On Thu, Jan 13, 2011 at 12:06 AM, Aviral Gupta <aviral....@gmail.com> wrote:
> we can do it when hcf(b,m)=1 , in that case find inverse of b by > extended euclidean mod m and then multiply it by a > Yes. And when m is prime, B(mulitplicative inverse of b) = b^(m-2) As b^(m-1)mod m = 1 if m is prime. > > Regards > Aviral > http://coders-stop.blogspot.com/ > > On Jan 12, 6:36 am, mittal <mohitm.1...@gmail.com> wrote: > > Somehelp with (a/b)modm expression. > > > > http://en.wikipedia.org/wiki/Modular_arithmetic > > i visited this link but found only for addition,subtraction and > > multiplication. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > Please access the attached hyperlink for an important electronic communications disclaimer: http://dce.edu/web/Sections/Standalone/Email_Disclaimer.php -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
