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If z is a positive integer, is z^(1/2) an integer?

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

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New post 24 Dec 2017, 01:44
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If z is a positive integer, is z^(1/2) an integer? [#permalink]

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Bunuel wrote:
If z is a positive integer, is \(\sqrt{z}\) an integer?

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


Statement 1: \(\sqrt{3z}=Integer\)

\(=>\sqrt{z}=\frac{Integer}{\sqrt{3}}=>\frac{Integer}{irrational}\).

This will not be an integer. sufficient

Statement 2: \(\sqrt{4z}=non{ }integer=>2\sqrt{z}=noninteger\)

\(\sqrt{z}=\frac{noninteger}{2}=>\frac{noninteger}{integer}\). This will not be an integer. Sufficient

Option D
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Re: If z is a positive integer, is z^(1/2) an integer? [#permalink]

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Answer is D.
1. says √3z = I (I =Integer)
That is 3z must be square no to become √3z Integer. For example, z = 3, 12 etc. SUFFICIENT
2. √4z is not an Integer
For that, √4z must not be a square no. SUFFICIENT
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Re: If z is a positive integer, is z^(1/2) an integer? [#permalink]

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Bunuel wrote:
If z is a positive integer, is \(\sqrt{z}\) an integer?

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


\(\sqrt{z}\) will only be an integer if \(z\) is a perfect square {4,9,16,25,36,49,etc...}.

(1) \(\sqrt{3z}\) is an integer. Then, z must be a multiple of 3 multiplied by a perfect square, which makes z not a perfect square and \(\sqrt{z}\) not equal to an integer, sufficient.

(2) \(\sqrt{4z}\) is not an integer. Then, z is not a perfect square, because if it was, then the \(\sqrt{4z}\) would be an integer, sufficient.

(D) is the answer.
Re: If z is a positive integer, is z^(1/2) an integer?   [#permalink] 24 Dec 2017, 16:00
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