TITLE: The remainder of the division of $2^{100}$ by $11$ is $1$?
QUESTION [5 upvotes]: $$2^{10}\equiv 1\;\text{mod}\;11\Longrightarrow(2^{10})^{10}\equiv1^{10}\;\text{mod}\;11\Longrightarrow2^{100}\equiv1\;\text{mod}\;11\;\;?$$$$$$Soon, the rest will be $1$, correct?

REPLY [4 votes]: Your computation is absolutely fine and most likely the shortest way. However, when writing this down, you should remark that you are applying Fermat's little theorem in the beginning.
In case you don't know Fermat's little theorem or weren't applying it but rather calculated $2^{10}\bmod 11$ by hand, this is slightly easier:
$$2^{100}=(2^{5})^{20}=32^{20}\equiv (-1)^{20}=1\pmod{11}$$