Question

[Nguyễn Gia Bảo]

chứng tỏ b^2 + b + 2022 ko chia hết cho 5

Answer 1

[Akai Haruma]

Lời giải:

Nếu $b\equiv 0\pmod 5$

$\Rightarrow b^2+b+2022\equiv 0^2+0+2022\equiv 2\pmod 5$

$\Rightarrow b^2+b+2022$ chia $5$ dư $2$

Nếu $b\equiv 1\pmod 5$

$\Rightarrow b^2+b+2022\equiv 1^2+1+2022\equiv 4\pmod 5$

$\Rightarrow b^2+b+2022$ chia $5$ dư $4$

Nếu $b\equiv 2\pmod 5$

$\Rightarrow b^2+b+2022\equiv 2^2+2+2022\equiv 3\pmod 5$

$\Rightarrow b^2+b+2022$ chia $5$ dư $3$

Nếu $b\equiv 3\pmod 5$

$\Rightarrow b^2+b+2022\equiv 3^2+3+2022\equiv 4\pmod 5$

$\Rightarrow b^2+b+2022$ chia $5$ dư $4$

Nếu $b\equiv 4\pmod 5$

$\Rightarrow b^2+b+2022\equiv 4^2+4+2022\equiv 2\pmod 5$

$\Rightarrow b^2+b+2022$ chia $5$ dư $2$
Từ các TH trên suy ra $b^2+b+2022$ không chia hết cho $5$.