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If a positive integer $n$ is congruent to $0$, $1$, $5$ or $6$ modulo $10$, there is some digit occuring in each of the powers of $n$. If the decimal expansion of $n$ ends in a $0$, $1$, $5$ or $6$ then the same is true for the decimal expansions of $n^k$ ($k \geq 1$).

My question is whether there is any other positive integer $n$ (not congruent to $0$, $1$, $5$ or $6$ modulo $10$) such that there is some digit that occurs in all powers of $n$.

A computer search shows that if such an $n$ were to exist, it should have at least $3$ digits. For example, $99$ does not have the desired property as $$99^{18} = 834513761450087614416078625185528201$$ does not contain a $9$. Perhaps a number such as $102$ will work, as $102^k$ starts with a $1$ for small $k$ and has a lot of digits for large $k$, so "probably" one of them will be a $1$. On the other hand, I do see no reason why there couldn't be some large integer $N$ for which $102^N$ does not contain a $1$ at all.

I would be interested to know if the answer on the question formulated above is known, and (if this is not the case) whether you think it will be true or not (of course, motivated by some argument).

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    $\begingroup$ Thogh mathematically not very natural, this is interesting in that one can imagine a computing-based solution. *Existence" seems a virtual certainty, even in the stronger sense that all digits occur in all positive powers. $\endgroup$ – André Nicolas Sep 2 '14 at 15:30
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I succeeded in answering my own question (André's comment put me on the right track).

We can take $n=1249$. Since $1249^2 \equiv 1 \mod 10000$, we have $$ 1249^k \equiv \begin{cases} 1 &\mod 10000 \mbox{ if $k$ is even}; \\ 1249 &\mod 10000 \mbox{ if $k$ is odd.} \end{cases} $$

Thus, each power of $1249$ contains $1$ as a digit. It remains to find an integer $n$ such that each power of $n$ contains all digits.

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