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I know that $2^n$ cannot $ = 1000\ldots002:$ because $3\mid 1000\ldots002$ but $3 \nmid\ 2^n.$

Also, $2^n$ cannot $ = 1111\ldots110:$ because $5\mid 1111\ldots110$ but $5 \nmid\ 2^n.$

Are there any number theory arguments for why $2^n \neq 111\ldots112?$

Adam Rubinson
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3 Answers3

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It seems that $16\nmid1112$, so $2^n$ cannot have this form for large $n$.

LittleBear
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    Ah, this is already be observed by the comment above~ – LittleBear Jul 07 '24 at 14:17
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    It beat you by nine seconds, so I wouldn't take that as having priority. That said, this is pretty thin as an answer---it would help to spell out the point more to avoid this being marked as "should be a comment". (In particular, why is this objection sufficient to prevent larger cases?) – Semiclassical Jul 07 '24 at 14:19
  • I don't get it. Why does $16\nmid1112$ mean that $2^n$ cannot have this form for large $n$? – Adam Rubinson Jul 07 '24 at 14:20
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    Ooooh, $16 \nmid 1112 \implies 16\nmid 1112 + 10000,$ because $16\mid 10000.$ And so on for larger $n$. – Adam Rubinson Jul 07 '24 at 14:23
  • @Semiclassical I post this as an answer since I don't get enough score to post a comment (before I post this answer)... – LittleBear Jul 07 '24 at 14:24
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\begin{align*} 111\ldots 112&=1+\sum_{k=0}^m 10^k =1+\dfrac{10^{m+1}-1}{10-1}=\dfrac{10^{m+1}+8}{9}=2^n\\[12pt] 9\cdot2^n&=2^3(1+2^{m-2}5^{m+1})\\[12pt] 9\cdot 2^{n-3}&=1+2^{m-2}5^{m+1} \end{align*} The RHS is odd for $m>2$ and the LHS is even for $n>3$ which rules out these cases. The remaining cases are $n\in \{0,1,2,3\}$ or $2^n\in \{1,2,4,8\}$ that are all of a different pattern, except $n=1$. $m\in \{0,1,2\}$ means $111\ldots 112\in \{2,12,112\}.$ Therefore $$ 2^1=2 $$
is the only solution.

Marius S.L.
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  • My cost for typing slowly and being wordy.... should I delete my answer as you beat me by a few minutes? – fleablood Jul 07 '24 at 14:58
  • @fleablood I have a trick. My keyboard is loaded with 67 LaTeX shortcuts, an AHK script. E.g. I type Alt+d and get \dfrac{}{} with the cursor placed within the first bracket. – Marius S.L. Jul 07 '24 at 15:04
  • ... and I have a tendency to got into wordiness and to think and rethink "am I explaining this in the most simple and explanatory method" – fleablood Jul 07 '24 at 16:00
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Well $N = \underbrace{111....1}_{k-2}2 = \underbrace{1111....11}_{k-1} + 1= \frac {10^k-1}9 +1$ so if this equal to $2^n$ then $9N= 9\cdot 2^n =(10^k-1) + 9= 10^k + 8$.

If we assume that $n$ and $k$ are both at least $3$ was can divide by $8$ to get

$\frac {9\cdot 2^n}8=\frac {9N}8\implies$

$9\cdot 2^{n-3} =\frac {10^k+8}8 = \frac {10^k}8 + 1=2^{k-3}\cdot 5^k + 1$ which is odd if $k> 3$.

We can rule out that if $k=3$ then $N=12$ is no a power of $2$. So if any $N$ is a power of $2$ we must have $9\cdot 2^{n-3}$ is odd which mean we must have $n=3$. But $2^3 = 8$ is not any $N$ of the form $1111....1$.

.....

Or to put it in a naive but more hands on way:

$2^n = 11111.....1112 \iff$
$2^n = 11111.....1111 + 1\iff$
$9\cdot 2^n = 99999....9999 + 9 \iff$
$9\cdot 2^n = 100000....0000 + 8 \iff$
$9\cdot 2^{n-3} = 100000...0\times \frac{1000}8 + 1\iff$
$9\cdot 2^{n-3} = 100000...0\times 125 + 1$

But LHS is odd and right hand side is even except for small values of $n$ and/or small numbers of ones and those cases can be easily checked.

fleablood
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