[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.lfdahl said:$n$ 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 said:[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.
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