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

Question

How can i prove that 100...00500...001 [with 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.

Answer 1

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.

Answer 2

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