262hmm.. got it now.
didnt know this theorem .
3 Answers
1By Fermat's Theorem
Since (a,p) = 1
a^{p-1}\equiv a \: \textup{mod}\: p
\Rightarrow 2^{100} \equiv 1\: mod \: 101
\Rightarrow 2^{200} \equiv 1^{2} \equiv 1 \: mod \: 101
\Rightarrow 2^{202} \equiv 4 \: mod \: 101
Hence remainder is 4 :)