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If x is a positive integer, is x^1/2 an integer
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13 Jan 2014, 00:51
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Re: If x is a positive integer, is x^1/2 an integer
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13 Jan 2014, 00:52
SOLUTION:If x is a positive integer, is \(\sqrt{x}\) an integer?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 \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, 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. Similar questions: https://gmatclub.com/forum/ifnpqpa ... 01475.htmlhttps://gmatclub.com/forum/ifdisapo ... 09384.htmlhttps://gmatclub.com/forum/ifdisapo ... 67950.htmlhttps://gmatclub.com/forum/ifzisapo ... 01464.htmlhttps://gmatclub.com/forum/ifyisapo ... 08287.htmlhttps://gmatclub.com/forum/ifsqrt4ai ... 06886.htmlhttps://gmatclub.com/forum/whatisthe ... 07195.htmlhttps://gmatclub.com/forum/issanodd ... 06562.html
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Re: If x is a positive integer, is x^1/2 an integer
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14 Jan 2014, 06:06
IMO A. 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) \sqrt{4x} = 2\sqrt{x} is also integer = hence sufficient. 2) if x = 27 then Yes, if x = 4 then no 2. hence not sufficient.
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Re: If x is a positive integer, is x^1/2 an integer
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14 Jan 2014, 10:49
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. Statement 1) root(4x) is an integer. root(4x) = either +2root(x) or 2root(x) . As in either case, the result is an integer, then root(x) has to be an integer as (+2 or 2) is integer. Hence Sufficient. Statement 2) root(3x) is an integer. root(3) is not an integer. There can be following two cases: 1) root(x) is an integer => root(3) * root(x) is not an integer as root(3) is not an integer. 2) root(x) is not an integer => root(3) * root(x) can never be an integer except in once case where x= 3. Either case it is not sufficient. Hence Option A)
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Re: If x is a positive integer, is x^1/2 an integer
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Updated on: 31 Jul 2014, 15:18
If x is a positive integer, is \(\sqrt{x}\) an integer?
(1)\(\sqrt{4x}\) is an integer. (2)\(\sqrt{3x}\) is not an integer.
Statement 1
\(\sqrt{4x} = k\), \(k\) is an integer Or,\(\sqrt{x}= k/2\)
If \(k/2\) is not an integer, then \((k/2)^2\) or \(k^2/4\) is also not an integer and this implies that x is not an integer as \(x = k^2/4\). But, it is given that \(x\) is an integer.
Therefore, \(k^2/4\) is an integer => \(k/2\) is an integer =>\(\sqrt{x}\) is an integer...........Sufficient....(B)(C)(E)
Statement 2
\(\sqrt{3x}=k\), \(k\) is not an integer Or, \(\sqrt{x} = k/\sqrt{3}\)
Thus, \(\sqrt{x}\) = integer, when \(k\) is a multiple of \(\sqrt{3}\) and \(\sqrt{x}\) is not an integer when \(k\) is not a multiple of \(\sqrt{3}\)
As the value of \(\sqrt{x}\) can not be uniquely determined, statement (2) is not sufficient.................(D)
Answer: (A)
Originally posted by arunspanda on 18 Jan 2014, 03:22.
Last edited by arunspanda on 31 Jul 2014, 15:18, 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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07 Jan 2018, 04:45
[quote="Bunuel"] The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. Given x>0 & x=Integer Find is \(\sqrt{x}\) an integer => let \(\sqrt{x}=k\) 'k' is integer => OR \(x=k^2\) Therefore Question can be rephrased as " IS 'x' A PERFECT SQUARE "Statement 1 \(\sqrt{4x}\) is an integer => let \(\sqrt{4x}=p\)  'p' is Integer => \(4x=p^2\) => \(x=\frac{p^2}{4}\) => OR \(x=(\frac{p}{2})^2\)  equ (1) => Now Since (given) x>0 & x=Integer=> Therefore 'p' HAS TO BE 2,4,6,8....OR any even number >0 => substituting 'p' in equ (1) makes \(x=1^2 or 2^2 or 3^2\) and so on.. => Therefore x=Perfect Square. => SUFFICIENT Statement 2 \(\sqrt{3x}\) is not an integer. => So \(x\neq{3, 3^3, 3^5...}\) => BUT x=1,2,4,5,6,9,.... => So 'x' can be Prefect Square ( when x=4,9,16....) and 'x' NOT a Perfect Square (when x=2,5,6...) => Therefore INSUFFICIENT Therefore 'A' Thanks Dinesh



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Re: If x is a positive integer, is x^1/2 an integer
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15 Apr 2018, 12:25
For the first prompt, I was able to simplify to get me to here: √4x > 2√x The obvious part to me is that since the above is an integer, then x is a perfect square, totally understand that. However, (1/2)*(1/2) = (1/4). To me, the square root of (1/4) would be (1/2), and 2*(1/2) = 1, an integer. I know Bunuel has said: Quote: The square root of any positive integer is either an integer or an irrational number. So, x√=integer‾‾‾‾‾‾‾√x=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 2‾√2, 3‾√3, 17‾‾‾√17, ...). Is there a steadfast rule that square roots of nonintegers is out of scope of the GMAT? Should I just never consider square roots of fractions going forward?



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Re: If x is a positive integer, is x^1/2 an integer
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15 Apr 2018, 21:36
jsheppa wrote: For the first prompt, I was able to simplify to get me to here: √4x > 2√x The obvious part to me is that since the above is an integer, then x is a perfect square, totally understand that. However, (1/2)*(1/2) = (1/4). To me, the square root of (1/4) would be (1/2), and 2*(1/2) = 1, an integer. I know Bunuel has said: Quote: The square root of any positive integer is either an integer or an irrational number. So, x√=integer‾‾‾‾‾‾‾√x=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 2‾√2, 3‾√3, 17‾‾‾√17, ...). Is there a steadfast rule that square roots of nonintegers is out of scope of the GMAT? Should I just never consider square roots of fractions going forward? Hello NO, there is NO rule that square roots of nonintegers is out of scope of GMAT. But the thing is, that in this particular question, we are specifically given that x is an integer; so for 2*√x to be an integer, x must be a perfect square integer. So we cannot plug and test the value of x as 1/4 in this particular question, because 1/4 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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16 Apr 2018, 05:13
amanvermagmat wrote: Hello
NO, there is NO rule that square roots of nonintegers is out of scope of GMAT. But the thing is, that in this particular question, we are specifically given that x is an integer; so for 2*√x to be an integer, x must be a perfect square integer. So we cannot plug and test the value of x as 1/4 in this particular question, because 1/4 is not an integer. Oh wow...I must have been getting fatigued while studying and missed that part of the prompt. Thank you!



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Re: If x is a positive integer, is x^1/2 an integer
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20 Apr 2018, 12:36
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf x is a positive integer, is \(\sqrt{x}\) an integer? (1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer. Target question: Is √x an integer?Given: x is a positive integer Statement 1: √(4x) is an integerIMPORTANT CONCEPT: If K is an integer, then √K will be an integer if the prime factorization of K has an even number of each prime. Some examples: √144 = 12 (integer), and 144 = (2)(2)(2)(2)(3)(3) [four 2's and two 3's] √1600 = 40 (integer), and 1600 = (2)(2)(2)(2)(2)(2)(5)(5) [six 2's and two 5's] √441 = 21 (integer), and 441 = (3)(3)(7)(7)[two 3's and two 7's] √12 = some noninteger, and 12 = (2)(2)(3)[two 2's and one 3's]  So, if √(4x) is an integer, then the prime factorization of 4x has an even number of each prime. Since 4x = (2)(2)(x) we can see that the prime factorization of x must have an even number of each prime. If the prime factorization of x has an even number of each prime, then √x must be an integer.Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: √(3x) is not an integer.There are several values of x that meet this condition. Here are two: Case a: x = 4. This means that √(3x) = √12, which is not an integer. In this case, √x is an integer.Case b: x = 5. This means that √(3x) = √15, which is not an integer. In this case, √x is not an integer.Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent
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