Bài 4 (1 điểm): a, Tìm số tự nhiên n sao cho 5n + 3 chia hết cho n + 2 b, Chứng minh rằng: $3^{5371}$ + $57^{2016}$ + $92^{2017}$ chia hết cho 10.

Answer 

`a,` Ta có:

`(5n + 3) \vdots (n + 2)`

`=> (5n + 10 - 7) \vdots (n + 2)`

`=> 7 \vdots (n + 2)`

Vì `n \in NN => n + 2 \in NN`

`=> n + 2 >= 2`

`=> n + 2 = 7`

`=> n = 5` (tm)

Vậy `n = 5`

`b, 3^{5371} + 57^{2016} + 92^{2017}`

`= (3^{4})^{1342} . 3^3 + (57^{4})^{504} + (92^4)^{504} . 92`

`= (...1)^{1342} . (...7) + (...1)^{504} + (...6)^{504} . (...2)`

`= (...1) . (...7) + (...1) + (...6) . (...2)`

`= (...7) + (...1) + (...2)`

`= (...0)`

Vậy `(3^{5371} + 57^{2016} + 92^{2017}) \vdots 10`