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805+ (Hard)|   Number Properties|   Roots|      
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Bunuel
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Is n the square of an integer? (1) n is the square root of an integer 01 Jul 2018, 21:47
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

21% (02:03) correct 79% (01:50) wrong based on 433 sessions

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Is n the square of an integer?

(1) n is the square root of an integer
(2) \(\sqrt{9n}\) is square of an integer
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C
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DavidTutorexamPAL
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Is n the square of an integer? (1) n is the square root of an integer Updated on: 02 Jul 2018, 00:13
Originally posted by DavidTutorexamPAL on 01 Jul 2018, 22:38.
Last edited by DavidTutorexamPAL on 02 Jul 2018, 00:13, edited 1 time in total.
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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
Show more

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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exc4libur
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Is n the square of an integer? (1) n is the square root of an integer 05 Sep 2019, 08:37
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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
Show more

is \(n=integer^2…\sqrt{n}=integer?\)

(1) n is the square root of an integer: insufic.
\(n=\sqrt{z}\)
if \(z=16…n=\sqrt{16}=4…\sqrt{4}=integer\)
if \(z=4…n=\sqrt{4}=2…\sqrt{2}=not.integer\)

(2) \(\sqrt{9n}\) is square of an integer: insufic.
\(\sqrt{9n}=x^2…x^4=9n…n=\frac{x^4}{3^2}\)
if \(x=3…n=\frac{3^4}{3^2}=3^2=9…\sqrt{9}=integer\)
if \(x=2…n=\frac{2^4}{3^2}=16/9…\sqrt{16/9}=4/3=not.integer\)

(1&2) \(n=\sqrt{z}\) and \(n=\frac{x^4}{3^2}\); so, \(z=\frac{x^8}{3^4}=integer\) and \(x\) is a multiple of \(3\);
therefore, \(n=\frac{(3m)^4}{3^2}=3^2m^4…\sqrt{n}=3m^2=integer\); sufficient.
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varundixitmro2512
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Is n the square of an integer? (1) n is the square root of an integer 01 Jul 2018, 22:58
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DavidTutorexamPAL
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

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 it is also an integer. Then \(\sqrt{n}\) must be an integer meaning that n must be a square of an integer.
Sufficient.

(B) is our answer.
Show more




Answer should be C

if n =4/9 statement 2 still holds.

Waiting for the OA
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DavidTutorexamPAL
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Is n the square of an integer? (1) n is the square root of an integer 02 Jul 2018, 00:13
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varundixitmro2512
DavidTutorexamPAL
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

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 it is also an integer. Then \(\sqrt{n}\) must be an integer meaning that n must be a square of an integer.
Sufficient.

(B) is our answer.




Answer should be C

if n =4/9 statement 2 still holds.

Waiting for the OA
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You're right! My bad, fixed.
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Is n the square of an integer? (1) n is the square root of an integer 10 Aug 2019, 21:21
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I don't understand something, maybe you can help me clarify some point.
When it's written :
(1) n is the square root of an integer; it means n = (int)^2 or n =(int)^1/2 ? For me it means n = (int)^1/2
Same question for (2) √(9n) is square of an integer, does it mean √(9n) = (int)^2 ? For me yes
But I think I am misunderstanding something (I am not a native speaker).

When you write :

" √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. " I don't understand why √n = k/3 and not k^2/3 ...

What am I getting wrong DavidTutorexamPAL, Bunuel ?
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Is n the square of an integer? (1) n is the square root of an integer 02 Sep 2019, 03:02
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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
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Hi Bunuel,

Could you please provide the detailed solution for this question?
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Is n the square of an integer? (1) n is the square root of an integer 03 Sep 2019, 00:08
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I don't understand how statement 2 is valid for n= 4/9.
for n=4/9, LHS is 2, as per the question 2 should be a square of some integer- which is not possible.
Can someone help me understand this?
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Is n the square of an integer? (1) n is the square root of an integer 05 Sep 2019, 09:58
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DavidTutorexamPAL
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

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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Hello, Can you help me understand "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" Why it has to be of the form k/3?
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Is n the square of an integer? (1) n is the square root of an integer 16 Mar 2021, 05:37
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can someone provide a detailed solution for this Bunuel ?
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Is n the square of an integer? (1) n is the square root of an integer 08 Dec 2022, 07:08
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