2012 AMC 12A Problem 20

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Concepts:polynomialpower of 2number base

Difficulty rating: 2220

20.

Consider the polynomial P(x)=k=010(x2k+2k)=(x+1)(x2+2)(x4+4)(x1024+1024).P(x) = \prod_{k=0}^{10}\left(x^{2^k} + 2^k\right) = (x+1)(x^2+2)(x^4+4)\cdots(x^{1024}+1024).

The coefficient of x2012x^{2012} is equal to 2a.2^a. What is a?a?

55

66

77

1010

2424

Solution:

Expanding the product, a term of degree 20122012 comes from choosing x2kx^{2^k} from some factors so that the exponents sum to 2012.2012. Since powers of two are distinct, this corresponds to the binary representation 2012=111110111002.2012 = 11111011100_2.

That representation is unique, so exactly one term gives x2012,x^{2012}, and its coefficient is the product of the constants 2k2^k from the remaining factors: those with k{0,1,5}.k \in \{0, 1, 5\}.

The coefficient is 202125=26,2^0 \cdot 2^1 \cdot 2^5 = 2^6, so a=6.a = 6.

Thus, the correct answer is B.

Problem 20 in Other Years