Stuck toward the end of the proof:
Prove: That $5\cdot10^n + 10^{n-1} + 3$ is divisible by $9$:
If $n=1$ then $5\cdot10^1 + 10^{1-1} + 3= 5\cdot10+10^0+3=54 $
$9$ surely divides $54$.
Assume, If $k$ is a natural number such that $9/5\cdot10^k + 10^{k-1} + 3$
then show that $$9/5\cdot10^{k+1}+5\cdot10^k + 3$$ $$9/10\cdot(5\cdot 10^{k+1}+5\cdot10^k + 3)$$ $$9/5\cdot10^{k+2}+5\cdot10^{k+1} + 10\cdot3$$
Stuck here I need to get each term to be divisible by $9$. I am stuck trying to get this.
Any help would be appreciated.
The difference \begin{align*} 5\cdot 10^{k+1} +10^k + 3 - (5\cdot 10^{k} +10^{k-1} + 3) &= 5\cdot 10^k(10-1) + 10^{k-1}(10-1) \\ &= 9(5\cdot 10^k + 10^{k-1}) \end{align*} is a multiple of 9. Since by inductive assumption, $5\cdot 10^{k} +10^{k-1} + 3$ is a multiple of 9, it follows that $5\cdot 10^{k+1} +10^k + 3 $ is also a multiple of 9.