This looked like an easy exercise, when a friend of mine asked me if I know a way to prove that the decimal representation of $3^k$ always contains a zero for $k\ge k_0$, but the more I think about this question the more I find it difficult.

After some experimentation (using Mathematica), I discovered that all the numbers $3^k$ for $69\le k\le 20000$ do have a zero in there decimal representation, and that this zero does appear among the first 100 digits. This suggests that maybe $k_0=69$.

I hope that some one can help in giving some insight on this question.

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    $\begingroup$ My advice is to try the last n decimal digits instead. The function is periodic modulo 10^n for any n. That means if you can find an n such that a long enough string have a zero among the last n digits, then you'll be done. $\endgroup$ Oct 10, 2014 at 12:15
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    $\begingroup$ @James Cranch: The $5\times 10^{n-2}$ values hit mod $10^n$ are arbitrary initial sequences with one of the $500$ allowed final $4$ digits. Since $3^8 = 6561$, we can find a power of $3$ that ends in $1111...1116561$. For example, $3^{195508}$ ends in $...1116561.$ $\endgroup$ Oct 10, 2014 at 12:38
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    $\begingroup$ Also, the fact that $\alpha = \log(3)/\log(10)$ is irrational, together with the fact that $\{ n \alpha \}$ is uniformly distributed mod $1$ means that for all positive integers $m$, there is an integer $n(m)$ so that the first $m$ digits of $3^{n(m)}$ are all nonzero. $\endgroup$ Oct 10, 2014 at 12:39
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    $\begingroup$ So this is a version of a known open problem. The version I heard is about 7's in base 10 expansions of $2^n$. You can with a bit of ingenuity prove that this holds for almost all $n$, but all $n$ is probably really hard. If you believe the digits are random (as they "should" be), then you expect the conjecture to be true by Borel-Cantelli type arguments. $\endgroup$ Oct 10, 2014 at 17:40
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    $\begingroup$ See also mathoverflow.net/questions/30357/… for a closely related question. $\endgroup$ Oct 10, 2014 at 17:42


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