What is the order of 2 modulo 101?
According to Euler's theorem, the order of 2 modulo 101 divides $100=2^{2} \cdot 5^{2}$. Let's calculate $2^{20}$ and $2^{50}$ modulo 101. To this end, note that $2^{10}=1024 \equiv 14 \bmod 101$. Thus,

$$
2^{20} \equiv 14^{2}=196 \equiv-6 \quad \bmod 101
$$

and

$$
2^{50} \equiv(-6)^{2} \cdot 14=504 \equiv-1 \quad \bmod 101
$$

Therefore, the order of 2 modulo 101 is 100.