$$(\sqrt[5]11+\sqrt[11]5)^{2015} $$

It is unnecesary to write binomial coefficents 
$$11^{\frac k{5}}5^{\frac {2015-k}{11}} $$ This is the general term of the sum , $k$ goes from $0$ to $2015$.
For a term to be a whole number it means that :
$$5\;|\;k\quad \wedge\; 11\;|\;2015-k $$
$2013$ is divisble by $11$ , so $11\;|\;2013+2-k=11\;|\;2-k$

$k$ should be divisible by $5$ and $k-2$ divisible by $11$. In the solutions it says that from this it follows that k should be of the form: $k=55l+35 $. **I dont understand this.**

On the other hand if we write the binomial expansion in different order things are simpler 
$$5^{\frac k{11}}11^{\frac {2015-k}{5}} $$
$k$ should be divisible by $ 11$ and by $5$ which makes k of the form $k=55l$.

These forms are different I don't know how to explain it. They both give the correct solution. There are 37 whole number terms ( $2015=55\*36+35$, and $k=0 $)

Edit: if someone could make the exponents look bigger