Question

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

$x^{2012}$ 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

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

解答：

其他年份的第 20 题