2006 AMC 12A Problem 23

Below is the professionally curated solution for Problem 23 of the 2006 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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Concepts:binomial theoremrecursion

Difficulty rating: 2400

23.

Given a finite sequence S=(a1,a2,,an)S = (a_1, a_2, \ldots, a_n) of nn real numbers, let A(S)A(S) be the sequence (a1+a22,a2+a32,,an1+an2) \left(\frac{a_1 + a_2}{2}, \frac{a_2 + a_3}{2}, \ldots, \frac{a_{n-1} + a_n}{2}\right) of n1n - 1 real numbers. Define A1(S)=A(S)A^1(S) = A(S) and, for each integer m,m, 2mn1,2 \le m \le n - 1, define Am(S)=A(Am1(S)).A^m(S) = A(A^{m-1}(S)). Suppose x>0,x \gt 0, and let S=(1,x,x2,,x100).S = (1, x, x^2, \ldots, x^{100}). If A100(S)=(1/250),A^{100}(S) = (1/2^{50}), then what is x?x?

1221 - \dfrac{\sqrt{2}}{2}

21\sqrt{2} - 1

12\dfrac{1}{2}

222 - \sqrt{2}

22\dfrac{\sqrt{2}}{2}

Solution:

Each application of AA averages adjacent terms, so after 100100 steps the single remaining term is 12100m=0100(100m)xm=(1+x)1002100. \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m = \frac{(1 + x)^{100}}{2^{100}}.

Setting this equal to 1250\dfrac{1}{2^{50}} gives (1+x)100=250,(1 + x)^{100} = 2^{50}, so 1+x=21/2=2.1 + x = 2^{1/2} = \sqrt{2}. Since x>0,x \gt 0, we get x=21.x = \sqrt{2} - 1.

Thus, the correct answer is B.

Problem 23 in Other Years