2022 AMC 12B 第 23 题

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23.

x0x_0x1x_1x2x_2\ldots 是一个数列,其中每个 xkx_k 都是 0011。对每个正整数 nn,定义 Sn=k=0n1xk2kS_n = \sum_{k=0}^{n-1} x_k 2^k\text{。} 假设对所有 n1n \ge 1 都有 7Sn1(mod2n)7 S_n \equiv 1 \pmod{2^n}。求下面这个和的值:x2019+2x2020+4x2021+8x2022x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}\text{?}

Let x0,x_0, x1,x_1, x2,x_2, \ldots be a sequence of numbers, where each xkx_k is either 00 or 1.1. For each positive integer n,n, define Sn=k=0n1xk2k.S_n = \sum_{k=0}^{n-1} x_k 2^k. Suppose 7Sn1(mod2n)7 S_n \equiv 1 \pmod{2^n} for all n1.n \ge 1. What is the value of the sum x2019+2x2020+4x2021+8x2022?x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?

66

77

1212

1414

1515

答案:A
知识点:模运算进制找规律
难度评级:2270
小提示:

SnS_n 是低 nn 位二元数;条件说明 Sn71(mod2n)S_n \equiv 7^{-1} \pmod{2^n}

SnS_n is the number formed by the low nn bits; the condition says Sn71(mod2n)S_n \equiv 7^{-1} \pmod{2^n}

大提示:

因此 xkx_k17\tfrac17 的二进制(即二进数)数字;找出其循环模式。

So the bits xkx_k are the binary (2-adic) digits of 17;\tfrac17; find their repeating pattern

解答:

因为 SnS_n 是由低 nn 位构成的整数,条件 7Sn17S_n \equiv 1 表示 Sn71(mod2n)S_n \equiv 7^{-1} \pmod{2^n} 对每个 nn 都成立。因此 xkx_k 是以 22 为底表示 17\tfrac17 时的数位,也就是它作为 22-进数时的数位。

22 进制长除可得 x0=x1=x2=1x_0 = x_1 = x_2 = 1,之后以 33 为周期重复:当 k1k \ge 1 时,恰在 kk33 的倍数时有 xk=0x_k = 0,否则 xk=1x_k = 1

因为 2019201920222022 都是 33 的倍数,而 202012020 \equiv 120212(mod3)2021 \equiv 2 \pmod 3,所以 x2019=0x_{2019} = 0x2020=1x_{2020} = 1x2021=1x_{2021} = 1x2022=0x_{2022} = 0。所求和为 0+2+4+0=60 + 2 + 4 + 0 = 6

所以正确答案是 A

Since SnS_n is the integer formed by the low nn bits, the condition 7Sn17S_n \equiv 1 means Sn71(mod2n)S_n \equiv 7^{-1} \pmod{2^n} for every n.n. Thus the digits xkx_k are the base-22 digits of 17\tfrac17 as a 22-adic number.

Long division in base 22 gives digits x0=x1=x2=1,x_0 = x_1 = x_2 = 1, and thereafter the block repeats with period 3:3: for k1,k \ge 1, xk=0x_k = 0 exactly when kk is divisible by 3,3, and xk=1x_k = 1 otherwise.

Since 20192019 and 20222022 are both divisible by 3,3, while 202012020 \equiv 1 and 20212(mod3),2021 \equiv 2 \pmod 3, we get x2019=0,x_{2019} = 0, x2020=1,x_{2020} = 1, x2021=1,x_{2021} = 1, x2022=0.x_{2022} = 0. The sum is 0+2+4+0=6.0 + 2 + 4 + 0 = 6.

Thus, the correct answer is A.

第 22 题#22
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