Find all $n\in\mathbb N$ such that by removing the last $3$ digits of $n$, $\sqrt[3]{n}$ is obtained.

Find all $n\in\mathbb N$ such that by removing the last $3$ digits of $n$, $\sqrt[3]{n}$ is obtained.

I found that $n=32768$ is a solution. Is there any other of $n$? It'll be best if no computer is used. Thank you.

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It should be pretty easy to figure out how many digits the answer can have, and from there it's a smallish amount of trial-and-error. – Gerry Myerson Nov 7 '12 at 12:09
@ Dan Brumleve: It means that the new number obtained is $\sqrt[3]{n}$. – ᴊ ᴀ s ᴏ ɴ Nov 7 '12 at 12:18

Focus instead on the cube root $m$. It must satisfy $$1000m\le m^3<1000(m+1).$$ From the first inequality, get $m^2\ge 1000$, i.e., $m\ge32$. Notice that $m=33$ violates the second inequality ($33^3=35937>34000$), and this gets worse with larger $m$. So $m=32$ is the only solution.
If you notice, he is referring to $m^{3}$ where $m=32$. That's the reverse of your notation. – GregRos Nov 7 '12 at 12:43
@GregRos: I don't understand your objection. Am I not saying the same thing? $m=\sqrt[3]{n}$, or $n=m^3$. I just find it easier to work with $m$ than with $n$. – Harald Hanche-Olsen Nov 7 '12 at 12:46
A cube root has a third as many digits and only $3$ are subtracted, so further candidates will be quickly outruled by a computational search.