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# If Z is a positive integer is root(Z) an integer Root(X*Z)

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If Z is a positive integer is root(Z) an integer Root(X*Z) [#permalink]  21 Sep 2010, 15:03
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42% (02:45) correct 57% (00:56) wrong based on 19 sessions
If Z is a positive integer is root(Z) an integer

Root(X*Z) is an integer

X = Z^3
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Re: Algebra DS [#permalink]  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]  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

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Re: Algebra DS [#permalink]  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]  21 Sep 2010, 23:23
Expert's post
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: Algebra DS   [#permalink] 21 Sep 2010, 23:23
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