$1996$ Austrian-Polish Number theory problem 
Let $k \ge 1$ be a positive integer. Prove that there exist exactly $3^{k-1}$ natural numbers $n$ with the following properties:

(i) $n$ has exactly $k$ digits (in decimal representation),
(ii) all the digits of $n$ are odd,
(iii) $n$ is divisible by $5$,
(iv) the number $m = n/5$ has $k$ odd digits
My work -
Let $n=10^{k-1}a_1+10^{k-2}a_2+...+a_k$
Now because all digits of n are odd and $5 | n$ we have $a_k =5$.
Now $m=n/5=2(10^{k-2}a_1+10^{k-3}a_2+...+a_{k-1})+1$
For $k=2$ I found that $n=55,75,95$ but not able to prove in general...
Hint says that all digit of m must be $1,5$, or $9$ and so there are $3^{k-1}$ choices for m hence n
but I am not able to understand why all digit of m must be 1,5, or 9 ???
Thankyou

The digits $3$ and $7$ are forbidden because they give an odd carry ($15$ and $35$) which leads to an even digit or to one digit more , if the first digit is $3$ or $7$.
Moreover, the first digit must be $1$, the others can be $1,5,9$. All such numbers give a valid $m$, hence $3^{k-1}$ is the number of possibilities.
