It is currently 20 Jan 2018, 16:39

### 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 June
Open Detailed Calendar

# M26-30

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139604 [0], given: 12794

### Show Tags

16 Sep 2014, 00:26
Expert's post
10
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

29% (01:09) correct 71% (01:27) wrong based on 48 sessions

### HideShow timer Statistics

If $$x$$, $$y$$, and $$z$$ are positive integers and $$xyz=2,700$$. Is $$\sqrt{x}$$ an integer?

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube.

(2) $$\sqrt{z}$$ is not an integer.
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139604 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139604 [0], given: 12794

### Show Tags

16 Sep 2014, 00:26
Expert's post
3
This post was
BOOKMARKED
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

_________________

Kudos [?]: 139604 [0], given: 12794

Intern
Joined: 11 Aug 2014
Posts: 5

Kudos [?]: 2 [0], given: 31

### Show Tags

15 Nov 2014, 23:21
1
This post was
BOOKMARKED
Bunuel wrote:
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1

Kudos [?]: 2 [0], given: 31

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139604 [0], given: 12794

### Show Tags

16 Nov 2014, 05:06
Bunuel wrote:
Official Solution:

Note: a perfect square, is an integer that can be written as the square of some other integer. For example $$16=4^2$$, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, $$27=3^3$$ is a perfect cube.

Make prime factorization of 2,700: $$xyz=2,700=2^2*3^3*5^2$$.

(1) $$y$$ is an even perfect square and $$z$$ is an odd perfect cube. If $$y$$ is either $$2^2$$ or $$2^2*5^2$$ and $$z=3^3= \text{odd perfect square}$$ then $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. But if $$z=1^3= \text{odd perfect cube}$$ then $$x$$ could be $$3^3$$ which makes $$\sqrt{x}$$ not an integer. Not sufficient.

(2) $$\sqrt{z}$$ is not an integer. Clearly insufficient.

(1)+(2) As from (1) $$\sqrt{z} \ne integer$$ then $$z \ne 1$$, therefore it must be $$3^3$$ (from 1), so $$x$$ must be a perfect square which makes $$\sqrt{x}$$ an integer: $$x=5^2$$ or $$x=1$$. Sufficient.

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1

$$z \ne 1$$ because if it were 1, then $$\sqrt{z}$$ would be an integer and that would violate the second statement.
_________________

Kudos [?]: 139604 [0], given: 12794

Intern
Joined: 30 Jun 2012
Posts: 13

Kudos [?]: 3 [0], given: 0

### Show Tags

07 Dec 2014, 15:21
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

Kudos [?]: 3 [0], given: 0

Board of Directors
Joined: 17 Jul 2014
Posts: 2719

Kudos [?]: 463 [0], given: 211

Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)

### Show Tags

07 Dec 2014, 15:57
rsamant wrote:
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.

Kudos [?]: 463 [0], given: 211

Math Expert
Joined: 02 Sep 2009
Posts: 43334

Kudos [?]: 139604 [0], given: 12794

### Show Tags

08 Dec 2014, 04:10
mvictor wrote:
rsamant wrote:
Why is the answer not D) Why is 2) not sufficient, if the square root of Z is not an integer that means that Z must be 3^ 3 and x must be either 5^2 or 2^2 and the square root of either of those number is an integer.

Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=2^2*3^3*5^2.

would like as well answer to this question.

There are other cases possible. For example, z=3, y=2*3*5^2 and x=2.
_________________

Kudos [?]: 139604 [0], given: 12794

Intern
Joined: 23 Apr 2016
Posts: 22

Kudos [?]: 10 [0], given: 39

Location: Finland
GPA: 3.65

### Show Tags

07 Nov 2016, 09:24
Really nice question Bunuel !

Kudos [?]: 10 [0], given: 39

Intern
Joined: 04 Jan 2017
Posts: 5

Kudos [?]: [0], given: 0

### Show Tags

10 Apr 2017, 08:19
Awesome question!

Kudos [?]: [0], given: 0

Intern
Joined: 24 Jun 2017
Posts: 15

Kudos [?]: 1 [0], given: 22

GMAT 1: 750 Q49 V44
GPA: 3.1

### Show Tags

30 Sep 2017, 08:21
Wow what a question! Very difficult without blatantly trying to trick you.

This one took me a while, but I'm very proud to say I got it right!

Kudos [?]: 1 [0], given: 22

Re: M26-30   [#permalink] 30 Sep 2017, 08:21
Display posts from previous: Sort by

# M26-30

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.