Prove that : $\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$ Prove that : ($n\in\mathbb{N}$)
$$\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$$ 
I know that from little theorem : 
$$10^{18}\equiv 1\pmod{19}$$ 
So : 
$$10^{18k}\equiv 1\pmod{19}$$ 
Now i will go to this step if $\operatorname{correct}$
$$10^{12}=100^{6}\equiv 5^{6}\pmod{19}$$
$$5^{6}=(5^{2})^{3}\equiv 6^{3}\pmod{19}$$ 
$$6^{3}\equiv 7\pmod{19}$$ 
So : 
$$10^{18k+12}-7\equiv 0\pmod{19}$$
But my be : 
$$\frac{10^{18n+12}-7}{3}\not\equiv 0\pmod{19}$$ 

Let $a=10^{18k+12}-7$.  
If $19|a$ and $3|a$, then $19|(\frac a3)3$, and $19\nmid3$, 
so, by Euclid's lemma, $19|\frac a3$.
