Prove that : $\frac{10^{18n+12}-7}{3}\equiv 0\pmod{19}$

Question

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 :

$$\frac{10^{18n+12}-7}{3}\not\equiv 0\pmod{19}$$

Answer 1

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, $19|\frac a3$.