a) B\(=\) 3 + 3² + 3³ + ... + 3⁶⁰ 

\(=\)(3+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

\(=\)3(1+3)+3³(1+3)+...+3⁵⁹(1+3)

\(=\)(3+1)(3+3³+...+3⁵⁹)

\(=\)4(3+3³+...+3⁵⁹)⋮4

⇒B⋮4

b) B\(=\)(3+3²+3³)+...+(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3⁰(3+3²+3³)+...+3⁵⁷(3⁵⁸+3⁵⁹+3⁶⁰)

\(=\)3+3²+3³(3⁰+3³+3⁶+...+3⁵⁷)

\(=\)39(3⁰+3³+3⁶+...+3⁵⁷)⋮13

⇒ B⋮13

A = ( 1 + 3^2) + (3^4 + 3^6) + ... + (3^2016 + 3 ^2018 ) + 3 ^ 2020

= 10 + 3^4(1+3^2) + .... + 3^2016.(1+3^2) + 3^2020

= 10.(1+3^4+...+3^2016) + 3^2020

Mà : 3^n có tận cùng là : 1,3,9,7

Do đó 3 ^2020 không chia hết cho 10

Lại có 10.(1+3^4+...+3^2016) chia hết cho 10

=> A không chia hết cho 10

A=(1+3²)+(3⁴+3⁶)+ ... + (3²⁰¹⁸+3²⁰²⁰)

  =(1+3²)+ 3⁴(1+3²)+....+3²⁰¹⁸(1+3²)

  =(1+3²) (1+3⁴+....+3²⁰¹⁸)

  =10 (1+3⁴+....+3²⁰¹⁸) ⋮10 ( do 10 ⋮10)

Vậy A=1+3²+3⁴+3⁶+ ... +3²⁰²⁰ ⋮ 10 (đpcm)

\(A=1+3^2+3^4+3^6+...+3^{2020}\\=(1+3^2)+(3^4+3^6)+(3^8+3^{10})+...+(3^{2018}+3^{2020})\\=10+3^4\cdot(1+3^2)+3^8\cdot(1+3^2)+...+3^{2018}\cdot(1+3^2)\\=10+3^4\cdot10+3^8\cdot10+..+3^{2018}\cdot10\\=10\cdot(1+3^4+3^8+...+3^{2018})\)

Vì \(10\cdot(1+3^4+3^8+...+3^{2018})\vdots10\)

nên \(A\vdots10\)

\(C=1+3+3^2+...+3^{11}\)

\(=\left(1+3+3^2\right)+...+3^9\left(1+3+3^2\right)\)

\(=13\cdot\left(1+...+3^9\right)⋮13\)

\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)

\(=13\left(1+...+3^7\right)⋮13\)

Đặt A = 3² + 3³ + 3⁴ + ... + 3⁹⁹

= 3² + 3³ + (3⁴ + 3⁵ + 3⁶) + (3⁷ + 3⁸ + 3⁹) + ... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 36 + 3⁴.(1 + 3 + 3²) + 3⁷.(1 + 3 + 3²) + ... + 3⁹⁷.(1 + 3 + 3²)

= 36 + 3⁴.13 + 3⁷.13 + ... + 3⁹⁷.13

= 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷)

Do 36 không chia hết cho 13

13.(3⁴ + 3⁷ + ... + 3⁹⁷) ⋮ 13

⇒ 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷) không chia hết cho 13

⇒ A không chia hết cho 13

Em xem lại đề nhé, có thể em viết thiếu số 3 rồi

a: (x-3)(y+1)=15

=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}

=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}

b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)

=>m chia 13 dư 4

\(m=1+3+3^2+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

\(=1+40\left(3+3^5+...+3^{97}\right)\)

=>m chia 40 dư 1

a: \(A=1+2+2^2+...+2^{41}\)

=>\(2A=2+2^2+2^3+...+2^{42}\)

=>\(2A-A=2^{42}-1\)

=>\(A=2^{42}-1\)

b: \(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{40}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{40}\right)⋮3\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{39}\left(1+2+2^2\right)\)

\(=7\left(1+2^3+...+2^{39}\right)⋮7\)