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
 A: Another short variant:


*

*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:
$$n=5(\underbrace{1+5+5^2}_{=31})(1+ 5^3+\dots+5^{147}).$$
A: $$n= 5(1 +5+ 5^2+ ...5^{149})  = 5 {5^{150}-1\over 4}$$
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.
A: Render $930=2×3×5×31$, which is square-free, and test for divisibility by each prime factor.
Divisibility by 2 $\to$ even number of odd terms, $\color{blue}{\text{passes}}$.
Divisibility by 3 $\to$ even number of terms with alternating residues $+1,-1\bmod 3, \color{blue}{\text{passes}}$
Divisibility by 5$\to$ each term is separately divisible, $\color{blue}{\text{passes}}$. 
Divisibility by 31 $\to$ terms cycle in residue:  $\overline{5,25,1}$, cycle adds up to zero and the number of terms is a multiple of the period (3); $\color{blue}{\text{passes}}$.
And we're good.
A: As $\sum_{r=1}^n5^r=\dfrac{5(5^n-1)}{5-1}$
Clearly it is divisible by $5$ as each term is divisible by $5$
Now the sum will be even for $n$ is even $=2m$(say)
As the denominator $(4)$ is relatively prime with $3\cdot31$
It's sufficient to establish $5^{2m}-1$ is divisible by $93$
As $(5,93)=1,$ the sufficient condition will be by https://en.m.wikipedia.org/wiki/Carmichael_function $$\lambda(93)=\cdots=30$$ should divide $2m$
So, it is sufficient to have $n=2m$ be divisible by $30$
A: $1+5+5^2+\dots+5^{150}=\dfrac{1-5^{151}}{1-5}$.
Now by lil Fermat, this is $1\pmod {2, 3, 31}$ respectively.
It's $1\pmod 5$ too.
Therefore by the Chinese remainder theorem,  it's $1 \pmod {930}$.
A: $N=5 + .... + 5^{150} = 5(1+...... + 5^{149} = 5\cdot \frac {5^{150}-1}{4}$.
$930| N \iff $
$4*930| 4*N = 5(5^{150}-1) \iff $
$\frac {4*930}5=\frac{4*93*10}5=4*93*2|5^{150}-1\iff$
$8*3*31|5^{150}-1\iff$
$8|5^{150}-1$ and $3|5^{150}-1$ and $31|5^{150}-1\iff$
$5^{150}\equiv 1\pmod {8,3,31}$.
And we can's Euler's Th|Fermat's Little Theorem to prove those.
$\phi(8=2^3) = 2^2 = 4$ so $5^4\equiv 1\pmod 8$.  But actually $5^2 =25 \equiv 1 \pmod 8$ so $5^{150}\equiv (5^2)^{75}\equiv 1\pmod 8$.
ANd $5^{2}\equiv 1\pmod 3$ so $5^{150}\equiv (5^2)^{75}\equiv 1 \pmod 3$.
$5^{30}\equiv 1 \pmod {31}$ so $5^{150}\equiv (5^{30})^5\equiv 1 \pmod{31}$.
That's it.
