Ta có :

\(A=\left(1+7+7^2\right)+\left(7^3+7^4+7^5\right)+...+\left(7^{2018}+7^{2019}+7^{2020}\right)\)

\(=\left(1+7+7^2\right)+7^3\left(1+7+7^2\right)+...+7^{2018}\left(1+7+7^2\right)\)

\(=\left(1+7+7^2\right)\left(1+7^3+7^6+...+7^{2018}\right)\)

\(=57\cdot\left(1+7^3+7^6+...+7^{2018}\right)\)

\(=19\cdot3\cdot\left(1+7^3+7^6+...+7^{2018}\right)⋮19\) (đpcm)

\(A=1+7+7^2+7^3+...+7^{2019}+7^{2020}\)

\(\Leftrightarrow A=\left(1+7+7^2\right)+\left(7^3+7^4+7^5\right)+....+\left(7^{2018}+7^{2019}+7^{2020}\right)\)

\(\Leftrightarrow A=\left(1+7+49\right)+7^3\left(1+7+49\right)+...+7^{2018}\left(1+7+49\right)\)

\(\Leftrightarrow A=57+7^3\cdot57+...+7^{2018}\cdot57\)

\(\Leftrightarrow A=57\left(1+7^3+....+7^{2018}\right)\)

\(\Leftrightarrow A=3\cdot19\left(1+7^3+...+7^{2018}\right)\)

=> A chia 19 dư 0

Ta có:A=1+(7+7²+7³)+...+(7²⁰¹⁸+7²⁰¹⁹+7²⁰²⁰)

         A=1+[7(1+7+49)+...+7²⁰¹⁸(1+7+49)]

         A=1+[57(7+7²+...+7²⁰¹⁸)]

         Do 57(7+7²+...+7²⁰¹⁸) chia hết cho 19 nên 1+[57(7+7²+..+7²⁰¹⁸)] chia 19 dư 1

\(A=1+2+2^2+...+2^{2020}+2^{2021}+2^{2023}\)

\(A=1+2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2020}\left(1+2+2^2\right)-2^{2022}+2^{2023}\)

\(A=1+2.7+2^4.7+...+2^{2020}.7-2^{2022}+2^{2023}\)

\(A=7\left(2+2^4+...+2^{2020}\right)+\left(2^{2022}+1\right)\left(1\right)\)

Ta có :

\(2^3=8\equiv1\) (mod 7)

\(\Rightarrow\left(2^3\right)^{674}\equiv1^{674}=1\) (mod 7)

\(\Rightarrow2^{2022}\equiv1\) (mod 7)

\(\Rightarrow2^{2022}+1\equiv1+1=2\)  (mod 7)

\(\Rightarrow2^{2022}+1\equiv2\) (mod 7)

mà \(7\left(2+2^4+...+2^{2020}\right)⋮7\)

\(\left(1\right)\Rightarrow A=7\left(2+2^4+...+2^{2020}\right)+\left(2^{2022}+1\right)\equiv2\) (mod 7)

Vậy số dư của A khi chia cho 7 là 2

\(A=1+2+2^2+...+2^{2019}+2^{2020}\)

\(A=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{2018}+2^{2019}+2^{2020}\right)\)

\(A=3+2^2\left(1+2+2^2\right)+...+2^{2018}\left(1+2+2^2\right)\)

\(A=3+2^2.7+....+2^{2018}.7\)

\(A=3+7\left(2^2+....+2^{2018}\right)\)

Vì 3 ko chia hết cho 7

=> A ko chia hết cho 7

=> A dư 3

\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)

\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)

\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)

7^1 + 7^2 + ... + 7^2013 
= ( 7^1 + 7^2 + 7^3 ) +.... + ( 7^2011 + 7^2012 + 7^2013 ) 
= 7^1 . ( 1 + 7 + 49 ) + .... + 7^2011( 1+ 7+ 49 ) 
= 7^1 . 57 + .... + 7^2011 . 57 
= 7^1 . 19 . 3 + ... + 7^2011 . 19 .3 
=> A chia cho 19 dư 0 
Tick nha

Lời giải:

Áp dụng định lý Fermat nhỏ thì:

$2020^6\equiv 1\pmod 7$

$\Rightarrow (2020^6)^{336}.2020^4\equiv 1^{336}.2020^4\equiv 2020^4\pmod 7$

Có:

$2020\equiv 4\pmod 7$

$\Rightarrow 2020^4\equiv 4^4\equiv 256\equiv 4\pmod 7$

$\Rightarrow A\equiv 2020^4\equiv 4\pmod 7$

Vậy $A$ chia $7$ dư $4$