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.

  • $\begingroup$ Do you mean to say, for a given $k$ there exists at least one $n$ such that its $k$th root has $c$ as the first digit after decimal? $\endgroup$ – gambler101 Sep 6 '16 at 8:19
  • 1
    $\begingroup$ Not just the $k$th root, but all roots from the $2$nd to the $k$th $\endgroup$ – pi66 Sep 6 '16 at 8:51

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 $c$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.