\(A=3^{2022}-2^{2022}+3^{2020}-2^{2020}\\=(3^{2022}+3^{2020})-(2^{2022}+2^{2020})\\=3^{2020}\cdot(3^2+1)-2^{2020}\cdot(2^2+1)\\=3^{2020}\cdot10-2^{2019}\cdot2\cdot5\\=3^{2020}\cdot10-2^{2019}\cdot10\)

Ta có: \(\left\{{}\begin{matrix}3^{2020}\cdot10⋮10\\2^{2019}\cdot10⋮10\end{matrix}\right.\)

\(\Rightarrow3^{2020}\cdot10-2^{2019}\cdot10⋮10\)

hay \(A⋮10\) (đpcm)

\(\text{#}Toru\)

`#3107.101107`

\(A = 2 + 2^2 + 2^3 + ... + 2^{2020} + 2^{2021} + 2^{2022}\)

\(= (2 + 2^2) + (2^3 + 2^4) + ... + (2^{2021} + 2^{2022})\)

\(=2(1+2) + 2^3(1 + 2) + ... + 2^{2021}(1 + 2)\)

\(=(1 + 2)(2 + 2^3 + ... + 2^{2021})\)

\(= 3(2 + 2^3 + ... + 2^{2021})\)

Vì \(3(2 + 2^3 + ... + 2^{2021})\) \(\vdots\) \(3\)

`\Rightarrow A \vdots 3`

Vậy, `A \vdots 3.`

A=(1+3+3²)+(3³+3⁴+3⁵)+...+(3²⁰¹⁹+3²⁰²⁰+3²⁰²¹)                                                  A=(1+3+3²)+3³.(1+3+3²)+...+3²⁰¹⁹.(1+3+3²)

A=13+3³.13+...+3²⁰¹⁹.13

A=13.(1+3³+...+3²⁰¹⁹)chia hết cho 13

=>A  chia hết cho 13

  S= 5+5²+5³+...+5²⁰²⁰+5²⁰²¹

 5S=5²+5³+5⁴+...+5²⁰²¹+5²⁰²²

 5S - S=4S=5²⁰²²-5

  Ta có: 4S+5=5²⁰²²

             =4S -5 +5 =5²⁰²²

              => 4S=5²⁰²²

\(C=4+4^2+4^3+...+4^{2021}+4^{2022}\)

\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)

\(=4.\left(1+4\right)+4^3.\left(1+4\right)+...+4^{2021}.\left(1+4\right)\)

\(=4.5+4^3.5+...+4^{2021}.5\)

\(=5.\left(4+4^3+...+4^{2021}\right)⋮5\)

Vậy \(C⋮5\)

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87476845+9

1000 chia hết cho 8 suy ra 10^3 chia hết cho 8

Suy ra:10^25x10^3 chia hết cho 8

và 8 chia hết cho 8

Suy ra 10^28+8 chia hết cho8(1)

Lại có 10^28+8=1000.....08

Suy ra:10^28+8 chia hết 9(2)

Lại vì:ƯCLN(8;9)=1(3)

Suy ra 10^2022+8 chia hết cho 72 

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