All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Repunit Rigor (Posted on 2011-10-25) Difficulty: 4 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
End with a repunit | Comment 4 of 5 |
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information