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Re: If x is a positive integer, is x^1/2 an integer ? [#permalink]

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15 Jan 2015, 17:09

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Bunuel wrote:

If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{36x}\) is an integer

(2) \(\sqrt{3x + 4}\) is an integer

Kudos for a correct solution.

Have started following you Bunuel and this is my first crack at any of your questions.

1. Sufficient: Can be written as \(\sqrt{36}\) * \(\sqrt{x}\). \(\sqrt{36}\) is an integer and therefore \(\sqrt{x}\) must be an integer given that x is a positive integer. 2. Insufficient: Let's take x to be 7 in which case \(\sqrt{3x + 4}\) = \(\sqrt{25}\), which is an integer, but root of 7 is not an integer. On the other hand, if you take x=4, then the statement holds true and root of x is also an integer.

Answer is A. Will I get my first Kudo from you? :D If the solution is incorrect, your feedback will be more valuable than a Kudo
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"Hardwork is the easiest way to success." - Aviram

One more shot at the GMAT...aiming for a more balanced score.

Re: If x is a positive integer, is x^1/2 an integer ? [#permalink]

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17 Jan 2015, 08:43

1

This post received KUDOS

If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{36x} is an integer

(2) \sqrt{3x + 4} is an integer Statement 1. \sqrt{36x}=6(x^(1/2)). Hence 6 is an integer then (x^1/2) should be integer too. Sufficient Statement 2. X can be 4 and then (x^1/2) is an integer. Or x=7 then (x^1/2) is not an integer. Insufficient. Answer A
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Re: If x is a positive integer, is x^1/2 an integer ? [#permalink]

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07 Feb 2015, 09:36

Well, I am not sure if I am off here, but this is what I though:

x>0, is \sqrt{x} an integer?

For SQRT of x to be an integer, then x=1, or x>=4 and a perfect square (otherwise the SQRT would not be an integer). Just some initial thoughts.

[1] says that SQRT of 36x is an integer. Based on my thinking above, 36x must be a perfect square, apparently greater than 36. For this to be true, x must be a perfect square. Any perfect square multiplied by 36 would have a SQRT that is an integer. So, if x is a perfect sqaure its SQRT will be an integer. So, [1] is sufficient.

[2] says that SQRT of 3x+4 is an integer. For x=4, we have 3*4+4=16, and SQRT 16 is 4, which is an integer. For x=5, we have 3*5+4=19, and SQRT 19 is not a integer. So, [2] is not sufficient. ANS A.

If x is a positive integer, is x^1/2 an integer ? [#permalink]

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11 Feb 2015, 01:29

Hi cg0588! We are told that \(\sqrt{36x}\) is an integer. Hence if you square out 6 then root of x should be integer (in order to \sqrt{36x} to be integer). Hence you can't say that x=2 since square root of 2 is not an integer. Hope it is clear
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