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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.

  • 2
    What about proving your formulas for "small" k, then lifting the count to k+6?2012-04-29
  • 1
    If $2\alpha + 3\beta = k$, then $2(\alpha+3) + 3(\beta-2)= ?$ Or do you mean *positive* integers when you say "whole numbers"?2012-04-29
  • 3
    @TMM Downvoting due to choice of terminology seems a bit extreme, esp. for new members.2012-04-29
  • 1
    @Bill: Especially when the terminology is not uncommon in U.S. primary and secondary schools.2012-04-29
  • 0
    @Bill: Alright, I'll remove the downvote. I had the same thoughts as Arturo above and Douglas [here](http://math.stackexchange.com/questions/138633/what-are-the-whole-numbers), that the only natural definition of "whole numbers" is integers (aren't negative numbers whole numbers too?). But apparently this bad terminology is often used in the US.2012-04-30

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