Left most Digits of $5^n$ 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.
 A: 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.
