Find the remainders of the division: a) $19^{10}$ by 6; b) $19^{14}$ by 70; c) $17^{9}$ by 48; d) $14^{14^{14}}$ by 100.
a) $19^{10} \equiv 1(\bmod 2), 19^{10} \equiv 1^{10}=1(\bmod 3)$. Moreover, by Fermat's little theorem $19^{10} \equiv 1(\bmod 11)$.

b) $19^{14} \equiv 1(\bmod 2), 19^{14} \equiv(-1)^{14}=1(\bmod 5), 19^{14} \equiv 19^{2} \equiv(-2)^{2}=4(\bmod 7)$.

c) $17^{9} \equiv 1^{9}=1(\bmod 16), 17^{9} \equiv(-1)^{9}=-1(\bmod 3)$.

d) $14^{14} \equiv 0(\bmod 4)$.

$\varphi(25)=20,14^{14} \equiv 0(\bmod 4), 14^{14} \equiv(-1)^{14}=1(\bmod 5)$, so $14^{14} \equiv 16(\bmod 20)$.

$14^{14^{14}} \equiv 14^{16}=(196)^{8} \equiv(-4)^{8}=(16)^{4} \equiv(-9)^{4} \equiv(6)^{2}=36(\bmod 25)$.

## Answer

a) 1 ; b) 11 ; c) 17 ; d) 36.