# HP1: The Sixth Chamber Solution

The following is a solution to Puzzle #19 – HP1: The Sixth Chamber. Take a look if you haven’t already – I mean it this time!

We have seven labelled bottles, three of which contain nettle wine, two of contain poison, one of which will allow Harry or Hermione to pass forward and one of which will allow them to pass backward. Our goal is to identify the contents of each bottle.

The hints given are mathematical, so we need a way of expressing their content in terms of equations. What are the relevant variables? They are $w, p, f$ and $b$. The variable $w$, for example, represents the numbers found on the bottles of wine, and so on. It is important to remember that, since $w$ can assume two values, any equation we form in terms of it must be quadratic. Similarly, as there are three poisonous potions, any equation involving $p$ must be cubic.

Let’s take each of the clues in turn and express them mathematically.

Clue 1

The first hint is:

First, observe the labels we’ve provided as a guide;

The bottles bear the numbers one to seven on the side.

This clue simply tells us that the bottles are distinguished by the numbers with which they are labelled (as opposed to, say, their relative positions and sizes, as was the case in the original text!).

Clue 2

The second clue is as follows:

Second, take the number found on either vintage white,

Sep’rately subtract that which thwarts flames as dark as night,

And that which sends you back; multiply these diff’rences –

For either of the nettle wines, three’s what the answer is.

The equation to which this corresponds to

$\left(w-f\right)\left(w-b\right)=3.$

As expected, the expression is quadratic, reflecting the two possible values $w$ can assume.

Clue 3

The third clue is

Third, work out the sum of those that shield you from the fire,

Divide by one of the number that will cause you to expire.

Next take the potion which will let you pass the flames behind,

Subtract its number from that of the poison said before,

Then square; the answers are identical, no less, no more.

The mathematical formulation of this clue is:

$\displaystyle \frac{f+b}{p}-1=\left(p-b\right)^2.$

By expanding the square bracket and multiplying through by $p$, the term $p$3 will appear on the right-hand side, verifying that the equation is in fact cubic in $p$.

Clue 4

Fourth, the number of the potion that will take you back

Is less than that on the potion that takes you through the black.

Simply put,

$f>b.$

$\$

These three clues are indeed enough to determine the values of $w,p,f$ and $b$, when combined with the pretty strong constraint that they must also be natural numbers between 1 and 7.

$\displaystyle \frac{f+b}{p}-1=\left(p-b\right)^2.$

What does this tell us? Clearly, as $p$ and $b$ are integers, $(p-b)$2 must be an integer. This can only mean that $\frac{f+b}{p}$ is also an integer, for all three values of $p$. This constrains the possible values that $f+b$ can take. In principle, it could assume any of the following values:

3 4 5 6 7 8 9 10 11 12 13

But, from the constraint above, $f+b$ must have (at least) three factors, which rules out all prime numbers:

4 6 8 9 10 12

We can do better than this, though. Our clue also tells us that $\frac{f+b}{p}$ must be one more than a (non-zero) square number for all three values of $p$. The only value of $f+b$ above for which this is possible is 10:

10/1 = 10 = 32 + 1

10/2 = 5 = 22 + 1

10/5 = 2 = 12 + 1

This also tells us, for free, the three possible values of $p$:

$p=1,2,5;$

$f+b=10.$

$\$

We’ve already made some good progress. Now let’s look at clue 2:

$\left(w-f\right)\left(w-b\right)=3.$

Fortunately, 3 is a prime number, which means there are only four feasible solutions to the equation above (two of which are correct):

$w-f = +3, w-b = +1;$

$w-f=+1, w-b=+3;$

$w-f=-3, w-b=-1;$

$w-f=-1, w-b=-3;$

We can now invoke clue 4, which says that $f>b$. This reduces the feasible solutions to

$w-f=+1,w-b=+3;$

$w-f=-3, w-b=-1;$

Both of these equations agree in telling us that $f=b+2$. When combined with $f+b=10$, we are forced to conclude that

$f=6,$

$b=4.$

By process of elimination, the values that $w$ can assume are

$w=3,7.$

You can verify that these values of $w$ solve the quadratic equation yielded from clue 2.

$\$

To summarise:

1. is poison;
2. is poison;
3. is nettle wine;
4. takes you backward;
5. is poison;
6. takes you forward;
7. is nettle wine.

This is just one means of attacking the problem. If you have found a more elegant solution, feel free to tell us in the comments!