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I'm wondering if some power of $2$ can be written in base $10$ as $2013$ followed by other digits.

Formally, does there exist $k,q,r \in \mathbb N$ such that $$2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r<10^q $$

I'm not sure if it's true or not. I would go for a 'no', but I can't prove it.

Thanks for your help.

Martin Sleziak
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Gabriel Romon
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1 Answers1

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Since $\dfrac{\log 2}{\log 10}$ is irrational, there are such $k$. We want

$$2.013 \cdot 10^m \leqslant 2^k < 2.014\cdot 10^m,$$

and taking the logarithm to the base $10$ that becomes

$$m + \frac{\log 2.013}{\log 10} \leqslant k\cdot\frac{\log 2}{\log 10} < m + \frac{\log 2.014}{\log 10}$$

or

$$\frac{\log 2.013}{\log 10} \leqslant \left\lbrace k\frac{\log 2}{\log 10}\right\rbrace < \frac{\log 2.014}{\log 10},$$

where $\{x\} = x - \lfloor x\rfloor$ is the fractional part. It is a well-known fact that for irrational $\alpha$ the sequence of $\{ k\alpha\}$ is dense in $[0,1]$.

Prelude> [n | n <- [1 .. 100000], take 4 (show (2^n)) == "2013"]
[1363,3499,5635,14664,16800,18936,30101,32237,34373,43402,45538
,47674,56703,58839,60975,63111,72140,74276,76412,85441,87577,89713]
Daniel Fischer
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  • I don't understand why the fact that ${k \alpha}$ is dense solves the problem. Can you explain me, please? Thank you very much! – g.pomegranate Nov 19 '17 at 19:00
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    That $\bigl{ k\alpha - \lfloor k\alpha\rfloor : k \in \mathbb{N}\bigr}$ is dense in $[0,1]$ means every interval $(a,b)$ with $0 \leqslant a < b \leqslant 1$ contains at least one of these values. [In fact, infinitely many.] In particular for $a = \frac{\log 2.013}{\log 10}$ and $b = \frac{\log 2.014}{\log 10}$, since $0 < a < b < 1$, the interval $(a,b)$ contains ${k\alpha}$ for infinitely many $k$. Pick any $k \geqslant 10$ such that $a < \bigl{ k\frac{\log 2}{\log 10}\bigr} < b$. Then, since $k \geqslant 10$ we have $m := \bigl\lfloor k\frac{\log 2}{\log 10}\bigr\rfloor \geqslant 3$ – Daniel Fischer Nov 19 '17 at 19:16
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    and hence $a + m < k\frac{\log 2}{\log 10} < b + m$. Then, exponentiating with base $10$, we obtain $$10^{a+m} = 2.013\cdot 10^m = 2013\cdot 10^{m-3} < 2^k < 10^{b+m} = 2.014\cdot 10^m = 2014\cdot 10^{m-3}.$$ Since $m \geqslant 3$, that means the decimal expansion of $2^k$ begins with $2013$. – Daniel Fischer Nov 19 '17 at 19:16
  • you are taking log with base $10 $ then why you not putting $log10$ as $1$? – CHOUDHARY bhim sen Mar 12 '19 at 03:06