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Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integer $n$?

In general, let $x_{k}$ denote the base-$k$ representation of the positive integer $x$. We say $x$ is $k$-powerful if $x^n_{k}$ contains all of the $k$ digits for all sufficiently large integers $n$.

For example, it's easy to check that 2 is 2-powerful, but 3 is not 3-powerful. The title question asks whether 2 is 10-powerful.

For any given $x$ and $k$, can we decide if $x$ is $k$-powerful?

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  • $\begingroup$ Methinks the label "k-powerful" is already taken: en.wikipedia.org/wiki/Powerful_number $\endgroup$
    – José Hdz. Stgo.
    May 25, 2022 at 5:20
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    $\begingroup$ An easy remark: $2^{68}$ contains all decimal digits, and it does so in its last $17$ digits. Now $2^{4\times 5^{16}}$ is $1$ mod $5^{17}$ by Euler's theorem. So if $n$ is congruent to $68$ mod $4\times 5^{16}$, then $2^n$ ends in the same $17$ decimal digits as $2^{68}$ and hence, contains all decimal digits. This shows that there are arbitrarily large $n$ such that $2^n$ contains all decimal digits (this is, of course, weaker than showing that all sufficiently large $n$ are such). $\endgroup$
    – Gro-Tsen
    May 25, 2022 at 9:56
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    $\begingroup$ But saying anything beyond this is probably an open problem, as this older answer to a related problem suggests. $\endgroup$
    – Gro-Tsen
    May 25, 2022 at 10:06
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    $\begingroup$ @Gro-Tsen: More generally, let $c > 0$ be any positive integer. Then there exist infinitely many powers of $2$ which have decimal expansion starting with $c$. $\endgroup$
    – spin
    May 25, 2022 at 14:00
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    $\begingroup$ $\{\dfrac{2^n}{10^m} :(m, n) \in \mathbb N^2\} $ is dense on $\mathbb R^+$ $\endgroup$
    – Dattier
    May 25, 2022 at 15:54

1 Answer 1

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Heuristically, one would expect the answer to be yes. There's an existing partially explicit version of this mentioned in Richard Guy's "Unsolved Problem in Number Theory" entry F24, which is that for $n> 86$, $2^n$ always contains a zero in its base 10 expansion. There are some related questions also in that entry and some other entries in the book as well. For example, there's a conjecture that every sufficiently large power of 2 contains 0s, 1s and 2s in its base 3 expansion. The reasonable generalization (which I have not seen stated explicitly but seems to be implicit in all of these), is that if $a$ and $b$ are relatively prime integers both greater than 1, then for all sufficiently large $n$, $a^n$ has in its base $b$ expansion all of $0, 1, \dotsc b-1$. It is also plausible that the sufficiently large is a not very fast growing function, something like being true for all $n \geq a^2b^2$. Edit: See Emil's comment below: as written this doesn't include the case $2$ and $10$; one wants that there's a prime $p$ such that $p$ divides exactly one of $a$ and $b$. That includes the relatively prime case and also cases like $2$ and $10$.

However, the set of $a$ and $b$ where we can prove anything like this is small and mostly relegated to when $b=2$. In that case, some of these results are implicitly very old, dating back to actually the middle ages. For example, any power of $3$ greater than $3$ must contain both $1$ and $0$ in its base $2$ expansion. This follows, since if not, $3^n$ would have to be of the form $2^k-1$, and Gersonide's theorem that $8$ and $9$ are the largest power of 2 next to a power of $3$ then applies. For larger bases, Mihailescu's proof of Catalan's conjecture shows that the same is essentially true if one has $b=2$ and $a$ is any number greater than $2$.

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    $\begingroup$ 2 and 10 are not relatively prime, so it’s not quite a generalization. $\endgroup$
    – Emil Jeřábek
    May 25, 2022 at 13:12
  • $\begingroup$ @EmilJeřábek Good point. The slightly more general statement that would include it is that there's at least one prime $p$ which divides $a$ or divides $b$ but doesn't divide the other one. $\endgroup$
    – JoshuaZ
    May 25, 2022 at 13:14
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    $\begingroup$ Heuristically I'd have expected the answer to be no! It goes like this: 1) I verified that for $d\le 10$ the lowest $d$ digits of $2^n$ are periodic with period $4\times 5^{d-1}$, starting at $n=d$; so I expect this to hold for any $d$, but have no idea how to prove it; 2) the probability of a $d$-digit string having a given digit is $1-0.9^d$, and for large $d$ that of having all digits is about $(1-0.9^d)^{10}$; 3) the probability of $4\times 5^{d-1}$ different $d$-digit strings each having all ten digits is about $(1-0.9^d)^{10\times 4\times 5^{d-1}}$, approaching $0$ very fast in $d$. $\endgroup$
    – Yaakov Baruch
    May 25, 2022 at 15:47
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    $\begingroup$ IIRC it’s a general fact that for any odd prime $p$ and integer $g$, if $g$ is a generator of $(\mathbb Z/p^2\mathbb Z)^\times$, then it is a generator of $(\mathbb Z/p^d\mathbb Z)^\times$ for each $d\ge2$. Thus, if the period is $20$ modulo $100$, it is indeed $4\cdot5^{d-1}$ modulo $10^d$ for each $d\ge2$. $\endgroup$
    – Emil Jeřábek
    May 25, 2022 at 16:14
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    $\begingroup$ The strings not being random most likely doesn't undermine my wrong heuristic. What certainly undermines it is that for each power the non-eventually-periodic upper part of the string becomes the increasingly vast majority of it. As the exponent grows the probability of not finding all digits in either that power or any subsequent one, drops exponentially to zero. $\endgroup$
    – Yaakov Baruch
    May 25, 2022 at 18:49

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