Stumbled across a question today and to me it makes very little sense and I have no idea as to how one would approach this question and thus solve it, anyone have any ideas? Would be much appreciated.
Question: If you can increase a natural number by the sum of its digits per single move, from the natural $1$, can you get to $2019\,2019\,2019\,2019$ after a finite number of steps?
The only idea I have is that I have heard before that $3$ has some significance with digit sums but I don't know how that links in.
Thanks in advance.
Best Answer
Note that $n$ and the the sum of its digits have the same remainder when divided by $3$ (why?). So if we start from $1$, the remainder of the division by $3$ of each term of the sequence would be $1,2,1,2,1,2,1\dots$ which means that we can not obtain any multiple of $3$ such has $2019201920192019$.