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If z is a positive integer is root(Z) an integer ?

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

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21 Sep 2010, 16:03
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If z is a positive integer is $$\sqrt{z}$$ an integer?

(1) $$\sqrt{xz}$$ is an integer

(2) x = z^3
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21 Sep 2010, 17:35
rxs0005 wrote:
If Z is a positive integer is root(Z) an integer

Root(X*Z) is an integer

X = Z^3

Z = int so is sqrt(z) = z^1/2 = int? for example, if z = 4 then 4^.5 = 2 and z =3 then 3^.5 != int

1. for this to be integer X has to be Z or their product is squarable - INSUFF

2. X =Z^3 - INSUFF by itself.

C. if we know both it can be Z^3 * Z =Z^4 Z^4 * .5 = Z^2. NOTE this is still insuff as we dont know Z - it could be anything...

E?????
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21 Sep 2010, 19:48
rxs0005 wrote:
If Z is a positive integer is root(Z) an integer

Root(X*Z) is an integer

X = Z^3

(1) Not sufficient. Eg. Z=25, X=25 ; Z=3, X=3
(2) X=Z^3. No relation to Z. Not sufficient

(1)+(2) Not sufficient. Eg. Z=25, X=25^3 ; Z=3, X=3^3

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21 Sep 2010, 21:24
E it is,,,... introduced X and then there are not enough info to solve for X or Z.
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22 Sep 2010, 00:23
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rxs0005 wrote:
If Z is a positive integer is root(Z) an integer

Root(X*Z) is an integer

X = Z^3

Question: is $$\sqrt{z}=integer$$?

(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.

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

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05 May 2014, 00:25
nandinigaur wrote:
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?

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 $$x=2$$ --> $$\sqrt{xz}=2=integer$$.

Hope it's clear.
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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?
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Re: If z is a positive integer is root(Z) an integer ? [#permalink]

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05 May 2014, 01:31
nandinigaur wrote:
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?

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

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

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Re: If z is a positive integer is root(Z) an integer ?   [#permalink] 17 Sep 2016, 01:26
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