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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

3

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.

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.

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