Chứng minh bằng phản chứng : 

Giả sử rằng tồn tại ít nhất một số tự nhiên n sao cho thỏa mãn \(n^2+7n+2014\) chia hết cho 9

Khi đó đặt n = 9k (k thuộc N)
 

Ta có \(n^2+7n+2014=\left(9k\right)^2+7.\left(9k\right)+2014=9.\left(9k^2+7k+223\right)+7\)

Từ đó ta thấy ngay điều giả sử sai, suy ra đpcm.

Ta có

A = n² + 7n + 2014 = (n + 2)(n + 5) + 2004

Giả sử A chia hết cho 9 thì A = 9k 

=> (n + 2)(n + 5) + 2004 = 9k (k tự nhiên)

Ta thấy 2004 chia hết cho 3 nên (n + 2)(n + 5) chia hết cho 3. Vậy 1 trong hai thừa số phải chia hết cho 3

Mà n + 5 - n - 2 = 3 chia hết cho 3 nên cả (n + 5) và (n + 2) đều chia hết cho 3.

Hay (n + 5)(n + 2) chia hết cho 9.

Mà A lại chia hết cho 9 nên 2004 chia hết cho 9 (vô lý)

Vậy không tồn tại số tự nhiên nào để A chia hết cho 9

ko có ai chả lời chán mà phải ko

Nhận thấy A = 3ⁿ + 4n +1 chia hết cho 2 với mọi n tự nhiên, để A chia hết cho 10 ta cần A chia hết cho 5 là đủ.

Nhận xét: 3⁴ \(\equiv\)1 (mod 5), ta sẽ xét các trường hợp: n = 4k, n = 4k+1, n = 4k+2, n = 4k+3 với k là số tự nhiên.

TH1: n = 4k.

A = 3^4k + 4.(4k) + 1 = 81^k + 16k +1 \(\equiv\)1 + k + 1 \(\equiv\)2+k (mod 5)

Để A chia hết cho 5 thì k phải có dạng 5h + 3, với h là số tự nhiên. Vậy n = 4.(5h+3) = 20h +12 thì A chia hết cho 10.

Tương tự với các trường hợp sau bạn giải tiếp nhé!

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