a) A  = 1+32+34+36+...+32006​.

2A= (32+32006)+(34+32004)+.....15988 cặp số..+2

= 32038.15988 + 2

= 512223546
Vậy tổng của A = 512223546
Số dư của A chia cho 113= 512223546 - 113.4532951=83 (Đây là cách tính số dư: Số chia - số bị chia x phần nguyên)

Nếu đúng là zậy thì mk biết làm.

A = 3 + 3² + 3³ + ...  + 3²⁰⁰⁴

A = (  3 + 3² + 3³ + 3⁴ ) + ... + ( 3²⁰⁰¹ + 3²⁰⁰² + 3²⁰⁰³ + 3²⁰⁰⁴ )

A = 3( 1 + 3 + 3² + 3³ ) + ... + 3²⁰⁰¹( 1 + 3 + 3² + 3⁹ )

A = 3.40 + ... + 3²⁰⁰¹.40

A = ( 3 + 3⁵ + ...  3²⁰⁰¹) . 40

=> A chia hết cho 40

A = 3 + 3² + 3³ +3⁴ + ... + 3²⁰⁰⁴ phải ko? 

sao lai la 3^9 vay bn

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 \(A\vdots10\)

Ta có:

\(A=3+3^2+3^3+3^4+3^5+3^6\)

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)\)

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

\(A=39+3^3.39\)

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

Vì \(39⋮13\) nên \(39.\left(1+3^3\right)⋮13\)

Vậy \(A⋮13\)

\(#WendyDang\)

Lời giải:
$A=(3+3^2+3^3)+(3^4+3^5+3^6)$

$=3(1+3+3^2)+3^4(1+3+3^2)=(1+3+3^2)(3+3^4)=13(3+3^4)\vdots 13$ 

Ta có đpcm.

`#3107.101107`

\(A=1+3+3^2+3^3+...+3^{101}\)

$A = (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^{99} + 3^{100} + 3^{101}$

$A = (1 + 3 + 3^2) + 3^3 (1 + 3 + 3^2)  + ... + 3^{99}(1 + 3 + 3^2)$

$A = (1 + 3 + 3^2)(1 + 3^3 + ... + 3^{99})$

$A = 13(1 + 3^3 + ... + 3^{99})$

Vì `13(1 + 3^3 + ... + 3^{99}) \vdots 13`

`\Rightarrow A \vdots 13`

Vậy, `A \vdots 13.`

\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)

Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)

nên \(A\vdots13\)

\(\text{#}Toru\)

Bài 1 :

a) \(a.b+b.19=713\) \(\left(a;b\inℕ^∗\right)\)

\(\Rightarrow b.\left(a+19\right)=713\)

\(\Rightarrow\left(a+19\right);b\in\left\{1;23;31;713\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-18;713\right);\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\left(a;b\inℕ^∗\right)\)

b) \(a.b-10.b=650\)

\(\Rightarrow b.\left(a-10\right)=650\)

\(\Rightarrow\left(a-10\right);b\in\left\{1;5;10;13;25;26;50;65;130;325;650\right\}\)

Bạn lập bảng sẽ tìm ra (a;b)...

Bài 2 :

a) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=3^4.40\)

b) \(B=1+3+3^2+3^3+...+3^{99}\)

\(\Rightarrow B=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)...+3^{96}.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow B=40+3^4.40...+3^{96}.40\)

\(\Rightarrow B=40\left(1+3^4...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

A = 8⁸ + 2²⁰

= (2³)⁸ + 2²⁰

= 2²⁴ + 2²⁰

= 2²⁰.(2⁴ + 1)

= 2²⁰.17 ⋮ 17

Vậy A ⋮ 17