2012 AMC 12A Problems/Problem 20

Revision as of 22:17, 9 February 2016 by WhaleVomit (talk | contribs) (Solution)

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 1

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 $6 \rightarrow \boxed{B}$.

Solution 2

The degree of $P(x)$ is $1024+512+256+\cdots+1=2047$. We want to find the coefficient of $x^{2012}$, so we need to omit powers of $2$ that add up to $2047-2012=35$. Because $35$ is odd, we know that one of these must be $2^0$. Then, we can test all cases (there are very few of them) and we find that only $2^0+2^1+2^5$ works. From here, we know that the answer is $2^0\cdot2^1\cdot2^5=2^6$. Therefore, the answer is $\boxed{(B)\:6.}$

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions

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