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M26-30

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M26-30  [#permalink]

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New post 16 Sep 2014, 01:26
20
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

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Re M26-30  [#permalink]

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New post 16 Sep 2014, 01:26
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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New post 16 Nov 2014, 00:21
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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
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New post 16 Nov 2014, 06:06
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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New post 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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M26-30  [#permalink]

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New post 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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New post 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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Re: M26-30  [#permalink]

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New post 07 Nov 2016, 10:24
Really nice question Bunuel !
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New post 10 Apr 2017, 09:19
Awesome question!
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New post 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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New post 29 Jul 2018, 03:43
I think this is a high-quality question and I agree with explanation. Awesome!!!
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New post 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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New post 28 Aug 2018, 07:56
Great Question, forgot to check the case for z= 1^3.
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New post 30 Aug 2018, 08:26
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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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New post 29 Sep 2018, 09:10
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.
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New post 23 Dec 2018, 13:26
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.
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Re: M26-30   [#permalink] 23 Dec 2018, 13:26
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