2012 AMC 12A 第 20 题

Question

考虑多项式

\begin{aligned} P(x) &= \prod_{k=0}^{10}\left(x^{2^k} + 2^k\right) \\ &= (x+1)(x^2+2)(x^4+4) \\ &\quad \cdots (x^{1024}+1024) \end{aligned}\text{。}

x^{2012}

的系数等于

2^a

。求

a

？

Consider the polynomial

\begin{aligned} P(x) &= \prod_{k=0}^{10}\left(x^{2^k} + 2^k\right) \\ &= (x+1)(x^2+2)(x^4+4) \\ &\quad \cdots (x^{1024}+1024). \end{aligned}

The coefficient of

x^{2012}

is equal to

2^a.

What is

a?

5

6

7

10

24

答案：B

知识点：多项式 2的幂进制

难度评级：2220

小提示：

每个因子贡献

x^{2^k}

项或常数项

2^k

；被选中的指数之和必须为

2012

Each factor contributes either its

x^{2^k}

term or its constant

2^k

; the chosen exponents must sum to

2012

大提示：

将

2012

写成若干个不同的二的幂之和只有一种方式，所以只有一项会产生

x^{2012}

。

There is only one way to write

2012

as a sum of distinct powers of two, so exactly one term produces

x^{2012}

解答：

展开乘积时，次数为

2012

的项来自某些因子中选取

x^{2^k}

，使指数和为

2012

。由于二的幂互不相同，这对应于二进制表示

2012 = 11111011100_2

。

这个表示唯一，所以恰好只有一项给出

x^{2012}

，它的系数是其余因子的常数项

2^k

的乘积：这些因子满足

k \in \{0, 1, 5\}

。

系数为

2^0 \cdot 2^1 \cdot 2^5 = 2^6

，所以

a = 6

。

因此，正确答案是 B。

Expanding the product, a term of degree

2012

comes from choosing

x^{2^k}

from some factors so that the exponents sum to

2012.

Since powers of two are distinct, this corresponds to the binary representation

2012 = 11111011100_2.

That representation is unique, so exactly one term gives

x^{2012},

and its coefficient is the product of the constants

2^k

from the remaining factors: those with

k \in \{0, 1, 5\}.

The coefficient is

2^0 \cdot 2^1 \cdot 2^5 = 2^6,

so

a = 6.

Thus, the correct answer is B.

2012 AMC 12A 第 20 题

答案：B

难度评级：2220

小提示：

大提示：

解答：

其他年份的第 20 题