If $a\pmod 3 \equiv 1$ and $b\pmod 3 \equiv 2$, then $ab \pmod 3 \equiv 2$

Question

I'm stuck on this this problem:

Let $a$ and $b$ be positive integers with $a\pmod 3 \equiv 1$ and $b\pmod 3 \equiv 2$. Prove that $ab \pmod 3 \equiv 2$.

I think the first step for the direct proof is:

$$ a \pmod 3 \equiv 1 \to \exists x \in \mathbb{Z},\,[3x + 1 = a] $$

I am unsure where to proceed from there.

Answer 1

Hint: By definition, $$c\mod 3=d$$ means that there is some integer $y$ such that $$c=3y+d.$$

Hence, by assumption, there exist integers $y,z$ such that $$a=3y+1$$ and $$b=3z+2.$$ What is $ab,$ then? Can you find some integer $w$ such that $ab=3w+2$?