What is the positive integer $n$ with a special property?

[lfdahl]

$n$ is a positive integer with the following property:

If the last three digits of $n$ are removed, $\sqrt[3]{n}$ remains.

Find with proof $n$.

Source: Nordic Math. Contest

 

[Opalg]

$n$ is a positive integer with the following property:

If the last three digits of $n$ are removed, $\sqrt[3]{n}$ remains.

Find with proof $n$.

Source: Nordic Math. Contest

[sp]Let $x = \sqrt[3]n$. We are told that $x^3 = n = 1000x + k$ (where $k$ is the number formed by the last three digits of $n$). Therefore $$x(x^2 - 1000) = k.$$ This implies that $x^2>1000$, and so $x\geqslant32$. But if $x = 33$ then $x^2 = 1089$ and $x(x^2-1000) = 33\times89 = 2937$, which is too big because $k$ must only have three digits.

So $32\leqslant x<33$, and the only possible value for $x$ is $32$. Then $n = 32^3 = 32\,768$. When the last three digits are removed, what is left is $32$, as required.

[/sp]

 

[lfdahl]

[sp]Let $x = \sqrt[3]n$. We are told that $x^3 = n = 1000x + k$ (where $k$ is the number formed by the last three digits of $n$). Therefore $$x(x^2 - 1000) = k.$$ This implies that $x^2>1000$, and so $x\geqslant32$. But if $x = 33$ then $x^2 = 1089$ and $x(x^2-1000) = 33\times89 = 2937$, which is too big because $k$ must only have three digits.

So $32\leqslant x<33$, and the only possible value for $x$ is $32$. Then $n = 32^3 = 32\,768$. When the last three digits are removed, what is left is $32$, as required.

[/sp]

Thankyou, Opalg, for an exemplary answer!