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# M26-30

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Math Expert
Joined: 02 Sep 2009
Posts: 49303

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16 Sep 2014, 01:26
18
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Difficulty:

95% (hard)

Question Stats:

24% (01:21) correct 76% (01:17) wrong based on 232 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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16 Sep 2014, 01:26
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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
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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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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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.
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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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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
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GMAT 1: 750 Q49 V44
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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!
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Joined: 01 May 2017
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29 Jul 2018, 03:43
I think this is a high-quality question and I agree with explanation. Awesome!!!
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29 Jul 2018, 10:33
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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28 Aug 2018, 07:56
Great Question, forgot to check the case for z= 1^3.
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30 Aug 2018, 08:26
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

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Re: M26-30 &nbs [#permalink] 30 Aug 2018, 08:26
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# M26-30

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