Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Number Prop DS [#permalink]
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.

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.
_________________

Re: Number Prop DS [#permalink]
31 Dec 2010, 01:53

Expert's post

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.

Re: Number Prop DS [#permalink]
08 Jan 2011, 20:59

Expert's post

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.
_________________

Re: Number Prop DS [#permalink]
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.