How can I prove the following:
Let $\mathbb{W}$ denote the set of non-negative integers. Then what is the cardinality of the set $$\bigl\{ (\alpha,\beta) \in \mathbb{W} \times \mathbb{W} \ | \ 2\alpha + 3\beta =k \bigr\}$$ I think it’s $\left\lfloor\frac{k}{6}\right\rfloor$ if $k \equiv 1 \ (\text{mod} \: 6)$ and $\left\lfloor\frac{k}{6}\right\rfloor + 1$ if $k \not\equiv 1 \ (\text{mod} \ 6)$.
But I am having trouble doing this. Can anyone provide me an answer, or thoughts on how to go about a solution.