Digit after decimal point of radicals Let $0\leq c\leq 9$ and $2\leq k$ be integers. 

Does there always exist a positive integer $n$ such that the decimal representation of each of $\sqrt{n},\sqrt[3]{n},\dots,\sqrt[k]{n}$ has the digit $c$ immediately after the decimal point?

For $c=0$ we can choose $n$ that is a perfect square, cube, ..., $k$th power simultaneously, while for $c=9$ we can choose a large $n$ that is one less than these perfect powers.
 A: Yes, such an $n$ does exist.  First, let me prove a lemma.

Lemma: Let $m>1$ and $\epsilon>0$.  Then there exists $N$ such that for all $x>N$, $$(x+\epsilon)^{\frac{m}{m-1}}-x^{\frac{m}{m-1}}>2.$$

Proof of Lemma: Let $\alpha=\frac{m}{m-1}$.  Since $m>1$, $\alpha>1$.  Writing $f(x)=x^\alpha$, $f'(x)=\alpha x^{\alpha-1}$ goes to infinity as $x$ goes to infinity.  In particular, we can choose $N$ such that $f'(x)>2/\epsilon$ for all $x>N$.  The desired result now follows by the mean value theorem.
Now we use the Lemma with $\epsilon=1/10$ and $m=2,\dots,k$ to choose $N$ such that for all $x>N$ and any of these values of $m$, $$(x+\epsilon)^{\frac{m}{m-1}}-x^{\frac{m}{m-1}}>2.$$
Now fix an integer $a_k>N$ and let $x_k=a_k+c/10$.  Then by the inequality above, there exists an integer $a_{k-1}$ such that setting $x_{k-1}=a_{k-1}+c/10$, we have $$x_k^{\frac{k}{k-1}}<x_{k-1}<x_{k-1}+1/10<(x_k+1/10)^{\frac{k}{k-1}}.$$ Raising this to the $(k-1)$st power, this is equivalent to $$x_k^k<x_{k-1}^{k-1}<(x_{k-1}+1/10)^{k-1}<(x_k+1/10)^k.$$
Note also that $x_{k-1}>x_k$, and in particular $x_{k-1}>N$.  Now we repeat with $k-1$ in place of $k$, and so on.  We thereby construct a sequence of numbers $x_k,x_{k-1},\dots,x_2$, all of which are $c/10$ more than an integer, satisfying $$x_k^k<x_{k-1}^{k-1}<\dots<x_2^2<(x_2+1/10)^2<\dots<(x_{k-1}+1/10)^{k-1}<(x_{k}+1/10)^k.$$
Finally, we can choose an integer $n$ between $x_2^2$ and $(x_2+1/10)^2$.  By the inequalities above, we have $x_m<\sqrt[m]{n}<x_m+1/10$ for $m=2,\dots,k$, and so the first digit of $\sqrt[m]{n}$ after the decimal place is $c$.
A: Yes. I'll aim to show it just for $k = 3$, but the idea should work in general.
Observe that there exists $m$ large enough that $\sqrt[3]{(m + 1)^2} - \sqrt[3]{m^2} < 0.01$ (for example). If you're familiar with calculus, we can do this by observing that $\frac{d}{dx}x^{2/3}$ converges to $0$ as $x \to \infty$; otherwise, just think of it as "$x^a$ grows more and more slowly for larger $x$ if $a < 1$."
Now, take $M > m$ so large that $(M + 0.a)^2$ and $(M + 0.b)^2$ differ by at least $1$ for each $a \neq b$. Take $M' > M$ least so that $\sqrt[3]{M'}$ has the desired first decimal. $\sqrt[3]{(M'+1)^2}$ is at most $0.01$ bigger, so has that same first digit. Take $n = \mathrm{floor}((M' + 0.c)^2)$.
The same idea should be perfectly adaptable to larger $k$; the rest should be a proof by induction.
