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.)
Solution Solution / Thoughts | Comment 1 of 7

There are an infinite number of trivial solutions. If we allow b to equal a! - 1, then c is equal to a!. For example, if a=3 and b=5, then since a!=6 and c!=5!*6=b!*a!, the solution is trivial. These trivial solutions get incalculable very quickly.

I don't believe there are any quick solutions if you eliminate that condition.


  Posted by Eric on 2004-08-16 09:21:19
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 (12)
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