> The other way is > > a^b mod m = a^(b mod phi(m)) mod m > > phi(65)=phi(5)*phi(13)=4*12 = 48 => a^64 mod 65 = a^16 mod 65. Unfortunately, this particular question says ``Pretend throughout the calculation that you do not know that 65 = 5x13.'' Peter ===== Peter Taylor pjt33@cam.ac.uk