Take any natural number $n_{0}$ , say $n_{0} = 928$. Form a new number $n_{1}$ by adding the digits of $n_{0} (n_{1} = 9 + 2 + 8 = 19)$. Iterate the procedure of adding digits, that is add the digits of $n_{1}$ to obtain $n_{2}$, add the digits of $n_{2}$ to obtain $n_{3}$, add the digits of $n_{3}$ to obtain $n_{4}$, and so on. (If $n_{1} = 19$ ,then $n_{2} = 1 + 9 = 10$, $n_{3} = 1 + 0 = 1$, $n_{4} = 1$,$n_{5} = 1$, ... ). In the above example the process of creating new numbers from $n_{0}$ has ended with $n_{3}$. This suggests two questions: (a) Is it true for every natural number $n_{0}$, that the above procedure of creating new numbers $n_{1}, n_{2}, n_{3}$, ... ends after a finite number of steps in a one-digit number $n_{k}$? (b) If the answer to question (a) is 'yes', discover the connection between $n_{k}$ and $n_{0}$. (In other words, find out for which numbers $n_{0}$ is $n_{k} = 1$, and for which no is $n_{k} = 2, 3, 4, ... ,9$).
Could someone give me any tips?
$$n_j = D_{j,0}+10\cdot D_{j,1}+10^2\cdot D_{j,2}+... (etc.)$$ where the digits of $n_j$ in reverse order are $D_{j,0} D_{j,1} D_{j,2} ... (etc).$ $$n_{j+1}=D_{j,0}+D_{j,1}+D_{j,2}+... (etc.)$$ So if $n_j\ge 10$ then $n_j>n_{j+1}.$
A strictly decreasing sequence of non-negative integers must stop after finitely many steps.