Can any base ten repunit, other than 1, be a perfect cube?

If so, give an example. Otherwise prove that no base ten repunit (other than 1) can be a perfect cube.

I'm not sure, but I think there are perfect cubes which end with an arbitrarily large number of ones.

I imagine this can be shown by writing k=(10n+1)^3=1000n^3+300n^2+30n+1, and figuring n must end in 7,

Then, repeat the process, writing (100n+71)^3=1,000,000n^3+2,130,000n^2+21,300n+357911, and noting n must end in 4.

Then, repeat with (1000n+471)^3 and so on. You get the following sequence

1
357,911
104,487,111
607,860,671,111

Posted by Gamer on 2011-10-25 16:01:09