How do I show that $\gcd(n,\frac{n}{k})=\frac{n}{k}$? 
$\gcd(n,\frac{n}{k})=\frac{n}{k}$

Let $n, k$ be positive integers.
This should be a trivial question but not having taken any classes in number theory I would like to be convinced with a simple proof that the above holds true.
Could someone kindly provide me with a proof?
Thanks in advance.
 A: $n/k$ is obviously divisible by $n/k$, and so is $n$, because $n = k \cdot (n/k)$. This is their greatest common divisor, because $n/k$ doesn't have any bigger divisor than $n/k$ itself.
A: Let dis the common divisor of n and n/k.So $d|n\;\;and\;\;d|{\frac{n}{k}}$.
So,we take the linear combination ${\implies}$ $nx+{\frac{n}{k}}y=d$.
We know that n=$dk_1$ and ${\frac{n}{k}}=dk_2$.Now we put the value of 
n=$dk_1$ in the equation.We get ${k_1}x+{k_2}y=1$.Hence GCD$(k_1,k_2)=1$
and $k_1\;\;and\;\;k_2 $are coprime to each other.We have $\frac{dk_1}{k}=dk_2$.
So $k_2=\frac{k_1}{k}$.Though $k_2$ is a multiple of $k_1$, $k_1$ does not 
divide $k_2$.Hence $\frac{k_1}{k}=1$ and $k_1=k$ and we get n=$dk_1$$\implies$
n=$dk$$\implies$d=$\frac{n}{k}$.Now you have to prove that d is the greatest.
Then you are done. 
A: The largest number which divides both the numbers $n/k$ and $n$ is $gcd(n,n/k)$.
Here, we can easily observe that $n/k$ divides both $n$ and $n/k$. Also, any number greater than $n/k$ cannot be $gcd(n,n/k)$, since any such number cannot divide $n/k$. Hence, $gcd(n,n/k)=n/k$.
