Infinitely many $n$ such that $2^n$ in base $10$ starts with $7777777$

Question

I'm attempting to prove that there are infinitely many $n$ such that the first $7$ digits in the base $10$ expressions of $2^{n}$ are $7777777$. However, I don't even know where to start. Apparently I'm supposed to use the fact that the set $$\{x_{n} = n\alpha-\lfloor n\alpha \rfloor \mid n \in \mathbb{N}\}$$ is dense in $[0,1]$, for a fixed irrational $\alpha \in \mathbb{R}$ (I've already proven this). Any help would be appreciated! (I'm more looking for a hint not a solution)

$2^n$ starts with $777777$ iff there is an integer $m$ such that $777777\cdot 10^m < 2^n < 777778\cdot 10^{m}$ — Winther, Jun 01 '18 at 13:34

score 2 · Accepted Answer · edited Jun 01 '18 at 15:26

As noted by @Winther, $2^n$ starts with $777777$ if and only if $777777\cdot 10^m \le 2^n < 777778\cdot 10^{m}$ for some $m \in \mathbb{N}$.

This can be rewritten as

$$m + \log_{10} 777777 \le n\log_{10}2 < m + \log_{10} 777778$$

Let $\alpha = \log_{10}2$ and conclude there are infinitely many $n \in \mathbb{N}$ such that $$n\log_{10}2- \left\lfloor n\log_{10}2\right\rfloor \in \big[\log_{10} 777777 - 5, \log_{10} 777778-5\big\rangle \subseteq [0,1]$$

For these $n$ we have

$$(\left\lfloor n\log_{10}2\right\rfloor - 5) + \lfloor n\log_{10}2\rfloor + \log_{10} 777777 \le n\log_{10}2 < (\left\lfloor n\log_{10}2\right\rfloor - 5) + \log_{10} 777778$$

so for them $2^n$ starts with $777777$.

@RossMillikan Thanks, fixed. – mechanodroid Jun 01 '18 at 14:36 — mechanodroid, Jun 01 '18 at 14:36

lhf · Answer 2 · 2018-06-01T15:24:10.420

1

The key fact is this (see here for a proof):

For every finite sequence of decimal digits, there is a power of $2$ whose decimal expansion begins with this sequence.

Now apply this to $7777777\cdots 7$, with an increasing number of $7$s.

edited Jun 01 '18 at 15:24

answered Jun 01 '18 at 13:35

lhf

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Infinitely many $n$ such that $2^n$ in base $10$ starts with $7777777$

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