17 Dec
2003
17 Dec
'03
11:17 p.m.
If (n!)mod x = 0 then there is a factor of x which is less than n. If you can solve modular factorials, then you can solve for the largest factor of x in logarithmic time. Obviously, nobody has found a method to do either.
Just some thoughts...
...
If x>n and x is not a prime then the result will again always be 0 since we can break x down into factors smaller than n and the previous argument removes the various factors.
This doesn't work--(x > n) & x not prime doesn't imply that x has a factor less than n. That's only true if sqrt(x) >= n.