Congruences doubt!

Question

What the rest of the division $2^{100}$ by $11$?

$$2^5=32\equiv10\equiv-1\pmod{11}\\(2^5)^{20}=2^{100}\equiv-1^{20}\;\text{or}\; (-1)^{20}$$??

Answer 1

We have \begin{align*} 2^{10} &= 2^5 \cdot 2^5 &\equiv (-1) \cdot (-1) &= (-1)^2 \pmod{11}\\ &\vdots\\ 2^{100} &=2^5 \cdots 2^5 &\equiv (-1) \cdots (-1) &= (-1)^{20} \pmod{11} \end{align*}

Answer 2

Since $(2,11)=1$, then by Fermat's Little Theorem $$2^{10}\equiv1\pmod{11}\\(2^{10})^{10}=2^{100}\equiv1\pmod {11}$$

Answer 3

Shorter, perhaps:

$$2^5=32=-1\pmod {11}\implies 2^{100}=(2^5)^{20}=(-1)^{20}=1\pmod {11}$$