Why is $2^{16}=65536$ the only power of $2$ less than $2^{31000}$ that doesn't contain the digits $1$, $2$, $4$ or $8$ in its decimal representation?

Question

$65536$ is the only power of $2$ less than $2^{31000}$ that does not contain the digits $1$, $2$, $4$ or $8$ in its decimal representation.

http://en.wikipedia.org/wiki/65536_%28number%29

Answer 1

A simple explanation turns on the apparent randomness of the base-$b$ digits of sufficiently large powers of two, in the sense that they tend to behave like random samples. (However, this leaves the apparent randomness unexplained.)

Thus, let $S_k$ denote the multiset of digits appearing in the numeral of $2^k$, and let $n_k$ be their number; i.e., $n_k = \lfloor 1 + k\cdot log_{b}2\rfloor $. If each $S_k$ were a simple random sample, then, for any $K$ and any subset $D\subset \{0,1,...,b-1\}$ (e.g., $D=\{1,2,4,8\}, \ b=10$),

$$\begin{align} P_K &= P(\text{at least one digit from D appears in *every* }S_K, S_{K+1}, S_{K+2},...)\\ &= P\left( \bigcap_{i=K}^\infty C_{i} \right)\\ &= \prod_{i=K}^\infty P(C_{i})\\ &= \prod_{i=K}^\infty(1-q^{n_i}) \end{align} $$

where

$C_i = \{S_i\cap D \ne \oslash\}$,

$q = P(\text{digit } \notin D) = 1 - \frac{|D|}{b}$.

Here are some computed cases (rounded) for $b=10$ and $|D|=4$:

\begin{array}{|c|c|} K & P_K \\ \hline 1&0.002\\ 10&0.304\\ 15&0.575\\ 20&0.780\\ 50&0.998\\ 100&0.999999\\ 200&0.9999999999999 \end{array}

As examples, I've verified that

• with $D=\{1,2,4,8\}$ and $b=10$, a digit from $D$ occurs among the digits of every $2^k$ for the range $17\le k\le 200000$
• with $D=\{1,2,3,4\}$ and $b=10$, a digit from $D$ occurs among the digits of every $2^k$ for the range $4\le k\le 200000$

Therefore, it seems extremely likely that both of these statements are true:

• $2^{16}$ is the only power of two that does not contain a digit from $\{1,2,4,8\}$.

• $2^3$ is the only power of two that does not contain a digit from $\{1,2,3,4\}$.

NB: These are examples of probably true and unprovable "Dyson statements".