A = (5+5^2+5^3)+(5^4+5^5+5^6)+...+(5^2020+5^2021+5^2022)

= 5(1+5+5^2)+5^4(1+5+5^2)+...+5^2020(1+5+5^2)

= 5.31+5^4.31+...+5^2020.31

= 31(5+5^4+...+5^2020) chia hết cho 31

\(A=5+5^2+5^3+5^4+...+5^{2022}\\ =\left(5+5^2+5^3+5^4+5^5+5^6\right)+....+5^{2016}\left(5+5^2+5^3+5^4+5^5+5^6\right)\\ =19530+....+5^{2016}.19530\\ =210.93+...+5^{2016}.210.93\\ =93.210.\left(1+...+5^{2016}\right)⋮93\left(ĐPCM\right)\)

Đặt A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰²²

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2²⁰²⁰ + 2²⁰²¹ + 2²⁰²²)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2²⁰²⁰.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2²⁰²⁰.7

= 7.(2 + 2⁴ + ... + 2²⁰²⁰) ⋮ 7

Vậy A ⋮ 7

\(A=8\left(1+8\right)+8^3\left(1+8\right)+...+8^{2021}\left(1+8\right)\)

\(=8.9+8^3.9+...+8^{2021}.9=9\left(8+8^3+...+8^{2021}\right)⋮9\)

Em xem lại đề nhé! Có xuất hiện dấu + không? Hay chỉ là dấu x

À em gấp quá nên ghi nhầm + thành x

\(B=2021\cdot1\cdot2\cdot3\cdot...\cdot2022\cdot\left(1+\dfrac{1}{2}+...+\dfrac{1}{2022}\right)⋮2021\)

Tham khảo

\(\text{+)}\)Ta có:\(5\equiv-1\left(mod3\right)\)

\(\Rightarrow5^{2022}\equiv\left(-1\right)^{2022}\left(mod3\right)\left(1\right)\)

\(\text{+)}\)Ta có:\(2\equiv-1\left(mod3\right)\)

\(\Rightarrow2^{2023}\equiv\left(-1\right)^{2023}\left(mod3\right)\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow5^{2022}+5^{2023}\equiv0\left(mod3\right)\)

Vậy...