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If x and y are positive integers, is (2xy)^(1/2) an integer?

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If x and y are positive integers, is (2xy)^(1/2) an integer?  [#permalink]

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New post 28 May 2017, 02:42
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A
B
C
D
E

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  25% (medium)

Question Stats:

77% (01:02) correct 23% (01:25) wrong based on 62 sessions

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Re: If x and y are positive integers, is (2xy)^(1/2) an integer?  [#permalink]

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New post 28 May 2017, 03:05
(1) says y=x and hence with x and y being positive integers the equation cannot be an integer (either x should have been 2y or y/2 ) hence sufficient

(2) substituting the value ..sufficient to answer

Hence D


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Re: If x and y are positive integers, is (2xy)^(1/2) an integer?  [#permalink]

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New post 29 May 2017, 03:44
Statement 1 gives x = y. So it gives us the answer that √(2xy) = x √(2) or y √(2) will not be integer. Sufficient.

Statement 2 gives x = 1/y. So it gives us the answer, √(2xy) = √(2) will not be integer. Sufficient.

Option D is correct.
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Re: If x and y are positive integers, is (2xy)^(1/2) an integer?  [#permalink]

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New post 04 Apr 2018, 11:58
Can any body explain this DS in a more detailed manner please? Thank you.
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Re: If x and y are positive integers, is (2xy)^(1/2) an integer?  [#permalink]

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New post 04 Apr 2018, 12:09
Tavo wrote:
Can any body explain this DS in a more detailed manner please? Thank you.


If x and y are positive integers, is \(\sqrt{2xy}\) an integer?

(1) y – x = 0. Rearrange to get x = y.

Hence, \(\sqrt{2xy}=\sqrt{2x^2}=x\sqrt{2}=integer*irrational\neq integer\) (x is an integer and \(\sqrt{2}\) is an irrational number. The product of a non-zero integer and an irrational number cannot be an integer). Sufficient.

(2) xy = 1.

Hence, \(\sqrt{2xy}=\sqrt{2}=irrational\neq integer\). Sufficient.

Answer: D.

Hope it's clear.
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Re: If x and y are positive integers, is (2xy)^(1/2) an integer?   [#permalink] 04 Apr 2018, 12:09
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