Observe that
$$20^n + 15^n + 8^n + 6^n = (5^n + 2^n)(4^n + 3^n) \quad (4)$$
Since $n = 10k + 5 = 5(2k + 1)$ for some integer $k$,
$$5^n + 2^n = 5^{5(2k+1)} + 2^{5(2k+1)}$$
and so $5^5 + 2^5$ divides $5^n + 2^n$. Similarly, $4^5 + 3^5$ divides $4^n + 3^n$. Now calculation shows that $5^5 + 2^5 = 7 \cdot 11 \cdot 41$ and $4^5 + 3^5 = 7 \cdot 181$ as products of primes. Since $2009 = 7^2 \cdot 41$, the prime factors of 2009 occur as factors of the RHS of equation (4) and the result follows.