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. 

Proof by induction:


*

*Base case: $10^{0+1}+3\cdot10^{0}+5=18$

*Assumption: $10^{n+1}+3\cdot10^{n}+5=9k$

*Inductive step:


$10^{n+2}+3\cdot10^{n+1}+5=$
$10^{n+2}+3\cdot10^{n+1}+50-45=$
$10(\color\red{10^{n+1}+3\cdot10^{n}+5})-45=$
$10(\color\red{9k})-45=$
$9(10k)-45=$
$9(10k-5)$
