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# M26-30

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Math Expert
Joined: 02 Sep 2009
Posts: 46291

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16 Sep 2014, 01:26
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Difficulty:

95% (hard)

Question Stats:

28% (01:14) correct 72% (01:26) wrong based on 54 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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Joined: 02 Sep 2009
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16 Sep 2014, 01:26
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.

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Joined: 11 Aug 2014
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16 Nov 2014, 00:21
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.

can you please explain 2) \sqrt{z} is not an integer. Clearly insufficient. and sqrt{z} ne integer then z ne 1
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Joined: 02 Sep 2009
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16 Nov 2014, 06:06
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.
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Joined: 30 Jun 2012
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07 Dec 2014, 16:21
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.
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07 Dec 2014, 16:57
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.
Math Expert
Joined: 02 Sep 2009
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08 Dec 2014, 05:10
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.
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Joined: 23 Apr 2016
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07 Nov 2016, 10:24
Really nice question Bunuel !
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Joined: 04 Jan 2017
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10 Apr 2017, 09:19
Awesome question!
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Joined: 24 Jun 2017
Posts: 15
GMAT 1: 750 Q49 V44
GPA: 3.1

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30 Sep 2017, 09:21
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!
Re: M26-30   [#permalink] 30 Sep 2017, 09:21
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# M26-30

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