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If sqrt(4a) is an integer, is sqrt (a) an integer

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If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink] New post 25 Dec 2010, 10:39
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Oct 2013, 05:15, edited 1 time in total.
Edited the OA.
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Re: Number Prop DS [#permalink] New post 25 Dec 2010, 11:42
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] New post 25 Dec 2010, 15:14
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] New post 30 Dec 2010, 13:24
I said the same thing than craky. Ans. D for me. But as previously mentioned the question is weirdly formulated.

Anyone cares to comment? :-D
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Re: Number Prop DS [#permalink] New post 30 Dec 2010, 17:36
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] New post 30 Dec 2010, 19:57
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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.

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Re: Number Prop DS [#permalink] New post 31 Dec 2010, 01:53
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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] New post 08 Jan 2011, 17:24
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] New post 08 Jan 2011, 20:59
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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 integer


I 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] New post 08 Jan 2011, 22:50
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] New post 27 Oct 2013, 05:12
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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] New post 27 Oct 2013, 05:15
Expert's post
vjns wrote:
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 :)!


Edited the OA. Thank you.
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink] New post 04 Feb 2014, 05:10
Wow very sneaky indeed.
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer   [#permalink] 04 Feb 2014, 05:10
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