a: \(G=8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

\(H=2+2^2+2^3+...+2^{60}\)

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

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

\(E=1+3+3^2+3^3+...+3^{1991}\)

\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)

\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)

\(A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\\ A=2\left(1+2^2+2^3\right)+...+2^{118}\left(1+2^2+2^3\right)\\ A=\left(1+2^2+2^3\right)\left(2+...+2^{118}\right)\\ A=7\left(2+...+2^{118}\right)⋮7\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7=7\left(2+2^4+...+2^{118}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=14+2^3\cdot14+...+2^{117}\cdot14\)

\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=62+2^5\cdot62+...+2^{115}\cdot62\)

\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=126+126\cdot2^6+...+126\cdot2^{114}\)

\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17

Bài 1

a, cm : A = 16⁵ + 2¹⁵ ⋮ 3

    A = 16⁵ + 2¹⁵

   A = (2⁴)⁵ +  2¹⁵

  A  = 2²⁰ + 2¹⁵

 A  =  2¹⁵.(2⁵ + 1)

 A = 2¹⁵. 33 ⋮ 3 (đpcm)

b,cm : B = 8⁸ + 2²⁰ ⋮ 17

    B = (2³)⁸ + 2²⁰ 

    B =  2¹⁶ + 2²⁰

    B = 2¹⁶.(1 + 2⁴)

    B = 2¹⁶. 17 ⋮ 17 (đpcm)

 

 

  

c, cm: C = 1 - 2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶ -...-2²⁰²¹ + 2²⁰²² : 6 dư 1

C=1+(-2+2²-2³+2⁴- 2⁵+2⁶)+...+(-2²⁰¹⁷+2²⁰¹⁸-2²⁰¹⁹+2²⁰²⁰-2²⁰²¹+2²⁰²²)

C = 1 + 42 +...+ 2²⁰¹⁶.(-2 + 2² - 2³ + 2⁴ - 2⁵ + 2⁶)

C = 1 + 42+...+ 2²⁰¹⁶.42

C = 1 + 42.(2⁰+...+2²⁰¹⁶)

42 ⋮ 6 ⇒ C = 1 + 42.(2⁰+...+2²⁰¹⁶) : 6 dư 1 đpcm

          

a, \(\overline{aaa}\) \(⋮\) 37

    \(\overline{aaa}\) = a x 111 = a x 3 x 37 ⋮ 37 (đpcm)

b, (\(\overline{ab}\) + \(\overline{ba}\)) ⋮ 11

  \(\overline{ab}\) + \(\overline{ba}\) = \(\overline{a0}\) + b + \(\overline{b0}\) + a = \(\overline{aa}\) + \(\overline{bb}\) = a x 11 + b x 11 = 11 x (a+b)⋮11

a) A chia hết cho 2 vì tất cả các số hạng của tổng đều chia hết cho 2.

b) Ta tách ghép các số hạng của A thành các nhóm sao cho mỗi nhóm xuất hiện thừa số chia hết cho 3. Khi đó:

4₁ A= 2 +2²³ +2 ² + + 220 a₁ A = 2₁ [1 + 2 +2²¹ +. +2¹2):2 Vay A chia hết choi b₁ A = 2 + 2² +2²+ + 220 (2 +2²) + (2 ² + 2 9) + . + (219+220) = 2₁ (1 + 2) + 2² (2+1). .. +2 19 (2+1) + = 2₁3 + 2³.3 + ..+ 219.3. = (2+2 ³+ + 219) 3:3 Vậy A chia hết cho 3 A = 2 + 2 ² + 2³ + 2ª +. 20 + 2.9+ +2 2+2 ³ + 2² +2²4 + + 218 + 720 +2²³ +2²+ +218 +220 2. (2 +2²) + 2² (1+2²) +.. + 218 ( 1 +2²) = 2 5 +2²5 + + 218 5. 12 +2° + 2 ... +218 ) 5 : 5. vậy A chia hết cho 5

A=\((1+2)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)

A=\(3.1+2^2\left(1+2\right)+...+2^{19}\left(1+2\right)\)

A=\(3.1+3.2^2+...+3.2^{19}\)

A=\(3\left(1+2^2+...+2^{19}\right)\)\(⋮3\)

Vậy A\(⋮3\)

A=(1+2)+(22+23)+...+(219+220)(1+2)+(22+23)+...+(219+220)

A=3.1+22(1+2)+...+219(1+2)3.1+22(1+2)+...+219(1+2)

A=3.1+3.22+...+3.2193.1+3.22+...+3.219

A=3(1+22+...+219)3(1+22+...+219)⋮3⋮3

NÊN  A⋮3