A formula involving the sum of digits and the number of digits greater or equal than $5$ For $m\in \mathbb{N}$ we denote by $L(m)$ the number of digits greater than or equal to $5$ of $m$ and by $s(m)$ the sum of digits of $m$.
Then I would like to show that $s(2k)=2s(k)-9L(k)$ for any $k\in \mathbb{N}$.
I think that this should be proved by induction on $k$. The base case $k=1$ is trivial, but I don't know how I should do the induction step $k \to k+1$. I tried to write the number $k=\overline{a_r a_{r-1}...a_0}$ in base $10$, but I don't know how to deal with the carry. I run into a lot of cases (whether $a_0$ is equal to $9$ or not and then the same discussion would be for $a_1, a_2, ...$), so I got stuck. How should this induction be approached?
 A: Let a number be $A=\overline{a_0a_1\ldots a_r}$.
Imagine computing $A+A$ by writing digits of $A$ in two rows one below the other. It is easy to see that there will be carries of $1$ whenever a digit is greater than or equal to $5$ (so their sum is $\ge 10$). Thus $L(A)$ denotes number of carries in performing $A+A$. Let us call $L(A)$, $C$ for simplicity.
Now a carry is actually a $10$ in their own column but it converts to a $1$ for the next column. For example,
$$\begin{array}{cccc}
&   & \small{1}  &   \\
& 1 & 2 & 8 \\
+ & 1 & 2 & 8 \\
\hline
& 2 & 5 & 6 
\end{array}$$
$8+8=16=\boxed{10}+6$ in its own column changes to $\boxed{1}+4=5$ for next column. In this example $C=1$.
Hence only $1\, C$ gets added to digits of $2A$ but $10\, C$ is added to digits of $A+A$. To equate these, we should write (by looking the addition table)
$$s(2A)=s(A)+s(A)-10C+C$$
$$\Rightarrow s(2A)=2s(A)-9C$$
which was to be proved.
This looks like in our example as
$$\begin{array}{ccccc}
& 1 & 2 & 8 &\\
+ & 1 & 2 & 8 &\\
\hline
& 2 & 4 & \boxed{10}+6 & =22\\
\hline
& 2 & 4+\boxed{1} & 6 &=13 
\end{array}$$
where we converted the $10$s to $1$s by doing $-10C+C$.
