2020 AMC 10B Problem 22

Let $m=2^{101}+2^{51}+1.$ We factor the numerator around this divisor: $$\begin{aligned} &2^{202}+202 \\ &\quad =(2^{101}+1)^2-(2^{51})^2+201. \end{aligned}$$ By the difference of squares, $$\begin{aligned} &(2^{101}+1)^2-(2^{51})^2 \\ &\quad =(2^{101}+2^{51}+1) \\ &\quad {}\cdot(2^{101}-2^{51}+1), \end{aligned}$$ which is a multiple of $m.$ Therefore $$2^{202}+202\equiv 201\pmod m.$$ Thus, the correct answer is D .