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

As given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.

(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2)\(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

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

(1) \(\sqrt{4x}\) is an integer. (2) \(\sqrt{3x}\) is not an integer.

Statement 1) root(4x) is an integer. root(4x) = either +2root(x) or -2root(x) . As in either case, the result is an integer, then root(x) has to be an integer as (+2 or -2) is integer. Hence Sufficient. Statement 2) root(3x) is an integer. root(3) is not an integer. There can be following two cases: 1) root(x) is an integer => root(3) * root(x) is not an integer as root(3) is not an integer. 2) root(x) is not an integer => root(3) * root(x) can never be an integer except in once case where x= 3. Either case it is not sufficient.

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

As given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.

(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2)\(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

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

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18 Jan 2014, 04:22

1

This post received KUDOS

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

(1)\(\sqrt{4x}\) is an integer. (2)\(\sqrt{3x}\) is not an integer.

Statement 1

\(\sqrt{4x} = k\), \(k\) is an integer Or,\(\sqrt{x}= k/2\)

If \(k/2\) is not an integer, then \((k/2)^2\) or \(k^2/4\) is also not an integer and this implies that x is not an integer as \(x = k^2/4\). But, it is given that \(x\) is an integer.

Therefore, \(k^2/4\) is an integer => \(k/2\) is an integer =>\(\sqrt{x}\) is an integer...........Sufficient....(B)(C)(E)

Statement 2

\(\sqrt{3x}=k\), \(k\) is not an integer Or, \(\sqrt{x} = k/\sqrt{3}\)

Thus, \(\sqrt{x}\) = integer, when \(k\) is a multiple of \(\sqrt{3}\) and \(\sqrt{x}\) is not an integer when \(k\) is not a multiple of \(\sqrt{3}\)

As the value of \(\sqrt{x}\) can not be uniquely determined, statement (2) is not sufficient.................(D)

Answer: (A)

Last edited by arunspanda on 31 Jul 2014, 16:18, edited 1 time in total.

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

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25 Feb 2016, 13:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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