AMC10 2020 B

AMC10 2020 B · Q22

AMC10 2020 B · Q22. It mainly tests Remainders & modular arithmetic, Powers & residues.

What is the remainder when $2^{202} + 202$ is divided by $2^{101} + 2^{51} + 1$?

求 $2^{202} + 202$ 除以 $2^{101} + 2^{51} + 1$ 的余数。

(A) 100 100

(B) 101 101

(D) 201 201

(E) 202 202

Answer

Correct choice: (D)

正确答案：（D）

Solution

Observe that $$\begin{align*}2^{202} + 202 &= (2^{101})^2 + 2 \cdot 2^{101} + 1 - 2 \cdot 2^{101} + 201 \\&= (2^{101} + 1)^2 - 2^{102} + 201 \\&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.\end{align*}$$ Therefore the remainder is $201$.

注意到 $$\begin{align*}2^{202} + 202 &= (2^{101})^2 + 2 \cdot 2^{101} + 1 - 2 \cdot 2^{101} + 201 \\&= (2^{101} + 1)^2 - 2^{102} + 201 \\&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.\end{align*}$$ 因此余数为 $201$。

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