Question

[22 Đinh Văn Thành]

cho biểu thức a=1+2+2 mũ 2+.......+2 mũ 2021 + 2 mũ 2022 a, tìm số dư khi a chia cho 21

Answer 1

[Akai Haruma]

Lời giải:

$A=1+2+2^2+...+2^{2021}+2^{2022}$

$2A=2+2^2+2^3+...+2^{2022}+2^{2023}$
$2A-A=2+2^2+2^3+...+2^{2022}+2^{2023}-(1+2+2^2+...+2^{2021}+2^{2022})$

$\Rightarrow A=2^{2023}-1$
Ta thấy:

$2\equiv -1\pmod 3\Rightarrow A=2^{2023}-1\equiv (-1)^{2023}-1\equiv 1\pmod 3(1)$
Mặt khác:

$2^3\equiv 1\pmod 7\Rightarrow 2^{2023}=(2^3)^{674}.2\equiv 1^{674}.2\equiv 2\pmod 7$

$\Rightarrow A=2^{2023}-1\equiv 2-1\equiv 1\pmod 7(2)$

Từ $(1); (2)$ mà $(3,7)=1$ nên $A\equiv 1\pmod {3.7}$ hay $A\equiv 1\pmod {21}$

Vậy $A$ chia $21$ dư $1$

Answer 2

[Đặng Hoàng Lâm]

Ta có: \(A=2^0+2^1+2^2+...+2^{2021}+2^{2022}\)

\(=>2A=2^1+2^2+2^3+...+2^{2022}+2^{2023}\)

\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}+2^{2023}\right)-\left(2^0+2^1+2^2+...+2^{2021}+2^{2022}\right)\)

\(A=2^{2023}-2^0=2^{2023}-1\)

Ta lại có: \(2^6=64\equiv1\left(mod21\right),2^{2023}=\left(2^6\right)^{337}.2\equiv1^{337}.2=1.2=2\left(mod21\right) =>2^{2023}-1=2-1=1\left(mod21\right)\)

=> A chia 21 dư 1. Vậy A chia 21 dư 1