what will bethe remainder when 4raised to power 101 is divided by 101.......?(power of 4 is 101)

akki ~~ unlucky forever ~~

$2^9\equiv7\ mod(101)\\ \\ \\ 2^{18}\equiv49\ mod(101)\\ \\ \\ 2^{19} \equiv 98\ mod(101),98 \equiv -3\ mod(101)\\ \\ \\ so,2^{19} \equiv -3\ mod(101)\\ \\ \\ 2^{190} \equiv 3^{10}\ mod(101)\\ \\ \\ 2^{12}.2^{190} \equiv 4.2^{10}.243^{2}\ mod(101)\\ \\ \\ 2^{10} \equiv 14\ mod(101),243^2 \equiv (41)^2 \ mod(101),41^2 \equiv 65\ mod(101)\\ \\ \\ => 2^{202} \equiv 4.2^{10}.3^{10}\ mod(101),4.2^{10}.3^{10} \equiv 4.14.65\ mod(101)\\ \\ \\ 4^{101} \equiv 3640\ mod(101) \equiv 4\ mod(101)\\ \\ \\ 4^{101} \equiv 4\ mod(101)\\ \\ \\ hence\ the\ remainder\ left\ is\ 4$

But, try to remember Fermat's Little Theorem, it's of good help.