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?
If a!b!=c!, then
a! * (1*2*3*...*b) = 1*2*3*...*b*(b+1)*...*c
Thus a! equals the product of terms from b+1 to c. The trivial solution has been pointed out to be b=a!1, in which case a! = c, the final single term of c!.
A nontrivial solution would be one where a! equals the product of several consecutive numbers. For instance,
6! = 720 = 8*9*10
thus 6!*7!=10!
Edited on August 16, 2004, 12:41 pm

Posted by Bryan
on 20040816 12:04:01 