GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Nov 2018, 14:10

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 07 Jun 2004
Posts: 600
Location: PA
If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

Updated on: 21 Nov 2017, 23:27
3
12
00:00

Difficulty:

75% (hard)

Question Stats:

54% (01:45) correct 46% (01:36) wrong based on 420 sessions

### HideShow timer Statistics

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

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

(2) x = z^3

_________________

If the Q jogged your mind do Kudos me : )

Originally posted by rxs0005 on 21 Sep 2010, 15:03.
Last edited by Bunuel on 21 Nov 2017, 23:27, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 50711
If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

21 Sep 2010, 23:23
7
2
rxs0005 wrote:
If Z is a positive integer is root(Z) an integer

Root(X*Z) is an integer

X = Z^3

If z is a positive integer is $$\sqrt{z}$$ an 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.
_________________
##### General Discussion
Current Student
Joined: 12 Jun 2009
Posts: 1766
Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

21 Sep 2010, 16:35
1
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?????
_________________

Retired Moderator
Joined: 02 Sep 2010
Posts: 769
Location: London
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

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

_________________
Manager
Joined: 20 Oct 2013
Posts: 54
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

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?
_________________

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

Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

04 May 2014, 23:25
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.
_________________
Manager
Joined: 20 Oct 2013
Posts: 54
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

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?
_________________

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

Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

05 May 2014, 00:31
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 $$\sqrt{z}=integer$$.
_________________
Manager
Joined: 20 Oct 2013
Posts: 54
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

05 May 2014, 01:01
1
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

Manager
Joined: 20 Feb 2017
Posts: 166
Location: India
Concentration: Operations, Strategy
WE: Engineering (Other)
Re: If z is a positive integer is root(Z) an integer ?  [#permalink]

### Show Tags

22 Nov 2017, 07:25
Easy E
from statement 1
x can be 8 and z can be 2
or x can be 9 and z can 4
hence insufficient
from statement 2
x=z^3
it can be anything since the nature of x is unknown
combining 1 and 2
we get \sqrt{z^4} = z^2 again insufficient z can be 2 or 4
_________________

If you feel the post helped you then do send me the kudos (damn theya re more valuable than \$)

Re: If z is a positive integer is root(Z) an integer ? &nbs [#permalink] 22 Nov 2017, 07:25
Display posts from previous: Sort by