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.
_________________

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

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

I said the same thing than craky. Ans. D for me. But as previously mentioned the question is weirdly formulated.

Anyone cares to comment?

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.

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!
_________________

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.
_________________

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.
_________________

Wow very sneaky indeed.

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
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________