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# If y is a positive integer, is square_root of y

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If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 03:46
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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.
[Reveal] Spoiler: OA

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Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 03:50
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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.

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Hope it helps.
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Re: If y is a positive integer, is square_root of y [#permalink]

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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.

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?
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Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 04:16
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.

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?

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]

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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]

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01 Jul 2012, 02:24
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 square_root of y [#permalink]

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12 Jun 2013, 04:27
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Re: If y is a positive integer, is square_root of y [#permalink]

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12 Aug 2015, 19:52
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Re: If y is a positive integer, is square_root of y [#permalink]

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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]

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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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If y is a positive integer, is square_root of y [#permalink]

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14 Aug 2015, 17:48
It's clear now. Completely forgot what the question was asking while looking at number 2. thanks

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Re: If y is a positive integer, is square_root of y [#permalink]

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06 Nov 2016, 08:27
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Re: If y is a positive integer, is square_root of y   [#permalink] 06 Nov 2016, 08:27
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