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Suppose the numbers $1,2,3,\dots,1986$ in any order are concatenated then prove that the number is not a perfect cube.

This problem gives me a feeling that here cubic residues can only help no other choice. Please help I am still trying.

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Hint: Any such number is a multiple of $3$ but not a multiple of $9$.

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"In any order" suggests we check the digit sum, which does not depend on order. Turns out ist is $27687\equiv 3\pmod 9$, hence any such number is $\equiv 3\pmod 9$. Perfect powers are either not divisible by $3$ or are divisible by $9$.

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    $\begingroup$ Technically, you don't need to compute the digit sum, you can just compute $1+2+\cdots+1986$ modulo $9$. $\endgroup$ – Thomas Andrews Apr 25 '15 at 12:03

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