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

thus 6!*7!=10!

Edited on August 16, 2004, 12:41 pm

Posted by Bryan on 2004-08-16 12:04:01