Use Fermat's Little Theorem to �find the remainder on division 5^120 by 19.

Answer 1

According to Fermat's Little Theorem:

5^18 = 1 Mod 19

This means that in the exponent you can reduce Mod 18.

120 Mod 18 = 20*6 Mod 18 = 2*6 Mod 18 = 12 Mod 18

5^3 Mod 19 = 125 Mod 19 =

(20*6+5) Mod 19 = 6+5 = 11

5^6 Mod 19 = 11^2 = (6*20+1) Mod 19 = 7

5^12 Mod 19 = 7^2 Mod 19 = 49 Mod 19 = 11