<s>[INST] 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
 [/INST] $N=5 + .... + 5^{150} = 5(1+...... + 5^{149} = 5\cdot \frac {5^{150}-1}{4}$.
$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.
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