|Jeroen Demeyer on Fri, 25 Nov 2005 14:01:14 +0100|
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|Re: modular exponentiation|
Henk Karssenberg wrote:
Dear M.,In PARI I try to calculate k = Mod(6682632708227277^28345232917, 72057594037927889) but this gives an overflow. Is there any option to calculate Mod(a^m,n) or a^m % n with huge numbers ?Thank you & kind regards.
Henk, You should do k = Mod(a,n)^m. This works because Mod(a,n) creates an object in Z/nZ.