From $3$ chocolate bars we can make at most $5$ chocolate rabbits. What is the greatest number of rabbits that can be made from $16$ bars? I'm using some questions from the 2019 International Math Contest to help my students prepare for the Mathcounts competition. While the IMC gives me the answers, I like to give my students the worked solutions so that they learn how to get to the answer. I'm stuck on this one, though:

From three chocolate bars we can make at most 5 chocolate rabbits and have some leftover. What is the greatest number of chocolate rabbits that can be made from sixteen chocolate bars (maybe with some leftover)?

The answer that was given by the IMC was $31$ rabbits. I just can't seem to get there.
I first set up a straight up proportion:
$$\frac{3\;\text{bars}}{5\;\text{rabbits}} = \frac{16\;\text{bars}}{x\;\text{rabbits}}$$ and wound up with $26+\frac23$ rabbits, which is not correct.
So, I thought they must be making more rabbits from the leftovers, but I'm still not sure how they are getting to $31$ rabbits.
Please help!
Given that it takes $3$ bars to make $5$ rabbits, how can you make $31$ rabbits from $16$ bars?
 A: FWIW the phrasing (as given here) isn't clear, and I'd argue that this is a badly written question.
The idea they are going for is that the infimum (but not minimum) amount of chocolate that a rabbit uses is 0.5 chocolate bar, and so with 16 bars, we can produce 31 rabbits.
If you're unfamiliar with the concept of infimum, then it's that

A chocolate rabbit uses $ 0.5 + x $ bars of chocolate, where $ 0 < x \leq 0.1$.
Hence, from 16 bars of chocolate, we can produce $\frac{16}{0.5 + x}$ rabbits, and we get that $ 26 \leq \lfloor \frac{16}{0.5+x} \rfloor \leq 31 $.
So the maximum number of rabbits is 31.

But obviously, we don't know what $x$ is, nor is it well defined, and I agree that any answer from 26 to 31 is reasonable (given that it's reasonable to assume $x$ should be some fixed value).

The way I would rephrase this question for clarity is:

From three chocolate bars, we can make at most 5 chocolate rabbits with some leftover. What is the greatest number of chocolate rabbits that we could have made from sixteen chocolate bars (maybe with some leftover)?

A: Say you need exactly  $x$ chocolate bars to make one rabbit.  What do we know about $x$?
Well, with $3$ bars we can make $\frac 3x$ rabbits and we know that we can make $5$ rabbits but not $6$.  Thus $$5≤\frac 3x<6$$
Which implies that  $$\frac 36<x≤\frac 35$$
If $x$ were to equal $\frac 36$ (which it can't quite) then with $16$ bars we could make $$16\times \frac 63=32$$
But, since $x$ must be at least slightly greater than $\frac 36$ we can't quite arrange to make $32$, so $31$ is the best we can manage.
To stress, we don't know that we can make $31$.  That's just an upper bound.  The lower bound, the number we know we can make, is $$\big \lfloor 16\times \frac 53\big \rfloor=26$$
A: I think the intended question is as follows. Each rabbit contains a certain amount $x$ of chocolate. Each bar contains a certain amount $y$ of chocolate. $3$ bars is enough chocolate for $5$ rabbits, but not $6$. For some value $k$, $15$ bars is enough for $k$ rabbits, but not $k+1$.
What is the largest possible value of $k$, given this information? In other words, how large can you make $k$ if you are allowed to choose the values $x$ and $y$, provided they have to satisfy the given condition?

 With this interpretation, you know $y<2x$ and choosing $y=1.99$ and $x=1$ means you have enough chocolate for $31$ rabbits. Since $y<2x$ you can't make $32$.

This is different from what it sounds like at first reading: what is the largest number of rabbits you can make for any such values of $x$ and $y$. (If someone else gets to choose $x$ and $y$, they can prevent you from making more than $26$ as you say.)
