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Re M2630
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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. Answer: C
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Re: M2630
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16 Nov 2014, 00:21
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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Re: M2630
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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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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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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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Re: M2630
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08 Dec 2014, 05:10



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Re: M2630
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07 Nov 2016, 10:24
Really nice question Bunuel !



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Re: M2630
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10 Apr 2017, 09:19
Awesome question!



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Re: M2630
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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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Re M2630
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29 Jul 2018, 03:43
I think this is a highquality question and I agree with explanation. Awesome!!!



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Re: M2630
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29 Jul 2018, 10:33



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Re: M2630
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28 Aug 2018, 07:56
Great Question, forgot to check the case for z= 1^3.



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Re: M2630
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30 Aug 2018, 08:26
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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