Show that $n^{23}+6n^{13}+4n^{3}$ is a multiple of $11$ I was checking the following Fermat's little theorem exercise:

Show that $n^{23}+6n^{13}+4n^{3}$ is a multiple of $11$ 

I've started by stating each congruence individually suposing that each $n,6n$ and $4n$ are primes with $11$, for the first one I have:
$$n^{10} \equiv 1 \mod {11}$$
$$ \equiv n^{3} \mod {11}$$
I've stated the second one this way
$$6n^{10} \equiv 1 \mod {11}$$
But honestly I don't know how to go ahead as long as I don't have a number to evaluate with $11$. 
Also I'm considering that I have:
$$n + 6n + 4n = 11n$$
This may have some relation with the proof but could be affected because of the powers of each one. Any help will be really appreciated.

Fermat's little theorem only applies when prime $p$ does not divide $a$ in $a^{p-1} \equiv 1 \pmod p$.
In your expression, if $n$ is a multiple of $11$ the theorem doesn't apply but the expression is trivially a multiple of $11$.
If $n$ is not a multiple of $11$ the theorem applies. Now note:
$n^{10} \equiv 1 \pmod{11}$
$n^{20} \equiv 1^2 = 1 \pmod {11}$
So modulo $11$, your expression reduces to $n^3 + 6n^3 + 4n^3 =11n^3$, which is clearly a multiple of $11$.
