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Bunuel
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Bunuel
Is n the square of an integer?

(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer


IMO Ans is C
both statements are insuff because of above mentioned reasons in earlier post.

combining n=√x (from 1)
thus √9*(√x) =square of number
so √x=9^a where a=any odd integer

thus n=√9^4 (for a=3)
thus n=x^2

suff

Ans C
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rohit8865
Nightfury14
Bunuel
Is n the square of an integer?

(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer

Statement 1)
n could be the square root of 4 => n=2 (n is not the square of an integer)
n could be the square root of 16 => n=4 (n is the square of an integer)
Not sufficient

Statement 2)
\(\sqrt{9n}\) translates to \(3\sqrt{n}\) => this expression could be square of an integer if n is
1. n=9 or \(3^2\) => \(3\sqrt{9}\) => 3*3 n is square of an integer.
2. n=9*9*9 or \(27^2\)=> \(3\sqrt{9*9*9}\) =>9*9 n is square of an integer.
Note n cannot be 9*9 as then the resultant expression would not be a square itself as defined by statement 2.
Sufficient

Ans. B

Nightfury14

what if n =1/9 ??

I agree your question is valid, the ans would then change to (C).
But i think there is some catch to this.
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Bunuel
Is n the square of an integer?

(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer


IMO Ans is C
both statements are insuff because of above mentioned reasons in earlier post.

combining n=√x (from 1)
thus √9*(√x) =square of number
so x=sq. of sq. of any number

thus n=√x^4
thus n=x^2

suff

Ans C

rohit8865 - Hi
Please review highlighted, try plugging in \(2^4\) or \(4^4\) and check whether it satisfies condition (2).
Let integer in Statement (2) be z, then \(z^2\) = \(\sqrt{9n}\)
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rohit8865
Bunuel
Is n the square of an integer?

(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer


IMO Ans is C
both statements are insuff because of above mentioned reasons in earlier post.

combining n=√x (from 1)
thus √9*(√x) =square of number
so x=sq. of sq. of any number

thus n=√x^4
thus n=x^2

suff

Ans C

rohit8865 - Hi
Please review highlighted, try plugging in \(2^4\) or \(4^4\) and check whether it satisfies condition (2).
Let integer in Statement (2) be z, then \(z^2\) = \(\sqrt{9n}\)


Edited !! but idea remains same....

Thanks
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