Prove $10^{n+1}+3\cdot 10^n+5$ is divisible by $9$? How do I prove that an integer of the form $10^{n+1}+3\cdot 10^{n}+5$ is divisible by $9$  for $n\geq 1$?I tried proving it by induction and could prove it for the Base case n=1. But  got stuck while proving the general case. Any help on this ? Thanks. 

Let $S(n)$ be the statement: $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$
Basis step: $S(1)$:
$\Rightarrow 10^{(1)+1}+3\cdot{10^{(1)}}+5=10^{2}+30+5$
$\hspace{45.5 mm}=100+35$
$\hspace{45.5 mm}=135$, which is divisible by $9$
Inductive step:
Assume $S(k)$ is true, i.e. assume that $10^{k+1}+3\cdot{10^{k}}+5$ is divisible by $9$
$\hspace{59 mm} \Rightarrow 10^{n+1}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10\cdot{10^{n}}+3\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 13\cdot{10^{n}}+5=9A$
$\hspace{59 mm} \Rightarrow 10^{n}=\dfrac{9A-5}{13}$
Then, $S(k+1)$: $10^{(k+1)+1}+3\cdot{10^{(k+1)}}+5$
$\hspace{23.5 mm} =10^{k+2}+3\cdot{10}\cdot{10^{k}}+5$
$\hspace{23.5 mm} =100\cdot{10^{k}}+30\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{10^{k}}+5$
$\hspace{23.5 mm} =130\cdot{\bigg(\dfrac{9A-5}{13}\bigg)}+5$
$\hspace{23.5 mm} =10\cdot{(9A-5)}+5$
$\hspace{23.5 mm} =90A-50+5$
$\hspace{23.5 mm} =90A-45$
$\hspace{23.5 mm} =9\hspace{1 mm}(10A-5)$, which is divisible by $9$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $10^{n+1}+3\cdot{10^{n}}+5$ is divisible by $9$.
