Lời giải:
Ta có:

$10\equiv -1\pmod {11}$

$\Rightarrow 10^{2022}\equiv (-1)^{2022}\equiv 1\pmod {11}$

$\Rightarrow A=10^{2022}-1\equiv 1-1\equiv 0\pmod {11}$

Vậy $A\vdots 11$

ok

A= 10^2022-1

Ta có thể thấy 10^2022=100000000...........0000000000 

 10000000.......0000000000-1 thì lúc nnày tổng bằng

9999999999999999........................999999999999999999999

mà 99999999999999999999999....................9999999999999999999chia hết cho 11 nên tổng này chia hết cho 11

Ta có: \(43^{43}=43^{40}.43^3\)

Lại có: 43⁴⁰tận cùng 1; 43³ tận cùng 7

=>43⁴³tận cùng 7(*)

           17¹⁷=17¹⁶.17

Mà 17¹⁶tận cùng 1

=>17¹⁷tận cùng 7(**)

Từ (*) và (**) suy ra 43⁴³-17¹⁷tận cùng 0

\(\Rightarrow43^{43}-17^{17}⋮10\)

\(43^{101}+23^{101}=43\cdot43^{100}+23\cdot23^{100}=\left(66-23\right)\cdot43^{100}+23\cdot23^{100}\)

\(=66\cdot43^{100}-23\cdot43^{100}+23\cdot23^{100}=66\cdot43^{100}-23\left(43^{100}-23^{100}\right)\)

\(=66\cdot43^{100}-23\left(43-23\right)\left(43^{99}+43^{98}\cdot23+43^{97}\cdot23^2+43^{96}\cdot23^3+...+43\cdot23^{98}+23^{99}\right)\)

\(=66\cdot43^{100}-23\cdot20\left(43^{98}\left(43+23\right)+43^{96}\cdot23^2\left(43+23\right)+...+23^{98}\left(43+23\right)\right)\)

\(=66\cdot43^{100}-460\left(4^{98}\cdot66+4^{96}\cdot23^2\cdot66+...+23^{98}\cdot66\right)\)

\(=66\cdot43^{100}-460\cdot66\left(4^{98}+4^{96}\cdot23^2+...+23^{98}\right)\)

\(=66\left(43^{100}-460\left(4^{98}+4^{96}\cdot23^2+...+23^{98}\right)\right)⋮66\Rightarrow43^{100}+23^{100}⋮66\)(đpcm)

cái chỗ \(43^{100}-23^{100}=\left(43-23\right)\left(43^{99}+43^{98}\cdot23+43^{97}\cdot23^2+43^{96}\cdot23^3+...+43\cdot23^{98}+23^{99}\right)\)

là áp dụng hđt \(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+a^{n-4}b^3+...+b^{n-1}\right)\)

4/ Chứng minh rằng :a.     76 +75 – 74 chia hết cho 11 . bạn nào giúp mình với (giải thích cho mình hiểu luôn nha các bạ... - Hoc24

\(7^6+7^5-7^4\)

\(=7^4\left(7^2+7-1\right)\)

\(=7^4\cdot55⋮11\)

bai re vai lam 30 giay

\(P=3^{10}+3^{11}+3^{12}\)

\(=3^{10}\cdot\left(1+3+3^2\right)\)

\(=3^{10}\cdot13\)

Vì \(13⋮13\) nên \(3^{10}\cdot13⋮13\)

hay \(P⋮13\)

Vậy ...

#\(Toru\)

P = 3¹⁰ + 3¹¹ + 3¹²

= 3¹⁰.(1 + 3 + 3²)

= 3¹⁰ . 13 ⋮ 13

Vậy P ⋮ 13

a) \(A=7^{13}+7^{14}+7^{15}+7^{16}+...+7^{100}\)

\(A=\left(7^{13}+7^{14}\right)+\left(7^{15}+7^{16}\right)+...+\left(7^{99}+7^{100}\right)\)

\(A=7^{13}\left(1+7\right)+7^{15}\left(1+7\right)+...+7^{99}\left(1+7\right)\)

\(A=7^{13}.8+7^{15}.8+...+7^{99}.8\)

\(A=8.\left(7^{13}+7^{15}+...+7^{99}\right)\)

⇒ \(A⋮8\)

Vậy A chia hết cho 8 (đpcm)

a) A = 7¹³ + 7¹⁴ + 7¹⁵ + 7¹⁶ + ... + 7⁹⁹ + 7¹⁰⁰

= (7¹³ + 7¹⁴) + (7¹⁵ + 7¹⁶) + ... + (7⁹⁹ + 7¹⁰⁰)

= 7¹³.(1 + 7) + 7¹⁵.(1 + 7) + ... + 7⁹⁹.(1 + 7)

= 7¹³.8 + 7¹⁵.8 + ... + 7⁹⁹.8

= 8.(7¹³ + 7¹⁵ + ... + 7⁹⁹) ⋮ 8

Vậy A ⋮ 8

b) B = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰

= 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + ... + 2¹⁹⁷ + 2¹⁹⁸ + 2¹⁹⁹ + 2²⁰⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁹⁷ + 2¹⁹⁸ + 2¹⁹⁹ + 2²⁰⁰)

= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + 2¹⁹⁶.(2 + 2² + 2³ + 2⁴)

= 30 + 2⁴.30 + ... + 2¹⁹⁶.30

= 30.(1 + 2⁴ + ... + 2⁹⁶)

= 5.6.(1 + 2⁴ + ... + 2¹⁹⁶) ⋮ 5

Vậy B ⋮ 5

\(B=2+2^2+2^3+...+2^{200}\)

\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{199}+2^{200}\right)\)

\(B=1.\left(2+2^2\right)+2^2.\left(2^{ }+2^2\right)+...+2^{198}.\left(2+2^2\right)\)

\(B=1.5+2^2.5+...+2^{198}.5\)

⇒\(B⋮5\)

Vậy B chia hết cho 5 (đpcm)

\(B=5.\left(1+2^2+...+2^{198}\right)\)

\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31