How many whole number terms does $(\sqrt[11]5+\sqrt[5]11)^{2015} $ have? $$(\sqrt[5]11+\sqrt[11]5)^{2015} $$
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
 A: ok, $11$ divides $k-2$ means that there is an $m$ such that $k=11m+2$. But $m$ can be of the form $5l,5l+1,5l+2,5l+3$ or $5l+4$. Substituting you get that $k$ is of the form $55l+2,55l+13,55l+24,55l+35$ or $55l+46$. So then it's not difficult to understand why $55l+35$ is the only possible solution. If you think about it and don't find why just ask.
A: There really is no reason that the formulas should be the same.  In the given solution, $k/5$ is the exponent of $11$.  In your solution, $k/11$ is the exponent of $5$.  It just happens that you've used the same letter.  If you wrote the exponent of $5$ as $j/11$ you wouldn't expect that $j=k$, now would you? 
A: The forms aren't different -- let's write the first one as 
$$11^\frac{k}{5}5^\frac{2015-k}{11}$$ and the second as 
$$5^\frac{j}{11}5^\frac{2015-j}{5}$$
As you said, the solutions for $k$ are $k\equiv35\mod55$, and for $j$ they are $j\equiv0\mod55$
How are $k$ and $j$ related? Well, $j=2015-k$. But if we look carefully, we see that $k\equiv35\mod55$ and $j=2015-k$ together imply that $j\equiv0\mod55$ so in fact the two equations in $j$ and $k$ are equivalent.
A: Given $5\mid k$ and $11\mid 2015-k$.
$5\mid k \quad\Longrightarrow\quad k\equiv 0\equiv\pmod{5}$
$11\mid 2015-k=11\mid 2-k \quad\Longrightarrow\quad k\equiv 2\equiv\pmod{11}$
Here you need to apply Chinese remainder theorem:
$k\quad\equiv\quad Chinese(0\equiv\pmod{5},2\equiv\pmod{11})\quad \equiv \quad35\equiv\pmod{55}$
gp-function:
? chinese(Mod(0,5),Mod(2,11))
%1 = Mod(35, 55)

