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13 Jan 2014, 01:51
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If x is a positive integer, is \(\sqrt{x}\) an integer?
(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.
(DS04474)
(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.
(DS04474)
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13 Jan 2014, 01:52
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SOLUTION:
If x is a positive integer, is \(\sqrt{x}\) an integer?
The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, 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 \(\sqrt{x}\) cannot 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.
Answer: A.
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If x is a positive integer, is \(\sqrt{x}\) an integer?
The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, 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 \(\sqrt{x}\) cannot 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.
Answer: A.
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https://gmatclub.com/forum/if-z-is-a-po ... 01464.html
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https://gmatclub.com/forum/is-s-an-odd- ... 06562.html
Updated on: 31 Jul 2014, 16:18
Originally posted by arunspanda on 18 Jan 2014, 04:22.
Last edited by arunspanda on 31 Jul 2014, 16:18, edited 1 time in total.
Last edited by arunspanda on 31 Jul 2014, 16:18, edited 1 time in total.
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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)
(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....
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.................
Answer: (A)
