Re: Fast Modular Factorial?

[Matthew J Ghio <mg5n+@andrew.cmu.edu>]

Jim choate <ravage@bga.com> wrote:

> 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.

And there will always be such a value for m equal to kx where k is an
integer less than n/x
If x is non-prime, there may be factors f and g such that f*g=x.  In
that case, if n>f and n>g then n=0, hence finding the smallest value of
n such that (n!)mod x =0, will yeild a factor of x.  In that case,
dividing by x would remove the factors f and g, yeilding a zero
remainder.

> 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.

Yes, but if x is not prime, and x>n, (n!)mod x will not necessarily be
zero, unless x>n>x/2


A few examples:

mod 7:
 n   1  2  3  4  5  6  7  8  9 10
 n!  1  2  6  3  1  6  0  0  0  0

mod 15:
 n   1  2  3  4  5  6  7  8  9 10
 n!  1  2  6  9  0  0  0  0  0  0


Note that for mod 15, n=>5 produces only zeros, revealing the factor 5.