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is n an integer? 1) n ^ 2 is an integer

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Intern
Joined: 11 Jul 2007
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is n an integer? 1) n ^ 2 is an integer [#permalink]  10 Aug 2007, 12:06
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hey guys,
this is question number 146 from the data sufficency section.

is n an integer?
1) n ^ 2 is an integer
2) square root of n is an integer

i don't quite get the OA for this one. it seems easy, but their answer makes no sense. thanks..
Director
Joined: 12 Jul 2007
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I would say B. Is that the OA?

1. n^2 is an integer

this isn't sufficient because n could be sqrt(10). yes, n^2 would be an integer (10 in this case) but sqrt(10) is not an integer

2. square root of n is an integer.

this means that n must be the product an integer multiplied by itself. any integer times another integer results in an integer.

so you get B for the answer
VP
Joined: 08 Jun 2005
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eschn3am wrote:
I would say B. Is that the OA?

1. n^2 is an integer

this isn't sufficient because n could be sqrt(10). yes, n^2 would be an integer (10 in this case) but sqrt(10) is not an integer

2. square root of n is an integer.

this means that n must be the product an integer multiplied by itself. any integer times another integer results in an integer.

so you get B for the answer

I second (B) - good answer !

Intern
Joined: 11 Jul 2007
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yup guys. OA is B. i never really thought of it that way. i get it now. i keep thinking in terms of whole numbers and fractions. at first i couldn't figure why isn't A sufficient enough, but it all makes sense now. thanks for your help guys.
Senior Manager
Joined: 14 Jun 2007
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Re: DS: integer question [#permalink]  10 Aug 2007, 12:55
hey guys,
this is question number 146 from the data sufficency section.

is n an integer?
1) n ^ 2 is an integer
2) square root of n is an integer

forgive me if others have said this... i am just going off the top of my head without looking at others explanations
i don't quite get the OA for this one. it seems easy, but their answer makes no sense. thanks..

I:
if n^2 is an integer, n could or couldnt be an integer. if n^2=4 yes, then n=2

if n^2 = 2, then no n= sqrt(2)

II:

sqrt n = some integer k. then n = k^2. any integer squared is an integer.

B
Re: DS: integer question   [#permalink] 10 Aug 2007, 12:55
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