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How can i prove that 100...500...1 [100 zeros in each group ( ... is 100 zeros)]is not a perfect cube? I tried symmetric features of the number but could not figure out anything related.any ideas please tell me.

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  • $\begingroup$ Look at the number mod n for a few values of n and see what you find out. $\endgroup$ – John Brevik Mar 1 '15 at 9:35
  • $\begingroup$ I don't understand what $100\ldots 500\ldots 0$ means. $\endgroup$ – TonyK Mar 1 '15 at 9:37
  • $\begingroup$ @Tony: The number is $10^{202}+5\cdot10^{101}+1$: a $1$ followed by $100$ zeroes, a $5$, another $100$ zeroes, and a $1$. $\endgroup$ – Brian M. Scott Mar 1 '15 at 9:43
  • $\begingroup$ Following up on John's suggestion. Try $n$ a prime that is congruent to $1\pmod 6$. Only one third of the residue classes are then cubic residues. Hint: a prime $<20$ will work! $\endgroup$ – Jyrki Lahtonen Mar 1 '15 at 9:53
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Hint : Every cube is either $0, 1,$ or $8 \mod 9$. Your number is equal to $7 \mod 9$

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well thanks to elaqqad i got an answer. Its just that the digital index of cubes are 1,8,9 in a pattern. ex- digital index of 1 cube = 1 "" "" 2 "" = 8 "" "" 3"" = 9 (27 - 2+7) "" "" 4 "" = 1 (64 - 6+4) "" "" 5 "" = 8 (1+2+5) "" "" 6 "" = 9 (2+1+6) but the digital index of the above number is 7 (1+5+1) so simply it's not a cube.

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