The equation a
! is trivial if we allow a
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 non-trivial solution would be one where a! equals the product of several consecutive numbers. For instance,
6! = 720 = 8*9*10
Edited on August 16, 2004, 12:41 pm
Posted by Bryan
on 2004-08-16 12:04:01