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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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13 Jan 2014, 00:51
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x is a positive integer, is $$\sqrt{x}$$ an integer?

(1) $$\sqrt{4x}$$ is an integer.
(2) $$\sqrt{3x}$$ is not an integer.

Data Sufficiency
Question: 31
Page: 155
Difficulty: 650

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

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13 Jan 2014, 00:52
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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.

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

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14 Jan 2014, 06:06
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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 [#permalink]

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14 Jan 2014, 10:49
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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18 Jan 2014, 03:22
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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.

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)

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 [#permalink]

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

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07 Jan 2018, 04:45
[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

If 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 re-phrased 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   [#permalink] 07 Jan 2018, 04:45
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# If x is a positive integer, is x^1/2 an integer

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