<s>[INST] Show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3\cdot4^{n+2} +5\right)$ using congruences Using congruence theory, show that $ \forall n \in \mathbb N$, $9\mid\left(10^n + 3 \cdot 4^{n+2} +5\right)$. The proof is quite simple with induction, but how can it be proved with congruences?
 [/INST] HINT:
As $10\equiv1\pmod 9, 10^n \equiv1^n\equiv1$
$\displaystyle 4^{n+2}=(1+3)^{n+2}$
$\displaystyle=1+\binom{n+2}13+\binom{n+2}23^2+\cdots+3^{n+2}\equiv1+(n+2)3\pmod 9\equiv3n+7$
Can you take it from here?

Another way 
$$10^n+3\cdot4^{n+2}+5=(10^n-1^n)+3\left(4^{n+2}-1^{n+2}\right)+1+3+5$$
Using congruence,
$\displaystyle4\equiv1\pmod 3\implies 4^{n+2}\equiv1^{n+2}\pmod3\equiv1$ 
and $10^n\equiv1\pmod9$
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