\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+7^4+...+7^{118}\right)⋮57\)

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

\(=57\left(7+...+7^{118}\right)⋮57\)

A =7(1+7+7₂)+7⁴(1+7+7₂)+...+7¹¹⁸(1+7+7²)A=7(1+7+7²)+74(1+7+7²)+...+7¹¹⁸(1+7+7²)

=57 (7+7⁴+...+71¹⁸)⋮57

\(A=7+7^2+7^3+...+7^{119}+7^{120}\)

\(\Rightarrow7A=7^2+7^3+7^4+...+7^{120}+7^{121}\)

\(\Rightarrow7A-A=\left(7^2+7^3+...+7^{120}+7^{121}\right)-\left(7+7^2+...+7^{119}+7^{120}\right)\)

\(\Rightarrow6A=7^2+7^3+...+7^{120}+7^{121}-7-7^2-...-7^{119}-7^{120}\)

\(\Rightarrow6A=7^{121}-7\)

\(\Rightarrow A=\dfrac{7^{121}-7}{6}\)

\(A=7+7^2+7^3+...+7^{120}\)

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

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

\(A=7.57+7^4.57+...+7^{118}.57\)

\(A=57\left(7+7^4+...+7^{118}\right)\)

\(\Rightarrow A⋮57\)

Sợ quá!

A = 7 + 72 + 73 + ... + 7119 + 7120

A = (71 + 72 + 73) + (74 + 75 + 76) + ... + (7118 + 7119 + 7120)

A = 7(1 + 7 + 72) + 74(1 + 7 + 72) + ... + 7118(1 + 7 + 72)

A = 7.57 + 74.57 + ... + 7118.57

A = 57(7 + 74 + ... + 7118)

Vì 57 ⋮ 57 nên 57(7 + 74 + ... + 7118) ⋮ 57

A=7+7²+7³+...+7²⁰¹⁶

=(7+7²)+(7³+7⁴)+...+(7²⁰¹⁵+7²⁰¹⁶)

=7.(1+7)+7³.(1+8)+...+7²⁰¹⁵.(1+7)

=7.8+7³.8+...+7²⁰¹⁵.8

=8.(7+7³+...+7²⁰¹⁵) chia hết cho 8 (đpcm)

A=7+7²+7³+...+7²⁰¹⁶

=(7+7²+7³)+...+(7²⁰¹⁴+7²⁰¹⁵+7²⁰¹⁶)

=7.(1+7+7²)+...+7²⁰¹⁴.(1+7+7²)

=7.57+...+7²⁰¹⁴.57

=57.(7+...+7²⁰¹⁴) chia hết cho 57 (đpcm)

\(A=7\left(1+7+7^2\right)+...+7^{88}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{88}\right)⋮57\)

\(=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{118}\right)⋮57\)

a/ 

\(\overline{aba}=101.a+10b=98a+3a+7b+3b=\)

\(=\left(98a+7b\right)+3\left(a+b\right)\)

\(98a+7b⋮7;\left(a+b\right)⋮7\Rightarrow3\left(a+b\right)⋮7\)

\(\Rightarrow\overline{abc}=\left(98a+7b\right)+3\left(a+b\right)⋮7\)

b/ xem lại đề bài

#Nguồn: Băng

Ta có: \(7^{100}+7^{99}+7^{98}\)

\(=7^{98}\left(1+7^1+7^2\right)\)

\(=7^{98}\times57\) chia hết cho \(57\)

Vậy \(\left(7^{100}+7^{99}+7^{98}\right)⋮57\left(đpcm\right)\)

A = 7¹⁰⁰ + 7⁹⁹ + 7⁹⁸

A = 7⁹⁸.7² + 7⁹⁸.7 + 7⁹⁸

A = 7⁹⁸.( 7² + 7 + 1)

A = 7⁹⁸.57 chia hết cho 57

=> 7¹⁰⁰ + 7⁹⁹ + 7⁹⁸ chia hết cho 57 (đpcm)

A = ( 7 + 7^2 + 7^3 ) + ( 7^4 + 7^5 + 7^6 ) + ... + ( 7^88 + 7^89 + 7^90 )

A = 7( 1 + 7 + 7^2 ) + 7^4 ( 1 + 7 + 7^2 ) + ... + 7^88( 1 + 7 + 7^2 ) 

A = 7 . 57 + 7^4 . 57 + ... + 7^88 . 57

A = 57( 7 + 7^4 + ... + 7^88 )

=> A chia hết cho 57

nhóm 3 số 1 rồi rút 7 ra là đc

cái đó biết rồi,muốn làm cách làm chi tiết cơ.