Divisibility of an expression Need some guidance
How to prove that $9\cdot n^9+7\cdot n^7+3\cdot n^3+n$ is divisible by $10$.
I've tried transforming the expression by adding $n^9$ and $-n^9$ in order to make a multiple of 10 but no use. I've even tried math. induction but got stuck, maybe this can not be proven. Are there any other methods?
Thanks in  advance

It is sufficient to check 2 and 5 divide this expression.
Since, you have an even number of terms with same parity, 2 divides it. 
$$9\cdot n^9+7\cdot n^7+3\cdot n^3+n \cong 4n^9 + 2n^7 + 3 n^3 +n\  mod\  5 $$ 
If $gcd(n, 5) =1 $, then by Fermat little theorem: 
$$n^ 4 \cong 1\  mod \ 5$$
Therefore, $4n^9 + 2n^7 + 3 n^3 +n \cong 4n + 2n^3 + 3 n^3 +n \cong 5n + 5n^ 2 \ mod \ 5$
Note: If $gcd(n, 5) \neq 1 \Rightarrow gcd(n, 5) = 5 $, which makes divisibility by 5 trivial.
