Difference between revisions of "2012 AMC 12A Problems/Problem 20"
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== Solution == | == 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 <math>x</math> or a power of 2 from each factor. | + | 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. |
| − | 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>. | + | 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>. |
| − | 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. | + | 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>. |
| − | 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>. | + | 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>. |
| − | Thus, the coefficient of the <math>x^{2012}</math> term is <math>32 \times 2 \times 1 = 64 = 2^6</math>. So | + | 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>. |
{{AMC12 box|year=2012|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2012|ab=A|num-b=19|num-a=21}} | ||
Revision as of 20:48, 20 February 2012
Problem
Consider the polynomial
![\[P(x)=\prod_{k=0}^{10}=(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\]](http://latex.artofproblemsolving.com/9/e/c/9ec81d75d7273f6b19ccee6b2771dc7f41a787a6.png)
The coefficient of 
is equal to 
. What is 
?
![\[\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24\]](http://latex.artofproblemsolving.com/f/f/0/ff00a33605701fb3b8adff21802c46d01211a2ed.png)
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 
or a power of 
from each factor.
Every number, including 
, 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. 
, meaning 
.
Thus, the term was made by multiplying 
from the 
factor, 
from the 
factor, and so on. The only numbers not used are 
, , and 
.
Thus, from the 
factors, , , and were chosen as opposed to 
, and .
Thus, the coefficient of the term is 
. So the answer is 
.
| 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 | |
