If n and k are greater than zero, is n/k an integer ?

Question

(1) n and k are both integers.
(2) n^ 2 and k^2 are both integers

uday1409 · Answer

1. say n=4;k=2; 
 so n/k=2. AND let n=3;k=2 So n/k becomes 3/2 which is not an integer.
Insufficeint

2. if n^2 and k^2 are integers then n and k are also integeres. Above case applies here. Insufficeint.

1+2 >> still makes n and k integeres. Above case applies here. Insufficeint

Ans E

WoundedTiger · Answer

Both n and k are integers greater than zero

St 1 n and k are integers..

Consider n=4,k=2, n/k=Integer
Considern=2,k=4 n/k = Not an Integer

St2 consider n^2 and k^2 are integer which is same n and k are integer...

No new information.

Ans E

beattheheat · Answer

If n and k are greater than zero, is n/k an integer ?

(1) n and k are both integers.
(2) n^ 2 and k^2 are both integers

Solution: Determine whether n is a multiple of k?

Statement 1: n and k are both integers
No relation is defined between n and k - Insufficient

Statement 2: n^ 2 and k^2 are both integers
Again no relation is defined between n and k - Insufficient

Even together statement 1 & 2 are insufficient. Answer E

Inten21 · Answer

Bunuel

Although I got the correct answer easily, looking at explanations got me a little curious about something.
I have a small query. Kindly correct me if I am wrong here.

I see almost every answer for statement 2 as "If n^2 and m^2 are integers, then n and m have to be integers. "

What if n^2 = 2 , then n=Root<2>
and m^2 = 3 ; then m=Root<3>
{Considering the fact that the only information we have been given is that n and m are positive.}

Is my reasoning correct here?
Thanks in advance.

Bunuel · Answer

Yes, just knowing that n^2 is an integer does not necessarily mean that n is an integer. For example,  n=\sqrt{2}  --> n^2 = 2 = integer but n is not an integer.

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If n and k are greater than zero, is n/k an integer ?

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If n and k are greater than zero, is n/k an integer ?

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