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

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

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New post 13 Nov 2008, 05:04
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N is a positive integer. Is 9 the factor of N?

(1) 18 is the factor of N^2

(2) 27 is the factor of N^3



Edit: Sorry, statement 2 should have N^3 rather than N^2

Last edited by tarek99 on 14 Nov 2008, 02:45, edited 2 times in total.

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Re: DS: Factor [#permalink]

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New post 13 Nov 2008, 05:45
From stmt1: N^2 has 3^2*2 as the factor. Not sufficient as N may have 3 as the factor.

From stmt2: N^2 has 3^3 as the factor and since N is a positive integer, 3^2 will be a factor of N....sufficient.

Hence, B.

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Re: DS: Factor [#permalink]

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New post 13 Nov 2008, 09:31
B

stmt1: eg: n = 6 - InSuff
stmt2: n^2 = 9 * 3x, x = 1,2,3,4,5...
for n to be positive, x can only be multiples of 3, so Suff

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Re: DS: Factor [#permalink]

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New post 13 Nov 2008, 11:06
Stmt 1 possible values of n^2= 36,144,324=> n=6,12,18=> 9 can be factor of N or can t be hence insufficient
stmt 2 possible value of n^2=81,324 => n=9,18 => 9 is always a factor of N hence sufficient
B is the answer

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Re: DS: Factor [#permalink]

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New post 14 Nov 2008, 02:44
wrong guys. The OA is E! So start explaining! :lol:

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Re: DS: Factor [#permalink]

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New post 14 Nov 2008, 03:19
I guess because of the typo , everyone has answered incorrectly. The answer seems to be E , as explained by hibloom ( I made few modifications to the original post )
Stmt 1 possible values of n^2= 36,144,324=> n=6,12,18=> 9 can be factor of N or can t be hence insufficient
stmt 2 possible values of n^3=81,729 => n=3, 9 => 9 can be factor of N or can t be hence insufficient

E is the answer

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Re: DS: Factor [#permalink]

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New post 15 Nov 2008, 15:53
is there a more generic equation that one can use to approach this problem rather than listing out all the different numbers?

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Re: DS: Factor [#permalink]

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New post 15 Nov 2008, 16:13
[quote="tarek99"]N is a positive integer. Is 9 the factor of N?

(1) 18 is the factor of N^2

(2) 27 is the factor of N^3




is n = 9x

from 1

n^2 = 3^2*2*a
n = 3*b, b is an even intiger.........insuff

from2

n^3 = 3^3*c

n = 3*f,f can be any intiger

both

n = 3*f = 3*b tell us that n is a multiple of 3 however we cant be sure that it is a multiple of 9........insuff

E

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Re: DS: Factor [#permalink]

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New post 15 Nov 2008, 16:28
I got E as well

N^2 is divisible by 18. N values can be any thing such as 18,24,36 (Y N Y for divisibility with N) Insuff

N^3 is divisible by 27. N values can be any thing such as 27, 33, 36 ( Y N Y for divisibility with N) Insuff)

Together N values can be any thing such as 36 and 48 ( Y, N) Insuff

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Re: DS: Factor [#permalink]

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New post 17 Nov 2008, 08:27
Thanks guys! The OA is E.

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Re: DS: Factor   [#permalink] 17 Nov 2008, 08:27
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N is a positive integer. Is 9 the factor of N? (1) 18 is the

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