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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 x^(1/2) an integer ?  [#permalink]

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New post 15 Jan 2015, 07:09
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Re: If x is a positive integer, is x^(1/2) an integer ?  [#permalink]

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New post 15 Jan 2015, 18:09
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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 ?  [#permalink]

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New post 17 Jan 2015, 09:43
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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 ?  [#permalink]

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New post 07 Feb 2015, 10:36
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 ?  [#permalink]

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New post 10 Feb 2015, 18:14
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 ?  [#permalink]

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New post 11 Feb 2015, 02:29
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 ?  [#permalink]

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New post 03 May 2016, 19:45
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 ?  [#permalink]

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New post 05 Oct 2017, 20:21
(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 ?  [#permalink]

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New post 16 Jul 2018, 00:31
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 ?  [#permalink]

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New post 16 Jul 2018, 00:44
sidsst

Quote:
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 suff

I 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 ?  [#permalink]

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New post 16 Jul 2018, 02:33
adkikani wrote:
sidsst

Quote:
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 suff

I 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 adkikani

yes 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

\(3x+4=I^2=>x=\frac{I^2-4}{3}\). Now take square root of both sides

\(\sqrt{x}=\sqrt{\frac{I^2-4}{3}}\), 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 [#permalink] 16 Jul 2018, 02:33
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