New post 25 Aug 2008, 09:37
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gmatcraze wrote:
N is a positive integer. Is 9 a factor of N?
1. 18 is the factor if n^2
2. 27 is the factor of N^3

How to approach these types of problems?


1. 18 is the factor if n^2

n^2 = 18*k= 9*2*k=3*3*2*k
when k=2 n=6 -- Is 9 a factor of N? No
when k=8 n=12 -- Is 9 a factor of N? No
when k=9*2 n=18 -- Is 9 a factor of N? yes

insuffciient

2. 27 is the factor of N^3
N^3 = 3^3*m
when m=1 n=3 Is 9 a factor of N? No
when m=27 n=9 Is 9 a factor of N? yes
insuffcient

combine

n^2 *n^3 = 18*27*k*m = 3^5*2*k*m
when k=4 m=4 n = 6 Is 9 a factor of N? No
when k=2^4 m=3^5 n=18 Is 9 a factor of N? yes

insuffcieint

E.
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Re: factor problems [#permalink]

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New post 25 Aug 2008, 09:45
I think the answer is E.
gmatcraze wrote:
N is a positive integer. Is 9 a factor of N?
1. 18 is the factor if n^2
2. 27 is the factor of N^3

How to approach these types of problems?


First, Figure out what positive integers 18 is a factor of. 18, 36, 54, 72, 90..Now you need remember that the statement says 18 is the factor of n^2, so that the list we just came up with are options for n^2. The first thing that strikes me is 36 that is a perfect square of 6. So n could = 6, and it's square is 18. If n = 6 is 9 a factor of 6? No. is it possible to find a value of n when squared has 18 as a factor? Yes. If n = 18, it's square will surely have 18 as a factor and then 9 will also be a factor of n. Insufficient because we have multiple answers.

Second Statement: Lets first say that n^3 = 27, so 27 is a factor. This would be mean n = 3. No way can 9 be a factor of 3. We can use the same concept here as we did with statement 1. let n = 27, so n^3 would surely have 27 as a factor, and if n = 27, then 9 is a factor of 27.

TOGETHER: Do the 2 statements together limit the problem so that only one value of n would satisfy both? No. If n = 18, n^2 = 324, and lets see if 27 will go into that. It does, so we know if 27 goes into n^2 when n=18, it certainly will go into 18^3. The reason for this is that 18^2 is 18 * 18..lets use prime factorization.

(2*3*3)(2*3*3). In order for the product to be divisible by 27, we need to find the prime factorization of 27.

3*3*3. All we need are 3-3's. We have 4, so 27 is a factor of 18^3. Now we can find a situation where n^2 is divisible by 18 and n^3 is divisible by 27 but n does not have 9 as a factor? Sure, lets use 6 again and determine this by prime factorization rather than brute force multiplication and division. It will save you time on G-Day also

6 * 6 = (2*3)*(2*3). 18 = 2*3*3. We need one 2 and 2-3's. Thats' present. 18 will go into 36 2 times. (Simple enough, we knew that and it's easy math, but the next one might not be, and the one you get on the real GMAT might not be as easy so...)

6*6*6 = 2*3*2*3*2*3. 27 = 3*9 = 3 * 3 * 3. There are 3-3s in 6^2, so 27 will be a factor.

Now, if n=6, we know 9 is not a factor of 6. So Together the statements are insufficient too.
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Re: factor problems   [#permalink] 25 Aug 2008, 09:45
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