Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that  $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?

$\begin{align} \bmod 10\!:\ f_n & = \ n+9n^9 +\ 3n^3\,+ 7n^7\\
 &\equiv  n(1\!-\!n^8) + 3n^3(1\!-\!n^4)\ \ \ {\rm by}\ \ \ 9\equiv -1,\,\ 7\equiv -3\\
 &\equiv  \color{#c00}{n(1\!-\!n^4)}\,g_n
\end{align}$
Notice that $\,2,5\mid \color{#c00}{n\,-\,n^5}\ $ by Fermat. 
Or, equivalently, $\,\color{#c00}{n^5\equiv n}\,\Rightarrow\, n^7\equiv n^3,\, n^9\equiv n\ $ so $\,f_n\equiv (1\!+\!9)n+(3\!+\!7)n^3\equiv 0$
