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let $a$ be a positive integer such that $$S(a^n+n)=1+S(n)$$ for any sufficiently large $n$ if and only if $a$ is a power of $10$

where $S(n)$ is digit sum of a positive integer $n$

if $a$ is a power of $10$,let $a=10^k$,for large $n$,then $10^{kn}>n$,so It is clear $$S(a^n+n)=S(10^{kn}+n)=1+S(n)$$ But for other hand,if $S(a^n+n)=1+S(n)$ I can't prove $a$ is a power of $10$

math110
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    Have you tried checking when $n$ is a power of $10$ itself? – John Omielan Mar 08 '22 at 05:31
  • Since $S(k) \equiv k\pmod 9$ for all $k\in\mathbb N$, we have $a^n \equiv 1\pmod 9$ for all sufficiently large $n\in \mathbb N$. This leads to $9\mid a-1$. For now, i don't know how to continue. – Jochen Mar 08 '22 at 07:56
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    +1 for the comment of John Omielan. If $n=10^k$, then the rhs of the equation is 2, so we get $a^{10^k}+10^k=20\cdots 0$ or $a^{10^k}+10^k=10\cdots 010\cdots 0$ for all $k$. – Jochen Mar 08 '22 at 08:12

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