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If x is a positive integer, is x^1/2 an integer
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If x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer
friends please help me in pointing out whee i am going wrong.
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Originally posted by sacmanitin on 10 Jan 2010, 00:55.
Last edited by Bunuel on 25 Nov 2017, 00:05, edited 2 times in total.
Renamed the topic, edited the question and added OA.




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Re: If x is a positive integer, is x^1/2 an integer
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09 Jan 2011, 09:52
fluke wrote: I didn't get the explanation:
if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.
So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.
What's wrong with my explanation?? You forgot that x is a positive integer, so \(\sqrt{x}\) cannot equal to \(\frac{integer}{2}\). Generally \(\sqrt{integer}\) is either an integer or an irrational number. Complete solution: If x is a positive integer, is sqrt(x) an integerIf \(x=integer\), is \(\sqrt{x}=integer\)? (1) \(\sqrt{4x}\) is an integer > \(2\sqrt{x}=integer\) > \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient. (2)\(\sqrt{3x}\) is not an integer > if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient. Answer: A. Hope it's clear.
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If x is a positive integer, is root(x) an integer?
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20 Feb 2010, 14:49
If x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. This is the question from GMAT Quant Review. My logic to solve this question: \sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient. \sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient. S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement. So I choose E as an answer. Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice.




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Re: If x is a positive integer, is x^1/2 an integer
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10 Jan 2010, 08:42
if x is a positive integer ,is sqrt (x) an integer
(1) sqrt(4x) is an integer .
(2) sqrt (3x) is not an integer .
the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer yes. sqrt(x) is an integer.
how i worked it out : sqrt(4x) is an integer ==> 2 * sqrt(x) is an integer
in order that the product of 2 and sqrt(x) be an integer, sqrt(x) must either be 1) an integer 2) exactly half of an integer. i.e a number like 0.5, 1.5, 2.5 etc etc you worked that out as well. sqrt(x) = integer/2
we also know x is an integer. is there any integer whose square root is a half of an integer ? no!
therefore the only other alternative is sqrt(x) is a whole integer.
hope that helped



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Re: If x is a positive integer, is x^1/2 an integer
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11 Jan 2010, 06:50
Hi Janani ,
Thanks for the explanation ,do we have a rule like we can't have a sqrt (x)=integer /2, or its by observation ...



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Re: Is OG Quant question answer wrong?
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20 Feb 2010, 15:14
alexBLR wrote: This is the question from GMAT Quant Review: If x is a positive integer , is \sqrt{x} an integer? 1) \sqrt{4x} is an integer. 2) \sqrt{3x} is not an integer. My logic to solve this question: \sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient. \sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient. S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement. So I choose E as an answer. Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice. IMO D... Ques: if x is a positive integer, is \(\sqrt{x}\) an integer? S1: \(\sqrt{4x}\) is an integer > \(2* \sqrt{x}\) is an integer > \(\sqrt{x}\) has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF S2: \(\sqrt{3x}\) is an integer > \(\sqrt{3}*\sqrt{x}\) > \(\sqrt{x}\) is not an integer as same could be a of a form of \(a\sqrt{3}\) where 'a' is a positive integer. Therefore SUFF
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If x is a positive integer, is root(x) an integer?
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20 Feb 2010, 15:57
jeeteshsingh wrote: alexBLR wrote: This is the question from GMAT Quant Review: If x is a positive integer , is \sqrt{x} an integer? 1) \sqrt{4x} is an integer. 2) \sqrt{3x} is not an integer. My logic to solve this question: \sqrt{4x}=2*\sqrt{x}, so \sqrt{x} can either be integer or not an integer (for example \sqrt{x}=2.5) and the 2*\sqrt{x} is still an integer. So Statement 1 is insufficient. \sqrt{3x}= \sqrt{3}*\sqrt{x}. As \sqrt{3} is not an integer, the \sqrt{x} can be either integer or non integer and the \sqrt{3}*\sqrt{x} will still be not integer. So Statement 2 is insufficient. S1 and S2 together is still insufficient as \sqrt{x}=2 and \sqrt{x}=2.5 both satisfy statements requirement. So I choose E as an answer. Is there a flaw in my reasoning? OG Quant review answer to this question is different from E. Please advice. IMO D... Ques: if x is a positive integer, is \(\sqrt{x}\) an integer? S1: \(\sqrt{4x}\) is an integer > \(2* \sqrt{x}\) is an integer > \(\sqrt{x}\) has to be an integer.. as x is a positive integer and hence cannot be a fraction. Therefore SUFF S2: \(\sqrt{3x}\) is an integer > \(\sqrt{3}*\sqrt{x}\) > \(\sqrt{x}\) is not an integer as same could be a of a form of \(a\sqrt{3}\) where 'a' is a positive integer. Therefore SUFF If x is a positive integer, is \(\sqrt{x}\) an integer?As given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number. (1) \(\sqrt{4x}\) is an integer > \(2\sqrt{x}=integer\) > \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient. (2) \(\sqrt{3x}\) is not an integer > if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient. Answer: A. jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements cannot give you TWO DIFFERENT answers to the question (as you've got). Hope it helps.
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Re: Is OG Quant question answer wrong?
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20 Feb 2010, 16:13
Bunuel wrote: jeeteshsingh, you should have spotted that there was something wrong with your solution as in DS two statements can not give you TWO DIFFERENT answers to the question (as you've got).
Hope it helps. My Bad.... overlooked it... Infact today I was telling this to someone over the forum that both the statements in DS would always be in sync.. Thanks Bunuel... for pointing this out.
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Re: Is OG Quant question answer wrong?
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20 Feb 2010, 16:42
When I assumed the case \sqrt{x}=2.5 I did not take into the account that x will not be an integer in this case(x=6.25). Thanks Bunuel



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Re: If x is a positive integer, is x^1/2 an integer
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Updated on: 29 Sep 2010, 09:39
If x is a positive integer, is \(\sqrt{x}\) an integer?
(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.
Originally posted by ezinis on 29 Sep 2010, 07:09.
Last edited by ezinis on 29 Sep 2010, 09:39, edited 1 time in total.



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Re: If x is a positive integer, is x^1/2 an integer
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29 Sep 2010, 07:29
ezinis wrote: If x is a positive integer, is \sqrt{x} an integer? (1) \(\sqrt{4x}\) is an integer4. (2) \(\sqrt{3x}\) is an integer.
I am not satisfied with the official explanation. Please give yours, thanks. I think (2) should be \(\sqrt{3x}\) is NOT an integer. If \(x=integer\), is \(\sqrt{x}=integer\)? (1) \(\sqrt{4x}\) is an integer > \(2\sqrt{x}=integer\) > \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient. (2)\(\sqrt{3x}\) is not an integer > if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient. Answer: A.
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Re: If x is a positive integer, is x^1/2 an integer
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29 Sep 2010, 08:10
ezinis wrote: If x is a positive integer, is \sqrt{x} an integer? (1) \sqrt{4x} is an integer4. (2) \sqrt{3x} is an integer.
I am not satisfied with the official explanation. Please give yours, thanks. (1) \(\sqrt{4x} = 2 * \sqrt{x}\) If this is an integer, then \(\sqrt{x}\) has to be an integer (2) \(\sqrt{3x} = \sqrt{3} * \sqrt{x}\) For this to be an integer, \(\sqrt{x}\) must be of the form \(\sqrt{3} * Integer\) So \(\sqrt{x}\) is not an integer I am not sure if the question is correct as (1) and (2) are contradicting. Is it supposed to say \(\sqrt{3x}\) is not an integer ?
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Re: If x is a positive integer, is x^1/2 an integer
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09 Jan 2011, 09:40
I didn't get the explanation: if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer. So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer. What's wrong with my explanation??
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Re: If x is a positive integer, is x^1/2 an integer
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10 Jan 2011, 01:32
good explanation Bunuel and you are right in saying that I carelessly overlooked the fact that x was a positive integer... thanks ~fluke
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Re: Integers
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Re: If x is a positive integer, is x^1/2 an integer
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19 Jul 2011, 06:29
Quote: then sqrt(x) can not be integer/2 I think sqrt(x) can be integer/2 as long as (integer/2) itself is an integer i.e. that integer is multiple of 2. I think that is what bunuel meant. And the answer remains same.
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Re: If x is a positive integer, is x^1/2 an integer
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03 Jan 2012, 13:06
Oh nice problem. Took quite some time to answer it but got A. Explanation by Bunuel is more than sufficient to understand the solution.
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Re: If x is a positive integer, is x^1/2 an integer
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03 Jan 2012, 14:31
Both equations and number plugging helps here. 1. sqrt(4x) = integer, that is 2 x sqrt(x) = integer. For this equation to be true, sqrt(x) has to be an integer. SUFFICIENT. If number plugging, use 4x4 and 4x9 combinations 2. sqrt(3x) = frac. Here, sqrt(3) x sqrt(x) = frac, that is frac x sqrt(x) = frac. Difficult to determine if this relationship can infer sqrt(x) as integer. So, let's go with number plugging. sqrt(3x4) = 12 satisfies, and sqrt(4) = integer. sqrt(3x5) = 15 satisfies, but sqrt(5) = frac. So, insufficient. +1 for A.
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Re: If x is a positive integer, is x^1/2 an integer
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15 Dec 2012, 11:21
Bunuel  can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!



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Re: If x is a positive integer, is x^1/2 an integer
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Re: If x is a positive integer, is x^1/2 an integer &nbs
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