Question

[nguyen the tuan quang]

CMR

11ⁿ⁺²+12²ⁿ⁺¹ ⋮133

Answer 1

[Trần Minh Hoàng]

Ta có:

11^{n + 2} + 12^{2n + 1}

= 11ⁿ . 11² + (12²)ⁿ . 12

= 11ⁿ . 121 + 144ⁿ . 12

= 11ⁿ . (121 + 12) + (144ⁿ - 11ⁿ) . 12

= 11ⁿ . 133 + (144ⁿ - 11ⁿ) . 12

Lại có: 144ⁿ \(\equiv\) 11ⁿ (mod 133)

\(\Rightarrow\) 144ⁿ - 11ⁿ \(⋮\) 133

\(\Rightarrow\) ​144ⁿ - 11ⁿ \(⋮\) 133

\(\Rightarrow\) (144ⁿ - 11ⁿ) . 12 \(⋮\) 133

\(\Rightarrow\) 11ⁿ . 133 + (144ⁿ - 11ⁿ) . 12 \(⋮\) 133

Answer 2

[Nguyễn Thành Trương]

Ta có:
$11ⁿ⁺² + 12²ⁿ⁺¹ + 12^2$
$= 11² x 11ⁿ + 12 x (12²)ⁿ + 12^2$
$= 121 x 11ⁿ + 12 x (12²)ⁿ + 144$
$= (133 - 12) x 11ⁿ + 12 x (12²)ⁿ + 144$
$= 133 x 11ⁿ - 12 x 11ⁿ + 12 x 144ⁿ + 144$
$= 133 x 11ⁿ + 12 x (144ⁿ - 11ⁿ) + 144$
Vì:
$133 x 11ⁿ ⋮ 133$
$12 x (144ⁿ - 11ⁿ) = 12 x (144 - 11) x P = 12 x 133 x P ⋮ 133$
$144$ chia cho $133$ dư $11$
Suy ra: $11ⁿ⁺² + 12²ⁿ⁺¹ + 12²$ chia cho $133$ dư $11$