Is this an open problem? I came up with the following question. I could not find a way to solve it, and haven't seen it anywhere also. Is this an open problem? 
Problem statement: Consider a prime number p. Consider all numbers having it as a factor. Repeatedly sum the digits of those numbers until you get a number less than or equal to p. In this sequence, will you ever get a p? Is it guaranteed that you get for various values of p (larger ones)? 
For example, a "sum" is defined as applying + together on all digits and get the resulting number. You apply "sum" again on it, if it is larger than p, of course. 
Thanks
Salahuddin
 A: Let $p$ be a prime other than $2$ or $5$. By Fermat's Theorem, $10^{p-1}\equiv 1\pmod{p}$.  Consider the sum 
$$N=1+10^{p-1}+10^{2(p-1)}+10^{3(p-1)}+\cdots +10^{(p-1)(p-1)}$$
(note there is a total of $p$ terms).
Each term in the sum is congruent to $1$ modulo $p$, so $N\equiv 0\pmod{p}$.  And the digit sum of $N$ is $p$.
A: Assume $p\notin\{2,5\}$, in which case $10$ is invertible mod $p$, and its multiplicative order $n$ modulo $p$ is (by definition) such that the power $10^n$ is $1$ modulo $p$ (or one can take $n=p-1$, which will do as well). Now the number
$$
  N=\sum_{i=0}^{p-1}10^{in}
$$
has sum of digits $p$ and is divisible by $p$. For $p\in\{2,5\}$ take $N=10p$.
More generally (for $p\notin\{2,5\}$), if $a_1,\ldots,a_p$ are distinct integers, one can take $N=\sum_{i=1}^{p}10^{na_i}$, for infinitely many solutions. Or you could have some repeated entries in the list $a_1,\ldots,a_p$, as long as no value occurs more than $9$ times, to make for a change of all those boring digits $1$ (the digits $0$ remain though). 
