Question

Cho A = 7 + 7^2 + 7^3 + ... + 7^119 + 7^120. chứng minh rằng a chia hết cho 57

Answer 1

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

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

Answer 2

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

\(=57\left(7+...+7^{118}\right)⋮57\)

Answer 3

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

Answer 4

`A=7+7^2+7^3+...+7^119+7^120`

`A=(7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^118+7^119+7^120)`

`A=7.(1+7+7^2)+7^4 . (1+7+7^2)+...+7^118.(1+7+7^2)`

`A=7.57+7^4 . 57+...+7^118 . 57`

`A=57.(7+7^4+...+7^118)\vdots57`

Vậy `A\vdots57`