The outside of a $180$-sheet roll of toilet paper is covered by two sheets; the inner cylinder, by one. What's wrong with how I counted the layers? Puzzle: A roll of toilet paper has 180 sheets on it. The outside is covered with exactly two sheets. The inside around the cardboard cylinder is covered by exactly one. Question of the puzzle: how many layers of toilet paper are on the roll of toilet paper?


The given solution: One way to solve this is by saying that the average round is covered by 1.5 sheets, so therefore the answer is $180\times\frac{2}{3}=120$

I tried a similar (but wrong) reasoning: "the average sheet makes an average of $\frac{3}{4}$ rounds (first sheet makes one round and the last sheet makes $\frac{1}{2}$ rounds), so the answer is $180\times\frac{3}{4}=$ 135"


  QUESTION: Apparently my answer is wrong. But since it seems analogical to the given solution I don't understand what error I made.


Possibly the growth of sheets per round is constant? While the (negative) growth of rounds per sheet is not constant? What are the related functions?

Put in another way: if $\frac{dSheets}{dRounds}=Constant$ isn't also $\frac{dRounds}{dSheets}=Constant$?

This question is linked to this question: Using differential equations to determine the number of rolls on a roll of toilet paper
There are twice as many sheets that make half a round than there are that make a single round, so it's inappropriate to simply average the numbers $1/2$ and $1$. A weighted average says the average sheet makes

$${2\cdot1/2+1\cdot1\over3}={2\over3}$$

of a round, which gives the answer $180\cdot2/3=120$ again.
I now understand that one cannot solve this puzzle using continuous functions. We can make an approximation however. This is how that would work:

$S$=total amount of sheets

$n$=number of completed rounds

We know that the growth of the amount of Sheets (S) per Round (n) is constant. Why? Because it is depending om the circumference which grows linearly with the radius with $2\pi r$.

Therefore the second derivative of $S(n)$ is constant: $S''(n)=C_1$

The first derivative is therefore: $S'(n)=C_1n+C_2$

In round $0$ the amount of sheets per round is $1$, so we can approximate: $S'(0)=C_2\approx 1$

In the last round (when we used up all $180$ sheets of toilet paper) the amount of sheets per round is $2$, so we can approximate: $S'(n_{180})=C_1 n_{180}+1\approx 2$

And we can conclude that $C_1=\frac{1}{n_{180}}$

So:

$$\bbox[5px,border:2px solid black]{S'(n)=\frac{1}{n_{180}}n+1}$$

We can integrate to find: $S(n)=\int S'(n)dn=\frac{n^2}{2n_{180}}+n+C_3$

$S(0)=C_3=1$

So:

$$\bbox[5px,border:2px solid black]{S(n)=\frac{n^2}{2n_{180}}+n+1}$$

Now we can take $S(n_{180})=\frac{n_{180}^2}{2n_{180}}+n_{180}+1=180$

To find that after $180$ sheets, the approximate amount of completed rounds is $n_{180}=\frac{358}{3}=119\frac{1}{3}$

Now we can also take the inverse to find that:

$$\bbox[5px,border:2px solid black]{n(S)=\sqrt{\frac{716}{3}S+14001\frac{7}{9}}-\frac{358}{3}}$$

With a derivative of:

$$\bbox[5px,border:2px solid black]{n'(S)=\frac{358}{3\sqrt{\frac{716}{3}S+14001\frac{7}{9}}}}$$

Now I see why $S'(n)$ is lineair, but $n'(S)$ is not lineair! Which was what I was trying to get my head around when asking this question.