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

 Factorials everywhere (Posted on 2004-08-16)
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: 2.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution / Thoughts | Comment 1 of 6

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

 Search: Search body:
Forums (0)