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(1) \(\sqrt{xz}=integer\), no info about \(x\), not sufficient. (2) \(x=z^3\) --> \(z^3\) equals to some number \(x\), clearly insufficient.

(1)+(2) \(\sqrt{xz}=\sqrt{z^4}=z^2=integer\) --> well, from the stem we know that \(z\) is an integer, so no wonder that \(z^2\) is also an integer, which means that we have no more info than at the beginning. Not sufficient.

Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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04 May 2014, 11:49

HI Bunnel

when you say \sqrt{xz}=integer and we already know that z is an Integer... I*I = I? i read in your other posts about irrational numbers: z cannot be an irrational number here as it is already given that it is a integer... therefore, shouldnt x also be an integer.. basically how would knowing anything about x help?
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Hope to clear it this time!! GMAT 1: 540 Preparing again

when you say \sqrt{xz}=integer and we already know that z is an Integer... I*I = I? i read in your other posts about irrational numbers: z cannot be an irrational number here as it is already given that it is a integer... therefore, shouldnt x also be an integer.. basically how would knowing anything about x help?

The stem says that \(z=integer\) and (1) says \(\sqrt{xz}=integer\). Is it necessary for x to be integer?

No. For example, consider \(z=2\) and \(x=\frac{9}{2}\) --> \(\sqrt{xz}=3=integer\).

Even if we knew that x were an integer the first statement still would not be sufficient: consider \(z=2\) (\(\sqrt{z}=\sqrt{2}\neq{integer}\)) and \(x=2\) --> \(\sqrt{xz}=2=integer\).

Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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05 May 2014, 01:18

hi

but in both examples z = sqrt 2 which is not an integer.... i am confused regarding the fact that y who we need to know about x... for example: if z=2 and x = 2, sqrt xz= Integer but sqrt z is not and integer. if z = 4 and x = 9/4, still sqrt xz = 3, Integer but sqrt z is an integer... so how is knowing the value of x even relevant for the qs?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

but in both examples z = sqrt 2 which is not an integer.... i am confused regarding the fact that y who we need to know about x... for example: if z=2 and x = 2, sqrt xz= Integer but sqrt z is not and integer. if z = 4 and x = 9/4, still sqrt xz = 3, Integer but sqrt z is an integer... so how is knowing the value of x even relevant for the qs?

Are you asking what should we know about x, so that the first statement would be sufficient? If for example, we were told that x is a perfect square, then the first statement would be sufficient to say that \(\sqrt{z}=integer\).
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Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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05 May 2014, 02:01

1

This post received KUDOS

so it means that:

case 1: if x = 9 (perfect sq) and z = cannot be any integer as it will not satisfy the equation sqrt xz = i... it has to be a perfect sq it self, such as 4... sqrt 9*4 = 6

Case 2: but if it x - not a perfect square say 3, then z will have to be 3 so satisfy the equation sqrt xz = Integer, but in that case sqrt z will not be an integer..

Case 3: if x is a reduced fraction: 5/2 then z will have to be 2 or 8 to satisfy the equation sqrt xz = Integer and even in this case sqrt z is not an integer

so the fact that \sqrt{z}is an integer depends on value of x.. therefore, Insufficient.

I hope i have finally understood.

st 2... in this x= z3 means nothing as z= cuberoot x... x can be 27 in this case z= 3 but \sqrt{z}= irrational number or x can be 64.. making z = 4 and \sqrt{z} and integer... Insufficient again

together (st 1 & st2): \sqrt{xz} = \sqrt{z^4} = z^2 = Integer.... but \sqrt{z} can be an integer or irrational number

i think i have got it now. Am i?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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15 Sep 2015, 23:12

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Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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17 Sep 2016, 01:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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