<s>[INST] Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:

$5*10^n+10^{n-1}+3$ is divisible by 9

Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$ 
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
 [/INST] Induction hypothesis:
$$9|5\times10^k+10^{k-1}+3$$
Thus,
$$9|10(5\times10^k+10^{k-1}+3)$$
$$9|5\times10^{k+1}+10^k+30$$
$$9|5\times10^{k+1}+10^k+3+27$$
But $9|27$. Hence,
$$9|5\times10^{k+1}+10^k+3$$
which is what had to be shown.
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