Irska

Observe that $$20^n+15^n+8^n+6^n=(5^n+2^n)(4^n+3^n)(4)$$ 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.