perplexus.info :: Just Math : Factorials everywhere

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?

Submitted by Federico Kereki

Rating: 3.0000 (6 votes)

Solution: (Hide)

If we take c=a! and b=(c-1), the equation is trivially satisfied.
Adding the condition 1<a<b<c-1 allows the solution a=6, b=7, c=10.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Solution	K Sengupta	2022-06-04 23:34:50
Partial Answer	K Sengupta	2007-08-13 03:46:11
re: More trivial solutions	Charlie	2004-08-16 12:25:32
Solution	Bryan	2004-08-16 12:04:01
More trivial solutions	Rajal	2004-08-16 11:17:58
SOLUTION++++	Ady TZIDON	2004-08-16 10:30:42
Solution / Thoughts	Eric	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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info