# Prove that 100…500…1 (100 zeros in each group) is not a perfect cube?

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.

• Look at the number mod n for a few values of n and see what you find out. – John Brevik Mar 1 '15 at 9:35
• I don't understand what $100\ldots 500\ldots 0$ means. – TonyK Mar 1 '15 at 9:37
• @Tony: The number is $10^{202}+5\cdot10^{101}+1$: a $1$ followed by $100$ zeroes, a $5$, another $100$ zeroes, and a $1$. – Brian M. Scott Mar 1 '15 at 9:43
• 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! – Jyrki Lahtonen Mar 1 '15 at 9:53

Hint : Every cube is either $0, 1,$ or $8 \mod 9$. Your number is equal to $7 \mod 9$