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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: 2.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: More trivial solutions | Comment 5 of 6 |
(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

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