2020 AMC 10B 第 22 题

Question

2^{202} +202

除以

2^{101}+2^{51}+1

的余数是多少？

What is the remainder when

2^{202} +202

is divided by

2^{101}+2^{51}+1?

100

101

200

201

202

尝试用平方差来分解分子。

Try to factor the numerator using a difference of squares

把

2^{202}+202

写成除数的倍数加上一个小余数。

Write

2^{202}+202

as a multiple of the divisor plus a small remainder

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令

m=2^{101}+2^{51}+1

。围绕除数分解：

\begin{aligned} &2^{202}+202 \\ &\quad =(2^{101}+1)^2-(2^{51})^2+201\text{。} \end{aligned}

利用平方差，

\begin{aligned} &(2^{101}+1)^2-(2^{51})^2 \\ &\quad =(2^{101}+2^{51}+1) \\ &\quad {}\cdot(2^{101}-2^{51}+1)\text{，} \end{aligned}

这是

m

的倍数。因此

2^{202}+202\equiv 201\pmod m\text{。}

所以正确答案是 D。

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 .

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