Ta có: a chia cho 3 dư 1 `=>` a = 3q + 1 ( q `in` N)
b chia cho 3 dư 2 1=>1 b = 3k + 2 (k `in` N)
A : b ( 3q + 1) (3k + 2) = 9qk + 6q + 3k + 2
Vì 9 `\vdots` 3 `=>` 9qk `\vdots` 3
6 `\vdots` 3 `=>` 6q `\vdots` 3
3`\vdots` 3 `=>` 3k `\vdots` 3
Vậy a.b = 9qk + 6q + 3k + 2= 3(3qk + 2q +k) + 2 chia cho 3 dư 2