Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 2)
 
Line 20: Line 20:
  
 
== Solution 2 ==
 
== 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 powers of <math>2</math> that add up to <math>2047-2012=35</math>. Because <math>35</math> is odd, we know that one of these must be <math>2^0</math>. Then, we can test all cases (there are very few of them) and we find that only <math>2^0+2^1+2^5</math> works. 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>
+
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>
  
 
==See Also==
 
==See Also==

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12A_Problems/Problem_20&oldid=101296"