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.

From what you shown, we have $$n^{10} \equiv n^{20} \equiv 1 \mod 11$$ and thus $$n^{23} \equiv n^3 \mod 11$$
We also have $$n^{10} \equiv 1 \mod 11$$ and thus we have $$6n^{13} \equiv 6n^3 \mod 11$$ 
Lastly, we have that the last term is: $$4n^3 \mod 11$$
Adding it all up, we have that this is equivalent to $$n^{23} + 6n^{13} + 4n^3 \equiv n^3 + 6n^3 + 4n^3 \equiv 0 \mod 11$$
