Yeah, sorry, there are extra steps here. If n= 43112609 and you have some 2n-bit number to reduce, you can split it into two n-bit words and simply add the words together, since 1 = 2^n (mod 2^n - 1). You can do this very quickly (ie you can add an arbitrary number of 2n -bit values together with only a small number of adds and shifts as overhead).