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If x is a positive integer, is square_root x [#permalink]
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Updated on: 26 May 2015, 06:24
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If x is a positive integer, is \((x^{0.5})^{0.5}\) an integer? (1) x is the square of an integer. (2) \(\sqrt{x}\) is the square of an integer. I agree with the OA. However, I have a doubt with (2):
\(\sqrt{x} = i^2\) , being \(i\) an integer.
Should we unsquare or use an additional square root on \(i\)? Is there a difference? I know that the square root only provide the positive root of the number.
Thanks!
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Originally posted by danzig on 29 Jul 2013, 16:18.
Last edited by Bunuel on 26 May 2015, 06:24, edited 2 times in total.
Edited the question.



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Re: If x is a positive integer, is square_root x [#permalink]
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02 Aug 2013, 19:12
danzig wrote: If x is a positive integer, is [(x)^(1/2)]^(1/2) ? (1) x is the square of an integer. (2) \(\sqrt{x}\) is the square of an integer.
I agree with the OA. However, I have a doubt with (2):
\(\sqrt{x} = i^2\) , being \(i\) an integer.
Should we unsquare or use an additional square root on \(i\)? Is there a difference? I know that the square root only provide the positive root of the number.
Thanks! danzig, my understanding since x is + ve, sq_rt(x) is +ve also i^2 is always +ve. since both are +ve, it doesnt matter if you unsquare of use additional sq rt. Hope I clarified your doubt.
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Re: If x is a positive integer, is square_root x [#permalink]
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03 Aug 2013, 08:39



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Re: If x is a positive integer, is square_root x [#permalink]
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03 Aug 2013, 08:46
bsahil wrote: The question asks if \(x^{\frac{1}{4}}\) is a positive integer (1) x is the square of an integer.So x can be \(4\), and \(4^{\frac{1}{4}}\) is NOT a positive integer. But x can be \(16\), and \(16^{\frac{1}{4}}\) is a positive integer. Not sufficient (2) \(\sqrt{x}\) is the square of an integer.So \(\sqrt{x}=n^2\), \(x^{\frac{1}{4}}=n^{2*\frac{1}{2}}=n\) where n is an integer. Sufficient Hope it's clear
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Re: If x is a positive integer, is square_root x [#permalink]
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Updated on: 22 May 2015, 03:51
Zarrolou wrote: (1) x is the square of an integer. So x can be \(4\), and \(4^{\frac{1}{4}}\) is NOT a positive integer.
How is \(4^{\frac{1}{4}}\) not a positive integer?? Edit: You can ignore this post...I get it.
Originally posted by raeann105 on 21 May 2015, 16:50.
Last edited by raeann105 on 22 May 2015, 03:51, edited 1 time in total.



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Re: If x is a positive integer, is square_root x [#permalink]
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21 May 2015, 20:04
Hi Bunuel, It looks like the prompt is incomplete. From the other posts, it appears that the question is supposed to ask "Is.......an integer?", but the original prompt is missing that language. The original poster also appears to be long gone, so could you edit the original post accordingly? Thanks. GMAT assassins aren't born, they're made, Rich
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Re: If x is a positive integer, is square_root x [#permalink]
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25 May 2015, 14:02
Experts, please critique if my approach...esp. in analyzing [1] is accurate.....Thanks in advance! I found that if I use the techniques taught by the Quant experts on GC, i arrive at the right solution. Solving [1] with my personal logic ... that X being the square of another integer (+ve or ve)....will make x a +ve integer (always), I arrived at the conclusion that [1] is also sufficient Your help is much appreciated! The way I understood this question is: We need to answer if √(√x) is a positive integer?
Given: [1] x is the square of an integer. So x = y^2 taking Sqrt on LHS and RHS √x = √(y^2) According to GMATCLUB math guide √(y^2) = y Therefore, √(√x) = √(y) Solving for y as (+y & y) we can't say if √(√x) is +ve or ve.....Not sufficient!
(2) √x is the square of an integer. So √x=y^2 taking Sqrt on LHS and RHS √(√x) = √(y^2) √(√x) = y...........hence sufficient!
Answer is B



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Re: If x is a positive integer, is square_root x [#permalink]
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25 May 2015, 18:02
rohitd80 wrote: Experts, please critique if my approach...esp. in analyzing [1] is accurate.....Thanks in advance! I found that if I use the techniques taught by the Quant experts on GC, i arrive at the right solution. Solving [1] with my personal logic ... that X being the square of another integer (+ve or ve)....will make x a +ve integer (always), I arrived at the conclusion that [1] is also sufficient Your help is much appreciated! The way I understood this question is: We need to answer if √(√x) is a positive integer?
Given: [1] x is the square of an integer. So x = y^2 taking Sqrt on LHS and RHS √x = √(y^2) According to GMATCLUB math guide √(y^2) = y Therefore, √(√x) = √(y) Solving for y as (+y & y) we can't say if √(√x) is +ve or ve.....Not sufficient!
(2) √x is the square of an integer. So √x=y^2 taking Sqrt on LHS and RHS √(√x) = √(y^2) √(√x) = y...........hence sufficient!
Answer is B
Never mind the analysis of [1]... I get it now. [1] is mainly to evaluate x as an integer or noninteger. Being the equivalent of a square of an integer x has to be positive.



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Re: If x is a positive integer, is square_root x [#permalink]
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26 May 2015, 06:25



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Re: If x is a positive integer, is square_root x [#permalink]
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24 May 2017, 19:19
Bunuel wrote: EMPOWERgmatRichC wrote: Hi Bunuel,
It looks like the prompt is incomplete. From the other posts, it appears that the question is supposed to ask "Is.......an integer?", but the original prompt is missing that language. The original poster also appears to be long gone, so could you edit the original post accordingly? Thanks.
GMAT assassins aren't born, they're made, Rich Done. Thank you for noticing this. With this question it easier to convert the exponents into fractions (x^.5)^.5 = (x^1/2)^1/2 x^1/4  translates to the 4th root of x (x^1/3 would be the cube root of x and x^2/3 would be the cube root of x^2  that's the pattern/rule) Statement 1 is insufficient because 2 is the square of 4 but the 4th root of 2 is not an integer Statement 2 is sufficient because it we take say 16, the square root of 16 is 4 and 4 is a perfect square, it is 2 squared so 16 =2^4




Re: If x is a positive integer, is square_root x
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