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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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28 May 2013, 00:26
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Hi all, I have a question solution from the official book that appears incorrect to me. I've added a scan as an attachment. The solution claims that A is sufficient. But you don't always get an integer, for example: If you take \sqrt{4x}=5 vs \sqrt{4x}=6. what do you think?
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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28 May 2013, 00:38



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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28 May 2013, 00:41



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Re: If x is a positive integer, is root(x) an integer? [#permalink]
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05 Jun 2013, 03:55



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Re: If x is a positive integer, is root(x) an integer? [#permalink]
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05 Jun 2013, 07:58
[quote="alexBLR"]If x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. (1) since \(\sqrt{4x}\) is an integer, it is the same as 2*\(\sqrt{x}\) that means \(\sqrt{x}\) also must be ans integer  sufficient (2) \(\sqrt{3x}\) is not an integer, in this case \(\sqrt{x}\) could be or couldn't be an integer. Case 1) x=4, where \(\sqrt{x}\) is an integer, then \(\sqrt{3*4}\)=2\(\sqrt{3}\) still not an integer Case 2) x=5, where \(\sqrt{x}\) is NOT an integer, then \(\sqrt{3*5}\)=\(\sqrt{15}\) still not an integer. This statement is not sufficient. Answer is A
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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29 Jun 2013, 06:03
The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). Check links provided here: ifxisapositiveintegerisx12aninteger88994.html#p1229717
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Re: If x is a positive integer, is root(x) an integer? [#permalink]
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27 Dec 2013, 06:37
alexBLR wrote: 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. From First statement we get 2 sqrt (x) is an integer Therefore sqrt (x) must be an integer Just to elaborate a bit further: The thing is that sqrt (x) can't be an integer/2, cause the square root of any positive integer (as stated in the givens) is always >=1. Therefore, sqrt (x) will have to be an integer. Suff From second setatement we get the sqrt (3x) is not an integer Therefore, 3x is not a perfect square or x is not perfect square that is multiple of 3 But it could very well be a perfect square Matter is we still don't know Just to clarify, sqrt (x) could or could not be an integer. If sqrt root (x) is an integer then sure 1.7 (integer) not an integer. But, sqrt (x) could also be say x=2, therefore sqrt (2)=1.4. Then 1.4*1.7 is clearly not an integer either. Answer A stands Insuff Hence A is our best choice here Hope it helps Let the kudos rain begin Cheers! J



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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12 Nov 2014, 01:36
Hello Easiest way i can think of: S1\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2  int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)
S2 \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But NonInteger * Integer OR Noninteger = Non integer So Insufficient (1E) or (AD)
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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05 Dec 2014, 00:00
deeuk wrote: Hello Easiest way i can think of: S1\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2  int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)
S2 \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But NonInteger * Integer OR Noninteger = Non integer So Insufficient (1E) or (AD)
thoughts? Bunuel could you please tell me if this approach is correct/incorrect?



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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05 Dec 2014, 04:28
Motivatedtowin wrote: deeuk wrote: Hello Easiest way i can think of: S1\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2  int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)
S2 \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But NonInteger * Integer OR Noninteger = Non integer So Insufficient (1E) or (AD)
thoughts? Bunuel could you please tell me if this approach is correct/incorrect? No. That's not correct. (Noninteger)*(integer) could yield an integer. For example, 1/2*2 = 1. (Noninteger)*(noninteger) could also yield an integer. For example, \(\sqrt{2}*\sqrt{2}=2\).
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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10 Dec 2014, 13:21
Thank you for the correction But now i am wondering since \(\sqrt{4}\) is 2, doesn't \(\sqrt{x}\) have to be an integer as well. I mean doesnt x have to be a perfect square?



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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11 Dec 2014, 03:35
my approach I try to be careful in such types of questions because of a trap where a fraction multiplied by an integer can lead to an integer. Statement 1 sqrt(4x)= +2 * sqrt(x) (2 not for gmat) so sqrt(4x)= 2*sqrt(x) here sqrt(x) has to be an integer or an irrational no. no matter what form x is in (complex nos out of scope so ignore) as per the statement 1 it will be an integer. so sufficient. now lets check the trap if sqrt(x) is of the form integer/2 then still statement 1 will hold true but since x is an integer then at best x will of the form 4*perfect square. so again this statement will also give us a positive answer. Statement 2 this basically gives us two opposite answers we need to compare x when it is a 1. perfect square (e.g. 25, sqrt(x)=5 and statement 2 is true) 2. not a perfect square (e.g 10 sqrt(10)=3.16 and statement 2 is still true) can't deduce anything
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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27 Sep 2015, 08:36
Bunuel wrote: 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. i think it does not matter whether statement 2 is integer or not, as in both the cases we are getting different solution. and answer will be one in both the cases.



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Re: If x is a positive integer, is root(x) an integer? [#permalink]
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24 Dec 2015, 04:30
sqrt (4x) is an integer implies that 2 * sqrt(x) is also an integer. That implies x is an integer. Statement 1 is sufficient sqrt (3x) is not an integer. For x = 2 it is possible. For a number such as 4 also it is. But sqrt () is not an integer while sqrt (4) is. Insufficient Hence A
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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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12 Feb 2017, 14:41
I am new to gmatclub so I apologize if I am using this forum incorrectly... I instinctively want to plug in numbers: The way i solved was: Sqrt(4*1)= 2= integer but sqrt(1) is not an integer Sqrt(4*4)= 4= integer and sqrt(4)=2=integer
so I would say statement 1 is not sufficient. Clearly my logic is flawed since statement 1 is sufficient, but can someone tell me why making up numbers for X does not work for this problem?



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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12 Feb 2017, 14:44



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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12 Feb 2017, 15:00
wow, that was a dumb mistake on my part. thanks Bunuel !!



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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
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