Prove that for every integer $m$ there is an integer $n$ such that the digits of $5^n$ start with $m$ (left most digits). For example for $m=156$, $n=6$ is the solution, because $5^6 = \color{red}{156}25$. I put a lot of thinking into it and I didn't even find a way to handle it.
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What exactly do you want to know? "$5^n$ digits starrt with $m$ "..what does that even mean? – Oct 08 '19 at 18:14
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It just means that $5^n$ starts with the digits of $m.$ For example, if $m=12$ then $5^3=125$ starts with the digits of $12.$ @HVxvejjw – Thomas Andrews Oct 08 '19 at 18:20
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3This is due to $\log_{10} 5$ being irrational. – Thomas Andrews Oct 08 '19 at 18:21
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@ErwinSchrodinger why did you "vandalise" the question? I restored it ... Don't worry if it will be closed, it happens. – rtybase Oct 08 '19 at 21:04
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What we want is $$m\underbrace{000..0}_k \leq 5^n \leq m\underbrace{999..9}_k$$ ($m$ is concatenated with $0$'s or $9$'s) which translates into $$m\cdot 10^k\leq 5^n \leq m\cdot 10^k + \underbrace{999..9}_k \iff \\ m\cdot 10^k\leq 5^n \leq m\cdot 10^k + 10^k-1 \iff \\ m\leq\frac{5^n}{10^k}\leq m\left(1+1-\frac{1}{10^k}\right) \iff $$ $$\log_{10}m\leq n\log_{10}5-k \leq \log_{10}m+\log_{10}\left(1+1-\frac{1}{10^k}\right) \tag{1}$$ since $\frac{1}{10^k}\to0, k\to\infty$ we will be ok to find a large enough $k$ such that $$\log_{10}m\leq n\log_{10}5-k \leq \log_{10}m+\log_{10}\left(1+\frac{1}{2}\right)\leq \\ \log_{10}m+\log_{10}\left(1+1-\frac{1}{10^k}\right)$$ or simply $$\log_{10}m\leq n\log_{10}5-k \leq \log_{10}m+\log_{10}\left(1+\frac{1}{2}\right) \tag{2}$$ But because $\log_{10}5$ is irrational then the set $\left\{n\log_{10}5-k \mid n\in\mathbb{N}, k\in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$. Thus, such $n$ and $k$ satisfying $(2)$ exist, for every given positive integer $m$. Actually, there are infinitely many of them with large enough $k$'s to cover $(1)$ as well.
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