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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
  
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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>
  
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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>?
  
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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>
  
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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 1==
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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 <math>2</math> from each factor.
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Every number, including <math>2012</math>, 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>.
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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 <math>32</math>, <math>2</math>, and <math>1</math>.
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Thus, from the <math>(x^{32} + 32), (x^2+2), (x+1)</math> factors, <math>32</math>, <math>2</math>, and <math>1</math> 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 the answer is <math>6 \rightarrow \boxed{B}</math>.
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== Solution 2 ==
 +
The degree of <math>P(x)</math> is <math>1024+512+256+\cdots+1=2047</math>. We want to find the coefficient of <math>x^{2012}</math>, so we need to omit the powers of <math>2</math> that add up to <math>2047-2012=35</math>. We find that <math>35=2^0+2^1+2^5</math>. From here, we know that the answer is <math>2^0\cdot2^1\cdot2^5=2^6</math>. Therefore, the answer is <math>\boxed{(B)\:6.}</math>
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==See Also==
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{{AMC12 box|year=2012|ab=A|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Latest revision as of 19:59, 5 February 2019

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 the powers of $2$ that add up to $2047-2012=35$. We find that $35=2^0+2^1+2^5$. 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 (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
All AMC 12 Problems and Solutions

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