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?
As pointed out, b=a!-1 is a family of trivial solutions. More generally, (b+1)(b+2)...=a! is a family of solutions. If these solutions are also elimited, another solution gets harder, still working on that one.
|
Posted by Rajal
on 2004-08-16 11:17:58 |