I'm curious about some properties of the powers of 5 $$5^2=25,\quad5^3=125,\quad 5^4=625,\quad 5^5=3125,\quad ...$$ Is it true that at least $50$% of the digits in the decimal representation of $5^n$ are non-zero? This seems pretty modest since assuming each digit will be equally likely, only about $10$% of the digits will be zero on average. The first zero occurs at $5^8=390625$ and the power with the largest percentage of zeros seems to be $$5^{45}=28421709430404007434844970703125$$ in which $\approx 22$% of the digits are zero. I checked up to $5^{1000}$.

The difficulty is the statement seems so obvious from a probabilistic perspective, yet I can't pin down any definite theorems!

Obviously the first and the last 3 digits will always be non-zero so we have at least 4 non-zero digits. But I'm hoping to prove some properties about $5^n$ in general which require the non-zero digits to be at least linear in $n$. So $50$% would be more than enough. Really any probability $\epsilon>0$ will do -- bigger the better though.

Maybe analyze $\langle 5 \rangle^\times$ in $\mathbb{Z}/10^k\mathbb{Z}$? Could probability theory produce the bound in question?

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    $\begingroup$ I suspect this is too hard for today's mathematics. It's probably not even known that for all sufficiently large values of $n$ there's at least one zero in the decimal expression for $5^n$. $\endgroup$ – Gerry Myerson Jul 1 at 2:41
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    $\begingroup$ This question has some information about the appearance of digits in powers. These seem like hard questions! math.stackexchange.com/questions/116026/… $\endgroup$ – Steve Kass Jul 11 at 0:24
  • $\begingroup$ recursive $5^n=25+100\cdot {5^{n-2}-1\over 4}\quad n>1$ $\endgroup$ – Roddy MacPhee Aug 12 at 20:23

A good picture is worth a thousand words. It seems that the number of zeroes is growing in an almost linear fashion with a fair amount of "noise" around the meadian line. Here are the results for a number of zeroes in $5^n$ up to $n=10,000$

enter image description here

Not much will change if you extend the range to $n=20,000$:

enter image description here

Linear fit gives the following approximation:

$$n_{zero}=0.0699383 n-0.606536$$

...which means that the number of zeroes is roughly around 7%.

Mathematica provides the following regression analysis:

$$\begin{array}{l|llll} \text{} & \text{Estimate} & \text{Standard Error} & \text{t-Statistic} & \text{P-Value} \\ \hline 1 & -0.606536 & 0.354669 & -1.71015 & 0.0872544 \\ x & 0.0699383 & 0.0000307141 & 2277.07 & 5.874\cdot 10^{-24155} \\ \end{array}$$

If you try to fit the data with a parabola you get the following approximation:

$$n_{zero}=0.0931693 + 0.0697284 n+1.0494\cdot 10^{-8}n^2$$

In the given range, the quadratic term is almost neglectable which supports the conjecture that the relation between the exponent and the number of zeroes is approximately linear.

EDIT: Fun fact: $5^{58}$ has not a single zero.

EDIT 2: Mathematica code to play with:

countZeros[n_] := Module[
   {m, cnt, d},
   m = n;
   cnt = 0;
   While[m > 0,
    d = Mod[m, 10];
    If[d == 0, cnt++];
    m = Quotient[m, 10];

analyzeExponents[limit_] := Module[
   {i, data, exp},
   i = 0;
   exp = 1;
   data = {};
   While[i <= limit,
    exp *= 5;
    AppendTo[data, {i, countZeros[exp]}];
    If[Mod[i, 100] == 0, Print["Reached i=", i]];

  • $\begingroup$ (+1) Would you mind sharing the code from Mathematica? $\endgroup$ – TheSimpliFire Jul 1 at 8:10
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    $\begingroup$ There will be about $n\log_{10}5=0.7n$ digits, so that would make it about ten percent of the digits. The first one and last three won't be, but that is a small effect. $\endgroup$ – Empy2 Jul 1 at 8:14
  • $\begingroup$ @TheSimpliFire I have updated the answer with the code. $\endgroup$ – Oldboy Jul 1 at 9:25
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    $\begingroup$ These calculations confirm our intuitions, but of course they prove nothing. $\endgroup$ – Gerry Myerson Jul 1 at 10:38
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    $\begingroup$ Not only this proves nothing, but it's barely related with the OP's goal: the question is not about the mean or approximate average number of zeroes, but about bounding the worst case. That is, is $z_n$ is the number of zeroes, we want to find some $\alpha$ such that $z_n/n \le \alpha $ for all $n$. In particular we want to know if $\alpha=0.5$ works . $\endgroup$ – leonbloy Jul 1 at 12:17

Turns out one good proof direction is hopeless -- but still informative.

Suppose we could bound the number of consecutive zeros that appear in $5^n$. So some statement like "No more than 4 consecutive zeros occur in the decimal expansion of $5^n$". That would give us our result since it would follow that at $1$ in every $5$ digits -- so at least $20$% -- are nonzero.

This turns out not to be the case. One can find arbitrarily long sequences of zeros in $5^n$. In fact, it is claimed that there exist arbitrarily long sequences of zeros in $5^{m+2^m+2}$ for sufficiently large $m$. For example, $$5^{2+2^2+2}=5^8=390625$$ $$5^{5+2^5+2}=5^{39}=\ ...30078125$$ $$5^{8+2^8+2}=5^{266}=\ ...10009765625$$ $$5^{12+2^{12}+2}=5^{4110}=\ ...100006103515625$$ $$5^{15+2^{15}+2}=5^{32785}=\ ...700000762939453125$$ $$5^{18+2^{18}+2}=5^{262164}=\ ...900000095367431640625$$ One can achieve this result by analyzing the $k$th digit of $5^n$ as $n$ varies. For example if we let $$5^n=\sum_{k=0}^\infty a_n(k)10^k$$ where each $a_n(k)$ is one of $0,...,9$. Denote the sequence of the $k$th digit as $$S_k=\{a_n(k)\}_{n=0}^\infty.$$ One can show that for all $k\ge1$ that $S_k$ begins a repeated cycle of length $2^{k-1}$ at $n=k+1$. The proof I found of this was tedious so I'll neglect to include it here. The important bit is this though: the first $\lfloor k\log_5 10\rfloor$ digits of $S_k$ must be zero due to the fact that $10^n$ grows faster than $5^n$. It follows that the repeated cycle of $S_k$, call it $C_k$, contains an increasingly longer prefix of zeros. Example, $$C_2=\{1,6\}$$ $$C_3=\{0,3,5,8\}$$ $$C_5=\{0,0,3,9,7,8,1,7,5,5,8,4,2,3,6,2\}$$ $$C_7=\{0,0,0,4,4,2,0,1,...\}$$ The values $n=m+2^m+2$ are simply the powers for which the prefixes of these cycles all line up again.

Some interesting corollaries also emerged from the proof:

$\bullet$ The distribution of digits in each $S_k$ tends to perfect equalibrium -- so $10$% of each digit -- as $k$ grows.

$\bullet$ Each $S_k$ obeys a sort of mirror image law in which there are equally many $0$'s and $5$'s, $1$'s and $6$'s, $2$'s and $7$'s, etc.


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