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If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]
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 25 Dec 2010, 11:39
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If \(\sqrt{4a}\) is an integer, is \(\sqrt{a}\) an integer (1) a is a positive integer (2) a = n^6, where n is an integer
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Last edited by Bunuel on 27 Oct 2013, 06:15, edited 1 time in total.
Edited the OA.
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Re: Number Prop DS [#permalink]
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25 Dec 2010, 12:42
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sqrt(4*a) = sqrt(4)*sqrt(a) = 2*sqrt(a) 1) The sign (+) doesnt really say anything since the function sqrt(a) is not defined for a<0 which we already knew. It could still be 3 => sqrt(3)*2 = not integer or 2*sqrt(4) = integer. 2) a = n^6 => a^(1/6) = n => (a^(1/2))^(1/3) = n = an integer a^(1/2) must be an integer if a^(1/6) is an integer. With numbers: 2^6 = 64 64^(1/2) = 8 64^(1/3) = 4 64^(1/6) = 2
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Re: Number Prop DS [#permalink]
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25 Dec 2010, 16:14
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This one is strange, because if If sqrt( 4a ) = integer then 2*sqrt(a) must be integer -- and that is integer when sqrt(a) in integer
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Re: Number Prop DS [#permalink]
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30 Dec 2010, 14:24
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I said the same thing than craky. Ans. D for me. But as previously mentioned the question is weirdly formulated. Anyone cares to comment? 
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Re: Number Prop DS [#permalink]
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30 Dec 2010, 18:36
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Don't know if I understand you guys but lets try with two numbers. If a = 1.5^2 = 2.25 sqrt(4a) = sqrt(4 * 2.25) = sqrt ( 9 ) = 3 = integer sqrt(a) = sqrt(2.25) = 1.5 = not integer If a= 4 sqrt(4a) = sqrt(16) = 4 = integer sqrt(4) = 2 = integer The questions is: IF sqrt(4a) is an integer, will sqrt(a) be an integer as well? The converse is true but that is not the question here.
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Re: Number Prop DS [#permalink]
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30 Dec 2010, 20:57
craky wrote: This one is strange, because if
If sqrt( 4a ) = integer
then
2*sqrt(a) must be integer -- and that is integer when sqrt(a) in integer Not necessarily true. Even if \(\sqrt{a} = \frac{1}{2}, 2\sqrt{a} = 1\), an integer. So if \(2\sqrt{a}\) is an integer, we cannot say whether \(\sqrt{a}\) is an integer. Sneaky GMAT tricks!
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Re: Number Prop DS [#permalink]
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rxs0005 wrote: If sqrt( 4a ) = integer is sqrt ( a ) an integer
(1) a is a positive integer
(2) a = n^6 where n is an integer Given: \(\sqrt{4a}=integer\) --> \(2\sqrt{a}=integer\): so either \(\sqrt{a}=integer\) (note that in this case \(a\) will be an integer, more it'll be a perfect square: 1, 4, 9, ...) or \(\sqrt{a}=\frac{odd \ integer}{2}\) (1/2, 3/2, 5/2, ... note that in this case \(a\) won't be an integer it'll equal to: 1/4, 9/4, 25/4, ...). Question asks whether we have the first case: is \(\sqrt{a}=integer\) (or as we can have only two cases question basically asks whether \(a\) is an integer)? (1) a is a positive integer --> we have the first case. Sufficient. Or: square root of an integer (\(\sqrt{a}\)) is either an integer or an irrational number, so we can not have the second case (if \(a=integer\) then \(\sqrt{a}\) can not equal to \(\frac{odd \ integer}{2}\)). Sufficient. (2) a = n^6 where n is an integer --> \(\sqrt{a}=n^3\) and as \(n=integer\) then \(\sqrt{a}=n^3=integer\). Sufficient. Answer: D.
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Re: Number Prop DS [#permalink]
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08 Jan 2011, 18:24
VeritasPrepKarishma,
Yes, but (1) tells us that A is a positive integer and therefore can't be 1/2 or 1/4? Therefore, wouldn't the answer be D?
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Re: Number Prop DS [#permalink]
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08 Jan 2011, 21:59
m990540 wrote: VeritasPrepKarishma,
Yes, but (1) tells us that A is a positive integer and therefore can't be 1/2 or 1/4? Therefore, wouldn't the answer be D? Yes definitely. My reply was for the quoted comment above ' This one is strange, because if (I think he implied that you don't need the statements to answer the question)
If sqrt( 4a ) = integer
then
2*sqrt(a) must be integer and that is integer when sqrt(a) in integerI had said that this is not true since 2*sqrt(a) can be an integer even if sqrt(a) is not. Implication: We need the statements to get to the answer, whatever the answer is. Think again and try to get the answer.
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Re: Number Prop DS [#permalink]
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08 Jan 2011, 23:50
rxs0005 wrote: If sqrt( 4a ) = integer is sqrt ( a ) an integer
(1) a is a positive integer
(2) a = n^6 where n is an integer 1. For \(sqrt(4a)\) to be an integer, a has to be a perfect square. therefore \(sqrt(a)\) will definitely be an integer. Sufficient 2. This statement says that a is a perfect square (of \(n^3\)). Sufficient. Answer D.
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Re: If sqrt( 4a ) = integer is sqrt ( a ) an integer (1) a is a [#permalink]
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27 Oct 2013, 06:12
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Bunuel wrote: rxs0005 wrote: If sqrt( 4a ) = integer is sqrt ( a ) an integer
(1) a is a positive integer
(2) a = n^6 where n is an integer Answer: D. If the OA is D, can somebody please correct the OA in the post? TIA  !
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Re: If sqrt( 4a ) = integer is sqrt ( a ) an integer (1) a is a [#permalink]
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27 Oct 2013, 06:15
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]
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04 Feb 2014, 06:10
Wow very sneaky indeed.
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]
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26 Jun 2016, 11:21
rxs0005 wrote: If \(\sqrt{4a}\) is an integer, is \(\sqrt{a}\) an integer
(1) a is a positive integer
(2) a = n^6, where n is an integer if a= 1/4, then \sqrt{4a} = 1 (Integer), but \sqrt{a} is not an integer. But if a is any +ve integer, then \sqrt{a} must be an integer in order to make \sqrt{4a} an integer. \sqrt{4a} = 2 \sqrt{a} (1) a is a positive integer. Sufficient (as mentioned above) (2) a = n^6, where n is an integer. this means a is a +ve integer. Sufficient. D is the answer
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer
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26 Jun 2016, 11:21
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