2012 AMC 12A Problems/Problem 20

Problem

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 $a$ ?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24$

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 $6 \rightarrow \boxed{B}$ .

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2012 AMC 12A Problems/Problem 20

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