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D
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Difficulty:
45%
(medium)
Question Stats:
61% (01:32) correct
39%
(01:33)
wrong
based on 622
sessions
History
Date
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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
(1) a is a positive integer
(2) a = n^6, where n is an integer
31 Dec 2010, 02:53
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rxs0005
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.
General Discussion
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
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
