Difference between revisions of "2012 AMC 12A Problems/Problem 20"

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Every term in the polynomial is formed from the multiplication of one thing from every term, either x to some power or an integer.
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== Problem ==
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Consider the polynomial
  
Every number, including 2012, has a unique representation by the sum of powers of two, and that representation can be showed by converting a number to its binary form. 2012 base 10 = 11111011100 base 2, meaning 2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4.
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<cmath>P(x)=\prod_{k=0}^{10}=(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)</cmath>
  
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.
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The coefficient of <math>x^{2012}</math> is equal to <math>2^a</math>. What is <math>a</math>?
  
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.
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<cmath>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24 </cmath>
  
Thus, the coefficient of the x^2012 term is 32 * 2 * 1 = 64 = 2^6. So, 6 is the right answer, and choice B was 6.
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== Solution ==
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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 <math>x</math> or a power of 2 from each factor.
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 +
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. <math>2012 = 11111011100_2</math>, meaning <math>2012 = 1024 + 512 + 256 + 128 + 64 + 16 + 8 + 4</math>.
 +
 
 +
Thus, the <math>x^{2012}</math> term was made by multiplying <math>x^1024</math> from the <math>(x^{1024} + 1024)</math> factor, <math>x^{512}</math> from the <math>(x^{512} + 512)</math> factor, and so on. The only numbers not used are 32, 2, and 1.
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Thus, from the <math>(x^{32} + 32), (x^2+2), (x+1)</math> factors, 32, 2, and 1 were chosen as opposed to <math>x^{32}, x^2</math>, and <math>x</math>.
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Thus, the coefficient of the <math>x^{2012}</math> term is <math>32 \times 2 \times 1 = 64 = 2^6</math>. So, 6 is the right answer, or B.

Revision as of 17:02, 16 February 2012

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, 6 is the right answer, or B.

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