Difference between revisions of "2012 AMC 12A Problems/Problem 20"
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Consider the polynomial | Consider the polynomial | ||
− | <cmath>P(x)=\prod_{k=0}^{10} | + | <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> |
The coefficient of <math>x^{2012}</math> is equal to <math>2^a</math>. What is <math>a</math>? | The coefficient of <math>x^{2012}</math> is equal to <math>2^a</math>. What is <math>a</math>? | ||
Line 8: | Line 8: | ||
<cmath>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24 </cmath> | <cmath>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24 </cmath> | ||
− | == Solution == | + | == 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 <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>. |
+ | |||
+ | == 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> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC12 box|year=2012|ab=A|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:59, 5 February 2019
Contents
Problem
Consider the polynomial
The coefficient of is equal to . What is ?
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 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 .
Solution 2
The degree of is . We want to find the coefficient of , so we need to omit the powers of that add up to . We find that . From here, we know that the answer is . Therefore, the answer is
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.