2021 AMC 12B Fall Problem 18

Attempt Problem 18 of the 2021 AMC 12B Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

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18.

Set

u_0 = \dfrac{1}{4},

and for

k \ge 0

let

u_{k+1}

be determined by the recurrence

u_{k+1} = 2u_k - 2u_k^2.

This sequence tends to a limit; call it

L.

What is the least value of

k

such that

|u_k - L| \le \dfrac{1}{2^{1000}}?

10

87

123

329

401

The limit is

L = \tfrac12

; substitute

v_k = 1 - 2u_k

The substitution gives

v_{k+1} = v_k^2,

v_k

squares repeatedly

The limit satisfies

L = 2L - 2L^2,

giving

L = \tfrac12.

Let

v_k = 1 - 2u_k.

Then

\begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2. \end{aligned}

Since

v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12,

we get

v_k = \left(\tfrac12\right)^{2^k},

|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}.

We need

2^k + 1 \ge 1000,

i.e.

2^k \ge 999.

The least such

k

10,

since

2^{10} = 1024.

Thus, the correct answer is A.

Problem 18 in other years