Lösung - Restklassen 04Modulares Potenzieren54^16 mod 55 =
2^269 mod 19 =
daher
oder mittels Algorithmus (Square and Multiply)
26910 = 1000011012
x = 1 mod 19 1
1 x = 2*1 mod 19 2
0 x = 2^2 mod 19 4
0 x = 4^2 mod 19 16 (-3)
0 x = 16^2 mod 19 9 (auch -3^2 mod 19 = 9)
0 x = 9^2 mod 19 5
1 x = 2*5^2 mod 19 12 (-7)
1 x = 2*-7^2 mod 19 3
0 x = 3^2 mod 19 9
1 x = 2*9^2 mod 19 10
Lösung = 10
3^333 mod 15 =
daher
Kontrolle mit DERIVE:
|