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

Home > Just Math
Factorials everywhere (Posted on 2004-08-16) Difficulty: 3 of 5
The equation a!b!=c! is trivial if we allow a or b to be 0 or 1.

However, adding that 1<a<b<c, also allows trivial solutions: which? What condition should be added to disallow such solutions?

Finally, adding that condition, can you find any solution to the problem?

See The Solution Submitted by Federico Kereki    
Rating: 3.0000 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: More trivial solutions | Comment 5 of 7 |
(In reply to More trivial solutions by Rajal)

If solutions of the form a!=(b+1)(b+2)... are considered trivial, then all solutions are trivial, as this must be the case for any a!b!=c!, as in Bryan's solution.
  Posted by Charlie on 2004-08-16 12:25:32

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 (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information