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 then (n!)modx will always be 0. Since n! is simply the product of the numbers 1...n and is always a integer product dividing by x simply removes the factor m such that we have the product of 1...m-1,m+1...n. 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. If x is prime and x>n then we will get a result that is non-zero. Take care.