# What is the highest number of digits so that this number of digits in a specific power of 2 are exactly 10%?

I accidentally found out that in $$2^{{10}^{6}}$$ (==$$2^{1000000}$$), there are exactly 10% digits of 6 (in the decimal form). And I would like to know - are there powers of 2 in which all the digits from 0 to 9 appear exactly 10% of the time? And if not, what is the highest number (from 1 to 10) c (for count), which c digits appear exactly 10% of the time in the decimal form of a specific power of 2 ($$2^{n}$$)?

c must be at least 5:

• $$2^{31}$$ == $$2147483648$$ (10 digits; 5 digits appear once)
• $$2^{166}$$ == $$93536104789177786765035829293842113257979682750464$$ (50 digits; 5 digits appear 5 times)

I made a small program in Python to calculate c. But I'm interested in the powers up to infinity.

for n in range(1, 10 ** 6 + 1):
a = 2 ** n
b = list(str(a))
s = []
for d in range(10):
s.append(sum([1 if i == str(d) else 0 for i in b]))
c = sum([1 if (j * 10 == len(b)) else 0 for j in s])
if (c >= 5):
print(c, n, len(b), s)


Addition: Someone noted to me that if a number contains exactly 10% digits of each digit from 0 to 9, then its sum is divisible by 9, so the number is also divisible by 9 and can't be a power of 2. But I checked also powers of 3, and also there I can't find numbers where there are more than 5 digits who appear exactly 10%. I did find solutions for 5 digits though (c==5), being $$3^{20}$$ and $$3^{124}$$, and of course also c==4 (4 powers, largest of them $$3^{292}$$). This while checking at least the first 200,000 powers of 3.

My hypothesis is that for powers of 2 and 3 c is 5, and c would change for other bases (and of course if the numbers are represented in another base, not base 10), but at least for base 10 I think c is never 10 for any power of a positive integer. At least not for big powers. If c is 10 then it would appear in the power of 1 ($$1234567890^{1}; 11223344556677889900^{1}$$) or at least powers up to 1000.

Here's a very rough heuristic argument (this is too long for a comment).

Let's assume that the digits of a power of $$2$$ are roughly "random" -- that is, if the number has $$\ell$$ digits, each of the $$\ell$$ digits is chosen uniformly at random from $$\{0,1,\dots,9\}$$. This is clearly not exactly correct, but it's a reasonable guess for at least the middle digits, which contribute most of the "arbitrariness" to the problem.

Let $$b=10$$, and pick some subset $$S\subset\{0,\dots,b-1\}$$ of digits. We'll calculate the probability that, for a number with $$bk$$ digits chosen randomly, for each digit $$d\in S$$, the number of occurrences of $$d$$ is exactly $$k$$ (i.e. $$d$$ occurs at exactly $$1/b$$ of the digits). Each of the $$b^{bk}$$ $$bk$$-digit numbers (including with leading $$0$$s) are equally likely to be chosen, so we need to count the number of such numbers with exactly $$k$$ occurrences of each $$d\in S$$ and divide that by $$b^{bk}$$.

Say $$|S|=c$$, and let $$S=\{d_1,\dots,d_c\}$$. There are $$\binom{bk}k$$ ways to choose the $$k$$ locations of $$d_1$$, then $$\binom{bk-k}k$$ ways to choose the locations of $$d_2$$, et cetera. So, the number of total choices for the digits in $$S$$ is $$\binom{bk}k\binom{bk-k}k\cdots \binom{bk-(c-1)k}k=\frac{(bk)!}{k!^c(bk-ck)!}.$$ There are $$(b-c)^{(b-c)k}$$ ways to choose the remaining digits, since they must all be in $$\{0,1,\dots,b-1\}\setminus S$$, with no other restriction. So, the desired probability is $$p_{b,c}(k):=\frac{\frac{(bk)!}{k!^c((b-c)k)!}(b-c)^{(b-c)k}}{b^{bk}}.$$

We'll now investigate, given fixed $$b$$ and $$c$$, how this quantity grows with $$k$$. For this, we'll use Stirling's approximation for factorials. This approximation only works with large inputs, so we'll deal with the case of $$c=b$$ (where we want all of the digits to occur exactly $$k$$ times) separately.

For $$c, Stirling's approximation gives the rather ugly computation \begin{align*} p_{b,c}(k) &\sim \frac{(b-c)^{(b-c)k}}{b^{bk}}\frac{\sqrt{2\pi bk}\left(\frac{bk}e\right)^{bk}}{\left(\sqrt{2\pi k}\left(\frac ke\right)^k\right)^c\sqrt{2\pi(b-c)k}\left(\frac{(b-c)k}{e}\right)^{(b-c)k}}\\ &=\sqrt{\frac{b}{(b-c)(2\pi k)^c}}, \end{align*} where $$\sim$$ means that the ratio between the two quantities tends to $$1$$ as $$k\to\infty$$. For $$c=b$$, we have $$p_{b,b-1}(k)=p_{b,b}(k)$$ (since if $$b-1$$ of the digits occur exactly $$k$$ times, the last must as well), and so $$p_{b,b}(k)\sim \sqrt{\frac{b}{(2\pi k)^{b-1}}}.$$

This means that, for fixed $$b$$ and $$c$$, and for large $$k$$, the probability that $$c$$ of the $$b$$ digits occur exactly $$k$$ times in a perfectly random length-$$bk$$ number is about some constant times $$k^{-c/2}$$, or $$k^{-(b-1)/2}$$ if $$c=b$$.

Now we'll apply this heuristic to the particular problem. The number of powers of $$2$$ of a given length in base $$10$$ is either $$3$$ or $$4$$, and on average it's about $$r:=\log_2(10)$$. Fix $$S\subset\{0,1,\dots,9\}$$ of size $$c$$. For large $$N$$, the expected number of powers of $$2$$ $$2^n$$ with $$n>N$$ with $$10k$$ digits for some $$k$$, for which every $$d\in S$$ occurs in $$2^n$$ exactly $$k$$ times, is very roughly $$r\sum_{k\geq \log_{10}(2^N)/10}\frac{\sqrt{\frac{10}{10-c}}}{(2\pi k)^{c/2}}=\frac{r\sqrt{\frac{10}{10-c}}}{(2\pi)^{c/2}}\sum_{k\geq rN/10}k^{-c/2}.$$

For $$c>2$$, this sum is finite, and should tend to $$0$$ quickly as $$N$$ grows. In particular, for $$N=10^6$$ (assuming it has been checked that there are no solutions $$2^n$$ with $$n\leq 10^6$$), and $$c=9$$ (acting as proxy for $$c=10$$), the sum is about $$3.6\times 10^{-23}$$. So, it's exceedingly unlikely, assuming our heuristic is close enough to the truth, that there are any powers of $$2$$ with all of the digits perfectly distributed.

For $$c=2$$, the sum diverges, as the harmonic series does. So, we expect infinitely many powers of $$2$$ to be exactly $$10k$$ digits long for some $$k$$, and to have some two digits which occur exactly $$k$$ times.

Some numerics to support the argument that these heuristics are reasonable:

I checked powers of $$2$$ up through $$2^{10^5}$$. Of these, exactly $$10^4$$ have a length which is exactly a multiple of $$10$$. Two of these numbers satisfied the condition for $$5$$ digits appearing exactly $$10\%$$ of the time: these are $$2^{31}$$ and $$2^{166}$$, of lengths $$10$$ and $$50$$, respectively. Two more, $$2^{30}$$ and $$2^{398}$$, satisfied the condition for $$4$$ digits.

Only $$21$$ numbers satisfied the condition for $$3$$ digits. These are fairly common for small $$k$$, and then tend to thin out, with the largest being $$2^{32685}$$. According to the presented heuristic, the probability that there's another such power out there is about $$0.31$$ (taking the value for $$c=3$$ and multiplying by $$\binom{10}3$$ to account for the number of different choices of $$S$$). (Edit: $$2^{112380}$$ works as well!)

$$114$$ numbers satisfied the condition for $$2$$ digits. These begin fairly common, but remain somewhat common throughout: the five largest such numbers I found were $$2^{90553}, 2^{92646}, 2^{93311}, 2^{96101}, 2^{98360}.$$ The fact that these approach the upper bound $$10^5$$ on the computation, with no signs of stopping, supports the heuristic's judgment that there should be infinitely many solutions with $$c=2$$.

• For $3$ digits, there's another one! The number $2^{112380}$ has $33830$ digits, and the digits $1$, $3$, and $9$ appears exactly $3383$ times each. Jan 4, 2022 at 13:54
• Although calculations are slow, it appears that there are possibly infinite powers of 2 where c==1, which means there is exactly one digit appearing exactly 10% of the time in the decimal form.
– Uri
Jan 4, 2022 at 17:20
• It's worth mentioning that some powers with c==1 are adjacent, for example $2^{999999}$ and $2^{1000000}$.
– Uri
Jan 4, 2022 at 17:22
• @Uri There should be infinitely many with $c=1$, and in fact these should be pretty common -- about $\Theta(\sqrt n)$ of the powers of $2$ up through $2^n$ should satisfy the property with $c=1$, while only about $\Theta(\log n)$ should satisfy it for $c=2$. I believe the heuristic would also give that about $\Theta(\log n)$ of the $c=1$ powers are consecutive, although it's definitely possible there's another reason the heuristic doesn't capture why these should be a bit more common. Jan 4, 2022 at 21:41