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Is it true that for all integers $k = 1, 2, 3, \cdots$ except powers of 10 ($1, 10, 100, \cdots, {10}^n, \cdots$),

there exist an integer $n$ such that $n^k$ include all digits from 0-9?

Example:

$$2 ^ {68} = 295147905179352825856\\ 3 ^ {39} = 4052555153018976267\\ 5 ^ {19} = 19073486328125$$

I can't seem to make any progress on this question, other than that if $n^{ab}$ includes all digits, then so must $(n^a)^b=(n^b)^a$... That's trivial though

Thank you

Gareth Ma
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    If $n$ is such that $\log_{10}n$ is irrational then there is a power of $n$ that starts with any given string of numbers. I'll look up a proof for the $n=2$ case... – lulu Mar 21 '20 at 13:52
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    @lulu Ah , yes, you are right. That is the proof :) – Peter Mar 21 '20 at 13:53
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    here is a proof for $n=2$. The general case I mentioned is no different. – lulu Mar 21 '20 at 13:53
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    A much stronger claim, which is much harder to prove, is that for each base $n$ only finitely many $k$ are such that $n^k$ does not have all the digits. As $k$ gets large, $n^k$ has lots of digits and it is hard to avoid any of them. http://oeis.org/A007377 gives the powers of $2$ that do not have a $0$ and states the conjecture that $86$ is the last. – Ross Millikan Mar 21 '20 at 14:11
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    @RossMillikan Are you saying that a proof of that is known? Certainly a natural conjecture... – lulu Mar 21 '20 at 14:14
  • @lulu: no, I do not believe it has been proven. The OEIS sequence says it is only a conjecture that $86$ is the last for $2$. But when you get to trillions of digits, how can you miss? – Ross Millikan Mar 21 '20 at 15:00
  • @RossMillikan Oh, I agree that a counterexample would be astonishing. And even the "$86$" seems like a stretch, but I thought that maybe finiteness was doable. Not that I have the slightest idea how to attack it. – lulu Mar 21 '20 at 15:07
  • @lulu: I'm sure the search has been run a long ways. – Ross Millikan Mar 21 '20 at 15:17
  • @RossMillikan: The OEIS entry says it's been checked up to $2^n$ for $n=10^{10}$. – joriki Mar 21 '20 at 23:23
  • @lulu: We should expect

    $$\frac{\log_210\cdot0.9^{n/\log_210}}{1-0.9}\approx33.2\cdot0.9^{n/3.32}$$ counterexamples for exponents above $n$. So $86$ is actually a bit lower than expected, since above that about $33.2\cdot0.9^{86/3.32}\approx2.2$ counterexamples would have been expected.

     – joriki Mar 21 '20 at 23:34

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