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?

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