On Fri, Jan 24, 2003 at 06:39:40PM -0500, Frank Anzalone wrote: > If I need m mod n and m is to big for my mod function, I believe I can add > a mod n to b mod n provided a+b=m. Is there an easier way? suppose n is to > big?
What do you mean by "is too big for your mod function"? Usually, in this case (at least if m is _way_ too big for your mod function), one would simply use another mod function. :-) But yes, (a mod n) + (b mod n) = (a + b) mod n, as you say -- I doubt it would help you that much, though. /* Steinar */ -- Homepage: http://www.sesse.net/ _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
