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If x is a positive integer, is x^(1/2) an integer ?
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15 Jan 2015, 06:09
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Re: If x is a positive integer, is x^(1/2) an integer ?
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15 Jan 2015, 17:09
Bunuel wrote: If x is a positive integer, is \(\sqrt{x}\) an integer?
(1) \(\sqrt{36x}\) is an integer
(2) \(\sqrt{3x + 4}\) is an integer
Kudos for a correct solution. Have started following you Bunuel and this is my first crack at any of your questions. 1. Sufficient: Can be written as \(\sqrt{36}\) * \(\sqrt{x}\). \(\sqrt{36}\) is an integer and therefore \(\sqrt{x}\) must be an integer given that x is a positive integer. 2. Insufficient: Let's take x to be 7 in which case \(\sqrt{3x + 4}\) = \(\sqrt{25}\), which is an integer, but root of 7 is not an integer. On the other hand, if you take x=4, then the statement holds true and root of x is also an integer. Answer is A. Will I get my first Kudo from you? :D If the solution is incorrect, your feedback will be more valuable than a Kudo
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Re: If x is a positive integer, is x^(1/2) an integer ?
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17 Jan 2015, 08:43
If x is a positive integer, is \sqrt{x} an integer? (1) \sqrt{36x} is an integer (2) \sqrt{3x + 4} is an integer Statement 1. \sqrt{36x}=6(x^(1/2)). Hence 6 is an integer then (x^1/2) should be integer too. Sufficient Statement 2. X can be 4 and then (x^1/2) is an integer. Or x=7 then (x^1/2) is not an integer. Insufficient. Answer A
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Re: If x is a positive integer, is x^(1/2) an integer ?
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07 Feb 2015, 09:36
Well, I am not sure if I am off here, but this is what I though:
x>0, is \sqrt{x} an integer?
For SQRT of x to be an integer, then x=1, or x>=4 and a perfect square (otherwise the SQRT would not be an integer). Just some initial thoughts.
[1] says that SQRT of 36x is an integer. Based on my thinking above, 36x must be a perfect square, apparently greater than 36. For this to be true, x must be a perfect square. Any perfect square multiplied by 36 would have a SQRT that is an integer. So, if x is a perfect sqaure its SQRT will be an integer. So, [1] is sufficient.
[2] says that SQRT of 3x+4 is an integer. For x=4, we have 3*4+4=16, and SQRT 16 is 4, which is an integer. For x=5, we have 3*5+4=19, and SQRT 19 is not a integer. So, [2] is not sufficient. ANS A.
Am I right...?



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Re: If x is a positive integer, is x^(1/2) an integer ?
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10 Feb 2015, 17:14
Hi Bunuel  I'm a little confused about the OA. If X = 1, then (1)^(1/2) is an integer but if X = 2, then 2^(1/2) is not an integer. Why is A correct?



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Re: If x is a positive integer, is x^(1/2) an integer ?
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11 Feb 2015, 01:29
Hi cg0588! We are told that \(\sqrt{36x}\) is an integer. Hence if you square out 6 then root of x should be integer (in order to \sqrt{36x} to be integer). Hence you can't say that x=2 since square root of 2 is not an integer. Hope it is clear
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Re: If x is a positive integer, is x^(1/2) an integer ?
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03 May 2016, 18:45
Bunuel wrote: If x is a positive integer, is \(\sqrt{x}\) an integer?
(1) \(\sqrt{36x}\) is an integer
(2) \(\sqrt{3x + 4}\) is an integer
Kudos for a correct solution. 1. sqrt(36x) = 6*sqrt(x) if this is an integer, then sqrt(x) is an integer. sufficient 2. if x=4, then yes. if x=7 then no. 2 alone 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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05 Oct 2017, 19:21
(1) 36x−−−√36x is an integer since x is a positive integer root[x] needs to be an integer (2) 3x+4−−−−−√3x+4 is an integer : if x is 7 then answer will be 5 hence not sufficient A it is
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Re: If x is a positive integer, is x^(1/2) an integer ?
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15 Jul 2018, 23:31
From statement 1 : \sqrt{36x}= 6\sqrt{x} Thus for \sqrt{36x} to be an integer, \sqrt{x} has to be an integer. Statement 1 : Sufficient. Statement 2 : \sqrt{3x+4} : putting x =1 , \sqrt{7}  > not an integer putting x =4 \sqrt{16} =4  > integer two contradictory answers. Hence not sufficient Thus Answer = A
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Re: If x is a positive integer, is x^(1/2) an integer ?
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15 Jul 2018, 23:44
sidsstQuote: Statement 2 :
\sqrt{3x+4} :
putting x =1 , \sqrt{7}  > not an integer
I am unsure if above approach is correct. We are given that \(\sqrt{3x+4}\) is an integer so if we take x=1 we are not satisfying this condition. Note that each statement by itself is suffI took x as 1,2,3 ..and so on, and found statement 2 satisfying for values of 4 and 7 niks18 pushpitkc Bunuel chetan2u pikolo2510 Any better approach?
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Re: If x is a positive integer, is x^(1/2) an integer ?
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16 Jul 2018, 01:33
adkikani wrote: sidsstQuote: Statement 2 :
\sqrt{3x+4} :
putting x =1 , \sqrt{7}  > not an integer
I am unsure if above approach is correct. We are given that \(\sqrt{3x+4}\) is an integer so if we take x=1 we are not satisfying this condition. Note that each statement by itself is suffI took x as 1,2,3 ..and so on, and found statement 2 satisfying for values of 4 and 7 niks18 pushpitkc Bunuel chetan2u pikolo2510 Any better approach? Hi adkikaniyes you are correct in your assessment. we need to find whether \(\sqrt{x}\) is an integer and not the validity of statement 2. if \(x=1\), then \(\sqrt{x}\) is an integer Statement 2: Given \(\sqrt{3x+4}=Integer\), square both sides to get \(3x+4=I^2=>x=\frac{I^24}{3}\). Now take square root of both sides \(\sqrt{x}=\sqrt{\frac{I^24}{3}}\), now this expression may or may not be an Integer. For eg, if I=2, then it will be an integer and if I=3, then it will be not an integer. This is just one way to analyze statement 2




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