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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.

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EricGunter7055
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1 Answers1

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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 $2019201920192019$.

Robert Z
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  • https://math.stackexchange.com/questions/341202/how-to-prove-the-divisibility-rule-for-3 – Ahmed Hossam Aug 09 '19 at 08:43
  • Thanks, this helps so much. One thing I'm struggling with is proving that the pattern of the remainder being 1,2,1,2,1,2... repeats forever. Otherwise I feel my solution could always be challenged in that I've just made an assumption that the pattern repeats infinitely. Is there anyway I can prove that this pattern repeats? – EricGunter7055 Aug 10 '19 at 08:10
  • If the remainder of $n$ is $1$ then $n=3a+1$, the sum of its digits is $s=3b+1$ and $n+s=3(a+b)+2$, that is the next remainder is $2$.

    If the remainder of $n$ is $2$ then $n=3a+2$, the sum of its digits is $s=3b+2$ and $n+s=3(a+b)+4=3(a+b+1)+1$, that is the next remainder is $1$.

    So the sequence of remainders $1,2,1,2,1,2...$ repeats forever.

     – Robert Z Aug 10 '19 at 08:16
  • Thanks so much! – EricGunter7055 Aug 10 '19 at 17:27