A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰

   = (2 + 2²) + (2³ + 2⁴) + ... + (2¹⁹ + 2²⁰)

   = 6 + 2² . (1 + 2) + ... + 2¹⁸ . (1 + 2)

   = 6 + 2² .3 + ... + 2¹⁸ . 3 chia hết cho 3

\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)

a) \(A=10^{100}+5\)

- Tận cùng A là số 5 \(\Rightarrow A⋮5\)

- Tổng các chữ số của A là \(1+5=6⋮3\Rightarrow A⋮3\) \(\)

\(\Rightarrow dpcm\)

b) \(B=10^{50}+44\)

- Tận cùng B là số 4 là số chẵn \(\Rightarrow B⋮2\)

- Tổng các chữ số của B là \(1+4+4=9⋮9\Rightarrow B⋮9\)

\(\Rightarrow dpcm\)

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

tích từ bài từng câu a , b , ... ra đi

\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)

\(=13+3^3.13+...+3^{96}.13\)

\(=13\left(1+3^3+...+3^{96}\right)⋮13\)

\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ A=13\left(1+3^3+...+3^{96}\right)⋮13\)

a: \(A=3+3^2+3^3+\cdots+3^{60}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{58}\right)\)

=>A⋮13

b: \(M=2+2^2+\cdots+2^{20}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+\cdots+2^{17}\right)\)

=>M⋮5

A=2+2^2+2^3+2^4+...+2^100

  =(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^99+2^100)

  =6+(2^2.2+2^2.2^2)+(2^4.2+2^4.2^2)+...+(2^98.2+2^98.2^2)

  =6+2^2.(2+2^2)+2^4(2+2^2)+...+2^98.(2+2^2)

  =6.1.2^2.6+2^4.6+...+2^98.6

  =6.(2^2+2^4+...+2^98)

Vì \(6⋮6\)

\(\Rightarrow\)\(6.\left(2^2+2^4+...+2^{98}\right)⋮6\)

     Hay \(A⋮6\)