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Math Expert V
Joined: 02 Sep 2009
Posts: 55276
M26-30  [#permalink]

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20 00:00

Difficulty:   95% (hard)

Question Stats: 25% (01:19) correct 75% (01:16) wrong based on 286 sessions

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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.

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Math Expert V
Joined: 02 Sep 2009
Posts: 55276
Re M26-30  [#permalink]

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1
5
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.

Answer: C
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Re: M26-30  [#permalink]

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1
1
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.

Answer: C

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1
Math Expert V
Joined: 02 Sep 2009
Posts: 55276
Re: M26-30  [#permalink]

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parameswaranprasad wrote:
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.

Answer: C

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.
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M26-30  [#permalink]

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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.
Board of Directors P
Joined: 17 Jul 2014
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M26-30  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 55276
Re: M26-30  [#permalink]

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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.
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Re: M26-30  [#permalink]

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Really nice question Bunuel !
Intern  Joined: 04 Jan 2017
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Re: M26-30  [#permalink]

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Awesome question!
Intern  B
Joined: 24 Jun 2017
Posts: 14
GMAT 1: 750 Q49 V44 GPA: 3.1
Re: M26-30  [#permalink]

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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!
Intern  S
Joined: 01 May 2017
Posts: 18
Location: United States (IL)
Concentration: Technology, Strategy
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Re M26-30  [#permalink]

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I think this is a high-quality question and I agree with explanation. Awesome!!!
Manager  S
Joined: 06 Nov 2016
Posts: 61
Location: Viet Nam
Concentration: Strategy, International Business
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Re: M26-30  [#permalink]

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I think this is a high-quality question and I agree with explanation.
Didn't think abt the case z=1^3 in statement 1 urgggg... Need to be more careful.
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Intern  B
Joined: 28 Aug 2018
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Location: India
GMAT 1: 650 Q49 V30 Re: M26-30  [#permalink]

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Great Question, forgot to check the case for z= 1^3.
Manager  S
Joined: 20 Jul 2018
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Re: M26-30  [#permalink]

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1
by prime factorization of 2700 = $$2^2 * 3^3 * 5^2$$

1) if y is perfect square and z is perfect cube then y could be $$2^2 * 5^2$$ or $$2^2$$ or $$5^2$$ and z could be $$3^3$$ or $$1^3$$. so if z is $$3^3$$ then $$\sqrt{x}$$ is an integer but if z is $$1^3$$ then $$\sqrt{x}$$ is not an integer. INSUFFICIENT

2) if $$\sqrt{z}$$ is not an integer, its clearly INSUFFICIENT

by combining both we know than z could only take value of $$3^3$$, so $$\sqrt{x}$$ would always be an integer. SUFFICIENT

Answer C
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Intern  B
Joined: 19 Nov 2012
Posts: 28
M26-30  [#permalink]

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Bunuel wrote:
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.

Hi Bunuel: shouldn't one additional 3 be in included in the factorization of y? I guess that's a typo.
Needless to say Brilliant question. even understanding the given solution takes lot of efforts; forget solving it correctly.
Manager  B
Joined: 01 Nov 2018
Posts: 74
GMAT 1: 690 Q48 V35 GPA: 3.88
Re: M26-30  [#permalink]

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Great Question. 50-51 level. Concepts isn't difficult but LOTS to think about, especially during a time crunch. Picked D because short on time but well though out question. Re: M26-30   [#permalink] 23 Dec 2018, 13:26
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