AMC12 2012 A

AMC12 2012 A · Q20

AMC12 2012 A · Q20. It mainly tests Polynomials, Primes & prime factorization.

Consider the polynomial \[P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\] The coefficient of $x^{2012}$ is equal to $2^a$. What is $a$? \[

考虑多项式 \[P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\] $x^{2012}$ 的系数等于 $2^a$。$a$ 是多少？ \[

(A) 5 5

(B) 6 6

(D) 10 10

(E) 24 24

Answer

Correct choice: (B)

正确答案：（B）

Solution

Every term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of $x$ or a power of $2$ from each factor. Every number, including $2012$, has a unique representation by the sum of powers of two, and that representation can be found by converting a number to its binary form. $2012 = 11111011100_2$, meaning $2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4$. Thus, the $x^{2012}$ term was made by multiplying $x^{1024}$ from the $(x^{1024} + 1024)$ factor, $x^{512}$ from the $(x^{512} + 512)$ factor, and so on. The only numbers not used are $32$, $2$, and $1$. Thus, from the $(x^{32} + 32), (x^2+2), (x+1)$ factors, $32$, $2$, and $1$ were chosen as opposed to $x^{32}, x^2$, and $x$. Thus, the coefficient of the $x^{2012}$ term is $32 \times 2 \times 1 = 64 = 2^6$. So the answer is $6 \rightarrow \boxed{B}$.

将该乘积展开时，每一项都是从每个因子中各取一项相乘得到的。因此，我们从每个因子中要么选取一个 $x$ 的幂，要么选取一个 $2$ 的幂。每个数（包括 $2012$）都能唯一表示为若干个 2 的幂之和，这可以通过写成二进制得到。$2012 = 11111011100_2$，即 $2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4$。因此，$x^{2012}$ 这一项是通过从 $(x^{1024}+1024)$ 中取 $x^{1024}$，从 $(x^{512}+512)$ 中取 $x^{512}$，依此类推得到的。唯一没有用到的幂是 $32$、$2$ 和 $1$。所以在 $(x^{32}+32)$、$(x^2+2)$、$(x+1)$ 这三个因子中，分别取 $32$、$2$、$1$，而不是取 $x^{32}$、$x^2$、$x$。因此 $x^{2012}$ 的系数为 $32\times 2\times 1=64=2^6$，所以 $a=6 \rightarrow \boxed{B}$。

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