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

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Math Expert
Joined: 02 Sep 2009
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If z is a positive integer, is z^(1/2) an integer?  [#permalink]

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24 Dec 2017, 01:44
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Difficulty:

45% (medium)

Question Stats:

61% (01:36) correct 39% (01:31) wrong based on 55 sessions

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

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

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

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24 Dec 2017, 02:59
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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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24 Dec 2017, 07:05
1
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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24 Dec 2017, 16:00
2
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.

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