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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] New post 21 Sep 2010, 15:03
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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
[Reveal] Spoiler: OA

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Re: Algebra DS [#permalink] New post 21 Sep 2010, 16:35
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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Re: Algebra DS [#permalink] New post 21 Sep 2010, 18:48
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

Answer is (E)
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Re: Algebra DS [#permalink] New post 21 Sep 2010, 20:24
E it is,,,... introduced X and then there are not enough info to solve for X or Z.
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Re: Algebra DS [#permalink] New post 21 Sep 2010, 23:23
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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.

Answer: E.

Hope it helps.
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Re: If z is a positive integer is root(Z) an integer ? [#permalink] New post 04 May 2014, 10:49
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] New post 04 May 2014, 23:25
Expert's post
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] New post 05 May 2014, 00:18
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] New post 05 May 2014, 00:31
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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] New post 05 May 2014, 01:01
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] 05 May 2014, 01:01
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