Question

Let $x_0, x_1, x_2, \ldots$ be a sequence of numbers, where each $x_k$ is either $0$ or $1.$ For each positive integer $n,$ define $S_n = \sum_{k=0}^{n-1} x_k 2^k.$ Suppose $7 S_n \equiv 1 \pmod{2^n}$ for all $n \ge 1.$ What is the value of the sum $x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

Solution:

Since $S_n$ is the integer formed by the low $n$ bits, the condition $7S_n \equiv 1$ means $S_n \equiv 7^{-1} \pmod{2^n}$ for every $n.$ Thus the digits $x_k$ are the base- $2$ digits of $\tfrac17$ as a $2$ -adic number.

Long division in base $2$ gives digits $x_0 = x_1 = x_2 = 1,$ and thereafter the block repeats with period $3:$ for $k \ge 1,$ $x_k = 0$ exactly when $3 \mid k,$ and $x_k = 1$ otherwise.

Since $3 \mid 2019$ and $3 \mid 2022,$ while $2020 \equiv 1$ and $2021 \equiv 2 \pmod 3,$ we get $x_{2019} = 0,$ $x_{2020} = 1,$ $x_{2021} = 1,$ $x_{2022} = 0.$ The sum is $0 + 2 + 4 + 0 = 6.$

Thus, the correct answer is A.

2022 AMC 12B Problem 23

Solution:

Problem 23 in Other Years