Bunuel wrote:
Is n the square of an integer?
(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer
Questions dealing with number properties can often be solved with very little calculations, using logic only.
We'll look for such a solution, a logical approach.
(1) Every positive integer is the square root of another integer (i.e x is the sqrt of x^2). But this doesn't mean that it also has to be the square of an integer!
Insufficient
(2) So \(\sqrt{9n}=3\sqrt{n}\) is the square of an integer meaning that \(\sqrt{n}\) can be written as k/3 for some integer k. That means that n = k^2/9 for some integer k. This is an integer only if k is divisible by 3, which it does not have to be.
Insufficient.
Combined:
So from (1) we know that n^2 must be an integer, and from (2) we know that n = k^2/9 for some integer k.
Combining, if n=k^2/9 then n^2 = k^4/81. Since this must be an integer then k^4 is divisible by 81 so k^2 is divisible by 9 and k is divisible by 3. Therefore n must be an integer.
(C) is our answer.
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20% (02:03) correct 
