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# Is n the square of an integer? (1) n is the square root of an integer

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Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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01 Jul 2018, 20:47
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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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Joined: 07 Dec 2017
Posts: 1153
Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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Updated on: 01 Jul 2018, 23:13
2
1
3
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.

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Originally posted by DavidTutorexamPAL on 01 Jul 2018, 21:38.
Last edited by DavidTutorexamPAL on 01 Jul 2018, 23:13, edited 1 time in total.
##### General Discussion
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Joined: 04 Apr 2015
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Re: Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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01 Jul 2018, 21:58
1
DavidTutorexamPAL wrote:
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 it is also an integer. Then $$\sqrt{n}$$ must be an integer meaning that n must be a square of an integer.
Sufficient.

if n =4/9 statement 2 still holds.

Waiting for the OA
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Joined: 07 Dec 2017
Posts: 1153
Re: Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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01 Jul 2018, 23:13
varundixitmro2512 wrote:
DavidTutorexamPAL wrote:
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 it is also an integer. Then $$\sqrt{n}$$ must be an integer meaning that n must be a square of an integer.
Sufficient.

if n =4/9 statement 2 still holds.

Waiting for the OA

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Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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10 Aug 2019, 20:21
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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Re: Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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02 Sep 2019, 02:02
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

Hi Bunuel,

Could you please provide the detailed solution for this question?
Intern
Joined: 20 Jan 2019
Posts: 4
Re: Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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02 Sep 2019, 23:08
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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Joined: 24 Nov 2016
Posts: 1545
Location: United States
Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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05 Sep 2019, 07:37
2
1
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

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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Joined: 18 Dec 2017
Posts: 1360
Location: United States (KS)
GMAT 1: 600 Q46 V27
Re: Is n the square of an integer? (1) n is the square root of an integer  [#permalink]

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05 Sep 2019, 08:58
DavidTutorexamPAL wrote:
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.

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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Re: Is n the square of an integer? (1) n is the square root of an integer   [#permalink] 05 Sep 2019, 08:58