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

Another short variant:


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*divisibility by $30$; group terms by consecutive pairs:
$$n= (\underbrace{5 + 5^2}_{=30})(1+ 5^2+\dots+5^{148}).$$

*divisibility by $31$: group by three consecutive numbers:
$$n=5(\underbrace{1+5+5^2}_{=31})(1+ 5^3+\dots+5^{147}).$$
