Ta có: \(a\) chia cho \(3\) dư \(1 \Rightarrow a=3q+1 (q \in \mathbb{N})\)
\(b\) chia cho \(3\) dư \(2\)\(\Rightarrow b=3k+2 (k \in \mathbb{N})\)
\(a.b=(3q+1)(3k+2)\)\(=9qk+6q+3k+2\)
Vì \(9\;⋮\;3\Rightarrow 9qk \;⋮\;3\)
\( 6\;⋮\;3 \Rightarrow 6q\;⋮\;3\)
\( 3\;⋮\;3 \Rightarrow 3k\;⋮\;3\)
Vậy \(a.b=9qk+6q+3k+2\)\(=3(3qk+2q+k)+2\) chia cho \(3\) dư \(2\).
