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If y is a positive integer is root(y) an integer?

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rxs0005
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If y is a positive integer is root(y) an integer? [#permalink] New post   26 Jan 2011, 06:17
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If y is a positive integer is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer
(2) \(\sqrt{5y}\) is an integer
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Bunuel
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If y is a positive integer is root(y) an integer? [#permalink] New post   26 Jan 2011, 06:37
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If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) cannot be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Answer: D.

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Re: If y is a positive integer is root(y) an integer? [#permalink] New post   03 Feb 2014, 12:26
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(1) \(\sqrt{4y}\) is not an integer.

That is \(2\sqrt{y}\) is not an integer => \(\sqrt{y}\) is not an integer. Sufficient

(2) \(\sqrt{5y}\) is an integer

let p be this integer.

\(\sqrt{5y} = p\)

then \(\sqrt{5}\sqrt{y}=p\)

or \(\sqrt{y} = \frac{p}{\sqrt{5}}\)

Now an integer divided by an irrational number cannot be integer. Sufficient

D is the answer.
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Re: Airthmetic DS [#permalink] New post   26 Jan 2011, 06:47
thanks for the explanation the approach i took was

if y = 25 then root( 5* y) root(125) is an integer is valid and root(25) is an integer

if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A

why is this approach wrong
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Re: Airthmetic DS [#permalink] New post   26 Jan 2011, 07:08
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rxs0005 wrote:
thanks for the explanation the approach i took was

if y = 25 then root( 5* y) root(125) is an integer is valid and root(25) is an integer

if y = 20 then root( 5* y) root(100) is still an integer BUT root(20) is not an integer so i chose A

why is this approach wrong


\(\sqrt{125}=5\sqrt{5}\approx{11.18}\neq{integer}\).
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Re: If y is a positive integer, is square_root of y [#permalink] New post   30 Jun 2012, 04:08
Bunuel wrote:
metallicafan wrote:
If y is a positive integer, is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer.
(2) \(\sqrt{5y}\) is an integer.


If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Answer: D.

Hope it helps.


Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt:
Ok, we know that \(x\sqrt{5}\).
But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make \(x\sqrt{5}\) a integer?
Maybe, I am forgeting a concept. Please, your help.
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Re: If y is a positive integer, is square_root of y [#permalink] New post   30 Jun 2012, 04:16
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metallicafan wrote:
Bunuel wrote:
metallicafan wrote:
If y is a positive integer, is \(\sqrt{y}\) an integer?

(1) \(\sqrt{4y}\) is not an integer.
(2) \(\sqrt{5y}\) is an integer.


If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

Answer: D.

Hope it helps.


Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt:
Ok, we know that \(x\sqrt{5}\).
But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make \(x\sqrt{5}\) a integer?
Maybe, I am forgeting a concept. Please, your help.


The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Hope it's clear.
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Re: If y is a positive integer, is square_root of y [#permalink] New post   30 Jun 2012, 18:22
Bunuel wrote:
The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Hope it's clear.


Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational?
Thanks!
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Re: If y is a positive integer, is square_root of y [#permalink] New post   01 Jul 2012, 02:24
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metallicafan wrote:
Bunuel wrote:
The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).

Hope it's clear.


Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational?
Thanks!


Since no prime can be a perfect square (or perfect cube, ...), then \(\sqrt{prime}=irrational\).
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Re: If y is a positive integer, is root y an integer? [#permalink] New post   Updated on: 15 May 2013, 08:25
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If y is a positive integer, is root y an integer?

(1) Root 4y is not an integer.
(2) Root 5y is an integer.

is y a perfect square?


1) 2 * sqrt y is not an integer .............. therefore sqrt y is not an integer and thus not a perfect square.......suff

2) y could be 1/5 or 5^3 for example still it can never be a perfect square .......suff

D

Originally posted by yezz on 15 May 2013, 08:19.
Last edited by yezz on 15 May 2013, 08:25, edited 1 time in total.
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Re: If y is a positive integer, is square_root of y [#permalink] New post   13 Aug 2015, 15:04
Bunuel

What if y=125?

125 is an integer.

sqrt of 5*125 = sqrt of 625 = 25

This would make B insufficient.

Am I missing something here?
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Re: If y is a positive integer, is square_root of y [#permalink] New post   14 Aug 2015, 17:11
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Mascarfi wrote:
Bunuel

What if y=125?

125 is an integer.

sqrt of 5*125 = sqrt of 625 = 25

This would make B insufficient.

Am I missing something here?


No, your calculations are fine and are consistent with the question. If you have y = 125, \(\sqrt{125}\) \(\neq\) Integer and thus get an unambiguous "no" making statement 2 sufficient. The original question is

"\(\sqrt{y}\) an integer?" So getting an unambiguous yes or no will be sufficient. Per Bunuel's solution, it is shown that y can not be a perfect square leading to \(\sqrt{y}\neq Integer\)

Hope this helps.
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Re: If y is a positive integer is root(y) an integer? [#permalink] New post   30 Apr 2023, 05:40
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Re: If y is a positive integer is root(y) an integer? [#permalink]
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