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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]

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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
[Reveal] Spoiler: OA

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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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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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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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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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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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30 Dec 2010, 20:57
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Expert's post
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 16909 [3], given: 230 Math Expert Joined: 02 Sep 2009 Posts: 41666 Kudos [?]: 124338 [2], given: 12077 Re: Number Prop DS [#permalink] ### Show Tags 31 Dec 2010, 02:53 2 This post received KUDOS Expert's post 7 This post was BOOKMARKED 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. _________________ Kudos [?]: 124338 [2], given: 12077 Intern Joined: 19 Dec 2010 Posts: 28 Kudos [?]: 18 [0], given: 4 Re: Number Prop DS [#permalink] ### Show Tags 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? Kudos [?]: 18 [0], given: 4 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7612 Kudos [?]: 16909 [0], given: 230 Location: Pune, India Re: Number Prop DS [#permalink] ### Show Tags 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 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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.

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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

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
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

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]

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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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26 Jun 2016, 05:16
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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.

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Re: If sqrt(4a) is an integer, is sqrt (a) an integer [#permalink]

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04 Jul 2017, 01:43
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Re: If sqrt(4a) is an integer, is sqrt (a) an integer   [#permalink] 04 Jul 2017, 01:43
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