Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. 
I'm thinking to show that $n$ is divisible by each of the prime factors of $930$, is that right? I'm stuck

$$n= 5(1 +5+ 5^2+ ...5^{149})  = 5 {5^{150}-1\over 4}$$
so $5\mid 4n\implies 5\mid n$. Now:
$$5^{150}-1 = 125^{50}-1 = (125-1)(125^{49}+...+125^2+125+1) $$
so $31\mid 4n \implies 31\mid n$ and $$5^{150}-1 = 25^{75}-1 = (25-1)(25^{74}+...+25+1) $$
so $24\mid 4n \implies 6\mid n$ and you are done.
