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

Question

Is n the square of an integer?

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

DavidTutorexamPAL · Answer

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.

varundixitmro2512 · Answer

Answer should be C

if n =4/9 statement 2 still holds.

Waiting for the OA

DavidTutorexamPAL · Answer

You're right! My bad, fixed.

Coggi · Answer

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  ?

aviejay · Answer

Hi  Bunuel ,

Could you please provide the detailed solution for this question?

deelakmi · Answer

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?

exc4libur · Answer

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.

TheNightKing · Answer

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?

sthahvi · Answer

can someone provide a detailed solution for this  Bunuel  ?

bumpbot · Answer

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

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