Đáp án: 21
Giải thích các bước giải:
Ta có:
* 2018 ≡ 2 (mod 12)
⇒ $2018^{2019}$ ≡ $2^{2019}$ (mod 12)
Mặt khác ta thấy: $2^{2n}$ = $4^{n}$ ≡ 4 (mod 12)
$2^{2n+1}$ = $4^{n}$.2 ≡ 4.2 ≡ 8 (mod 12)
⇒ $2^{2019}$ ≡ 8 (mod 12) ⇒ $2018^{2019}$ chia 12 dư 8 (1)
* 2019 ≡ 3 (mod 12)
⇒ $2019^{2020}$ ≡ $3^{2020}$ (mod 12)
Mặt khác ta thấy: $3^{2n}$ = $9^{n}$ ≡ 9 (mod 12)
$3^{2n+1}$ = $9^{n}$.3 ≡ 9.3 ≡ 3 (mod 12)
⇒ $3^{2020}$ ≡ 9 (mod 12) ⇒ $2019^{2020}$ chia 12 dư 9 (2)
* 2020 ≡ 4 (mod 12)
⇒ $2020^{2021}$ ≡ $4^{2021}$ (mod 12)
Mặt khác ta thấy: $4^{n}$ = 4 (mod 12)
⇒ $4^{2021}$ ≡ 4 (mod 12) ⇒ $2020^{2021}$ chia 12 dư 4 (3)
Từ (1), (2), (3) suy ra: $2018^{2019}$ + $2019^{2020}$ + $2020^{2021}$ chia 12 dư: 8 + 9 + 4 = 21.
