a: \(8^8+2^{20}\)

\(=\left(2^3\right)^8+2^{20}\)

\(=2^{24}+2^{20}=2^{20}\left(2^4+1\right)=2^{20}\cdot17\) ⋮17

b: \(A=10^{28}+8=10\ldots08\) (Với 28 chữ số 0)

A có tổng các chữ số là 1+0+0+...+0+8=9

=>A⋮9

Ta có: \(10^{28}=10^3\cdot10^{25}=1000\cdot10^{25}=8\cdot125\cdot10^{25}\) ⋮8

8⋮8

Do đó: \(10^{28}+8\) ⋮8

=>A⋮8

mà A⋮9

và ƯCLN(8;9)=1

nên A⋮8*9

=>A⋮72

c: \(T=2+2^2+2^3+\cdots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{57}\left(1+2+2^2+2^3\right)\)

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

=>T⋮15

mà 15⋮3

nên T⋮3

Ta có: \(T=2+2^2+2^3+\cdots+2^{60}\)

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

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

\(=7\left(2+2^4+\cdots+2^{58}\right)\) ⋮7

A=2+2^2+2^3+...+2^60

=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)

=2(1+2)+2^3(1+2)+...+2^59(1+2)

=3(2+2^3+...+2^59) chia hết cho 3

A=2+2^2+2^3+...+2^60

=(2+2^2+2^3)+...+(2^58+2^59+2^60)

=2(1+2+2^2)+...+2^58(1+2+2^2)

=7(2+...+2^58) chia hết cho 7

A=2+2^2+2^3+...+2^60

=(2+2^2+2^3+2^4)+...+(2^57+2^58+2^59+2^60)

=2(1+2+2^2+2^3)+...+2^57(1+2+2^2+2^3)

=15(2+...+2^57) chia hết cho 15

A=2(1+2)+2^3(1+2)+...+2^59(1+2)

A=2.3+2^3.3+...+2^59.3

A=3(2+2^3+...+2^59) chia hết cho 3

Vậy a chia hết cho 3

A=2.(1+2+4)+...+2^58(1+2+4)

A=2.7+...+2^58.7

A=7.(2+..+2^58) chia hết cho7

Vậy A chia hết cho 7

A=2(1+2+4+8)+...+2^57(1+2+4+8)

A=2.15+...+2^57.15

A=15.(2+...+2^57) chia hết cho 15 

Vậy A chia hết cho 15 

Vậy A chia hết cho 3,7,15

a) Ta có: \(\overline{abcdeg}=\overline{ab}.1000+\overline{cd}.100+\overline{eg}\)

                               \(=\overline{ab}.999+\overline{cd}.99+\overline{ab}+\overline{cd}+\overline{eg}\)

                               \(=\left(\overline{ab}.999+\overline{cd}.99\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)

Vì \(\left(\overline{ab}.999+\overline{cd}.99\right)⋮11\)

và \(\left(\overline{ab}+\overline{cd}+\overline{cd}\right)⋮11\left(gt\right)\)

\(\Rightarrow\overline{abcdeg}⋮11\left(đpcm\right)\)

b) \(\cdot A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+...+\left(2^{50}+2^{60}\right)\)

\(A=2.3+...+2^{50}.3\)

\(A=3\left(2+..+2^{50}\right)⋮3\)

các trường hợp còn lại tự lm nhé!!

+) A = 2 + 2² + 2³ + ... + 2⁶⁰ = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2⁵⁹ + 2⁶⁰ )

= 2( 1 + 2 ) + 2³( 1 + 2 ) + ... + 2⁵⁹( 1 + 2 ) = 2 . 3 + 2³ . 3 + ... + 2⁵⁹ . 3

= ( 2 + 2³ + ... + 2⁵⁹ ) . 3 chia hết cho 3 ( ĐPCM )

+) A = 2 + 2² + 2³ + ... + 2⁶⁰ = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2⁵⁸ + 2⁵⁹ + 2⁶⁰ )

= 2( 1 + 2 + 2² ) + 2⁴( 1+ 2 + 2² ) + ... + 2⁵⁸( 1 + 2 + 2² )

= 2 . 7 + 2⁴ . 7 + ... + 2⁵⁸ . 7 = ( 2 + 2⁴ + ... + 2⁵⁸ ) . 7 chia hết cho 7 ( ĐPCM )

+) A = 2 + 2² + 2³ + ... + 2⁶⁰ = ( 2 + 2³ ) + ( 2² + 2⁴ ) + ... + ( 2⁵⁸ + 2⁶⁰ )

= 2( 1 + 2² ) + 2²( 1 + 2² ) + ... + 2⁵⁸( 1 + 2² ) = 2 . 5 + 2² . 5 + ... + 2⁵⁸ . 5

= ( 2 + 2² + ... + 2⁵⁸ ) . 5 chia hết cho 5

Mà A cũng chia hết cho 3 và ( 3, 5 ) = 1 => A chia hết cho 15 ( ĐPCM )

Cảm ơn bạn nhiều !!!

MBĐTTBRĐ

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2²) + (2³ + 2⁴) + ... + (2⁵⁹ + 2⁶⁰)

= 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁵⁹.(1 + 2)

= 2.3 + 2³.3 + ... + 2⁵⁹.3

= 3.(2 + 2³ + ... + 2⁵⁹) chia hết cho 3

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰)

= 2.(1 + 2 + 2²) + 2⁴.(1 + 2 + 2²) + ... + 2⁵⁸.(1 + 2 + 2²)

= 2.7 + 2⁴.7 + ... + 2⁵⁸.7

= 7.(2 + 2⁴ + ... + 2⁵⁸) chia hết cho 7

A = 2 + 2² + 2³ + ... + 2⁶⁰

= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰)

= 2.(1 + 2 + 2² + 2³) + 2⁵.(1 + 2 + 2² + 2³) + ... + 2⁵⁷.(1 + 2 + 2² + 2³)

= 2.15 + 2⁵.15 + ... + 2⁵⁷.15

= 15.(2 + 2⁵ + ... + 2⁵⁷) chia hết cho 15

bạn có nhầm đề ko ? Xem lại 2⁵⁰ +2⁶⁰ 

a)A= 2+2^2+2^3+...+2^59+2^60

A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^59+2^60)

A=2.(1+2)+2^3.(1+2)+2^5.(1+2)+...+2^59.(1+2)

A=2.3+2^3.3+2^5.3+...+2^59.3

A=3.(2+2^3+2^5+...+2^59)

=>A chia hết cho 3

Vậy A chia hết cho 3

b)A= 2+2^2+2^3+..........................+2^59+2^60

A=(2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)+...+(2^58+2^59+2^60)

A=2.(1+2+2^2)+2^4.(1+2+2^2)+2^7.(1+2+2^2)+...+2^58.(1+2+2^2)

A=2.7+2^4.7+2^7.7+...+2^58.7

A=7.(2+2^4+2^7+...+2^58)

=>A chia hết cho 7

Vậy A chia hết cho 7

c)A= 2+2^2+2^3+..........................+2^59+2^60

A=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+(2^9+2^10+2^11+2^12)+...+(2^57+2^58+2^59+2^60)

A=2.(1+2+2^2+2^3)+2^5.(1+2+2^2+2^3)+2^9.(1+2+2^2+2^3)+...+2^57.(1+2+2^2+2^3)

A=2.15+2^5.15+2^9.15+...+2657.15

A=15.(2+2^5+2^9+...+2^57)

=>A chia hết cho 15

Vậy A chia hết cho15