Question

Sea $x_0,x_1,x_2,\dotsc$ una sucesión de números, donde cada $x_k$ es $0$ o $1.$ Para cada entero positivo $n,$ define $S_n = \sum_{k=0}^{n-1} x_k 2^k$ Supón que $7S_n \equiv 1 \pmod{2^n}$ para todo $n \geq 1.$ ¿Cuál es el valor de la suma $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}$ ?

Let $x_0,x_1,x_2,\dotsc$ 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 $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1.$ What is the value of the sum $x_{2019} + 2x_{2020} +$ $4x_{2021} + 8x_{2022}?$

$6$

$7$

$12$

$14$

$15$

Solución:

Nota primero que $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$ $2x_{2020} + 4x_{2021} + 8x_{2022}.$ Por lo tanto, debemos intentar encontrar $S_{2019}, S_{2023}.$ Además, nota que $0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$

Ahora, como $7S_n \equiv 1 \pmod{2^n},$ sabemos que $7S_n = m2^n +1 \implies$ $S_n = \dfrac {m2^n+1}{7}$ para algún entero $m.$ Además, como $0 \leq S_n < 2^n,$ sabemos que $0 \leq \dfrac {m2^n+1}7 < 2^n.$ Esto significa que $0 \leq m < 7.$ Ahora, buscamos $m$ tal que $m2^n + 1$ sea divisible por $7.$ Esto hace que $m2^n + 1 \equiv 0 \mod 7.$

Si $n=2019,$ entonces $0 \equiv m2^{2019} + 1 \equiv$ $m(2^3)^{673}+1 \equiv$ $m8^{673} + 1\equiv m+1,$ así que $m=6.$ Esto hace que $S_{2019} = \dfrac {6(2^{2019})+1}{7}.$

Si $n=2023,$ entonces $0 \equiv m2^{2023} + 1$ $\equiv m(2^3)^{674}\cdot 2+1$ $\equiv m8^{674}\cdot 2 + 1$ $\equiv 2m+1,$ así que $m=3.$ Esto hace que $S_{2023} = \dfrac {3(2^{2023})+1}{7}.$

Nuestra respuesta es $\begin{gathered} \frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{1}{2^{2019}} \\ {}\cdot \small \left(\frac{3\cdot2^{2023}+1}{7}-\frac{6\cdot2^{2019}+1}{7}\right) \\ = \frac{42\cdot2^{2019}}{7\cdot2^{2019}} = 6. \end{gathered}$

Así, la respuesta correcta es A.

Note first that $\dfrac{ S_{2023} - S_{2019}}{2^{2019}} = x_{2019} +$ $2x_{2020} + 4x_{2021} + 8x_{2022}.$ Therefore, we should attempt to find $S_{2019}, S_{2023}.$ Also, note that $0 \leq S_n \leq \sum_{k=0}^{n-1} 2^k < 2^n.$

Now, since $7S_n \equiv 1 \pmod{2^n},$ we know $7S_n = m2^n +1 \implies$ $S_n = \dfrac {m2^n+1}{7}$ for some integer $m.$ Also, since $0 \leq S_n < 2^n,$ we know $0 \leq \dfrac {m2^n+1}7 < 2^n.$ This means $0 \leq m < 7.$ Now, we find $m$ such that $m2^n + 1$ is divisible by $7.$ This makes $m2^n + 1 \equiv 0 \mod 7.$

If $n=2019,$ then $0 \equiv m2^{2019} + 1 \equiv$ $m(2^3)^{673}+1 \equiv$ $m8^{673} + 1\equiv m+1,$ so $m=6.$ This makes $S_{2019} = \dfrac {6(2^{2019})+1}{7}.$

If $n=2023,$ then $0 \equiv m2^{2023} + 1$ $\equiv m(2^3)^{674}\cdot 2+1$ $\equiv m8^{674}\cdot 2 + 1$ $\equiv 2m+1,$ so $m=3.$ This makes $S_{2023} = \dfrac {3(2^{2023})+1}{7}.$

Our answer is $\begin{gathered} \frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{1}{2^{2019}} \\ {}\cdot \small \left(\frac{3\cdot2^{2023}+1}{7}-\frac{6\cdot2^{2019}+1}{7}\right) \\ = \frac{42\cdot2^{2019}}{7\cdot2^{2019}} = 6. \end{gathered}$

Thus, the correct answer is A .

2022 AMC 10B Problema 25

Solución:

El Problema 25 en otros años