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?

Question

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

Answer 1

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.

Answer 2

Suppose the inner cardboard cylinder has radius $r$, and the entire roll has radius $R$. The inner circumference then, is $2\pi r$. Furthermore this is equal to the length of a single sheet, $\ell$. The entire roll has circumference $2\pi R$, which we are told is equal to $2\ell$. Some simple algebra tells us that $R=2r$. So the distance between the inner roll and the outside paper is $r$. Now suppose the thickness of the toilet paper is $t$. The number of layers, $L$, of toilet paper should be equal to $r/t$.

We'll split our interval of length $r$ between the inside and the outside into $L$ intervals of thickness $t$. We can add up the circumferences around all these intervals and divide the expression by $\ell$ to get the approximate number of sheets. So, $$\sum ^{L}_{k=0} 2\pi ( r+kt) =N\ell$$ Where $N$ is the total number of sheets. Substituting $r/L$ for $t$, we'll re-express this sum as $$\frac{2\pi r}{\ell }\sum ^{L}_{k=0}\left( 1+\frac{k}{L}\right) =N$$ But of course $\ell=2\pi r$, so $$\sum ^{L}_{k=0}\left( 1+\frac{k}{L}\right) =N$$ We are told in the problem that there are $180$ sheets, so the goal is to find a value of $L$ that satisfies the equation using $N=180$. Let's re-express this sum as $$L+\frac{1}{L}\sum ^{L}_{k=0} k=N$$ And finally, $$L+\frac{L(L+1)}{2L}=N$$ $$\frac{3L}{2}+\frac{1}{2}=N$$ Setting $N=180$, we find $L=119.666...$ which, rounding up, is $120$.

Answer 3

Consider the cross-section of the roll.

It might be easier to think of it as a continuous spool of paper that's later going to be divided lengthwise equally into sheets.

The circumference of the cross section at a radial distance $r$ from the centre is $2\pi r$. The thickness of one layer is $t$. The circumference one layer outward will be $2\pi (r+t) $, which is $2\pi t$ more. All this is not necessary, it's just to show that adding a layer adds a constant to the cross sectional circumference.

If we now think in terms of sheets, we can say the initial cross-sectional circumference is $1$ sheet while the final is $2$ sheets. You're adding a constant each time. This is an arithmetic progression.

The sum of an arithmetic series can be given by different formulas. The easiest one to use here is $S(n) = \frac n2 (a+l) $, where $n$ is the number of terms (equal to number of layers, and this is what you need to solve for). $a$ is the first term ($1$ here) and $l$ is the final term ($2$ here). You can also think of it like $n$ times the average term, which allows you to relate it to your given solution.

So $\frac n2 (1+2) = 180 \implies n =120$.

Answer 4

To solve more rigorously, call $L$ the length of a single sheet, $r$ and $R$ the inner and outer radius of the roll, respectively and $n$ the number of layers.

From the information given, we know that $R = 2\pi L = 2r$, and also $$ \begin{cases} (n-1)\Delta R = R - r = r\\ 180 L = 2\pi\sum_{k=0}^{n-1} R_k, \end{cases} $$ where $\Delta R$ is the reduction in radius after a layer and $ R_k $ is the radius of the roll after $k$ layers, which we can assume to be constant on that layer provided $\Delta R \ll \min_k R_k = r$ (which we can check at the end).

Then $ R_k = R-k\Delta R $ and the sum gives us $$ 180 L = \pi n (R+r), $$ and applying all the identities between these quantities this simplifies to $$ n = 120. $$

As a final consistency check, notice that $ \Delta R = r/(n-1) = r/119 \ll r $ as required.

Answer 5

None of the math in this Answer is rigorous, since neither sheets or rounds are continuous variables. However, I'm using this weird continuous model to try to align with what the OP is asking... Hope this helps as opposed to confuses even further!

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Let $s(r)$ be the number of sheets in round $r$. Let $R =$ total no. of rounds, i.e. the value we're seeking. We have:

• $s(0) = 1$

• $s(R) = 2$

• $ds/dr = constant = {2 - 1 \over R - 0} = 1/R$

• So: $s(r) = 1 + r/R$

  • Technically you find this by integrating $ds/dr$ to find $s = r/R + C$ and then finding the integration constant via the boundary conditions... but it is also obvious.

So far so far... Now we know total no. of sheets is $180$, which in this weird continuous model would be:

$$180 = \int_0^R s(r) dr = [r + r^2/2R]^R_0 = R + R/2 = \frac32 R \implies R = 120$$

which is the correct answer. At this point the OP seems confused. Option #1 is to consider:

• $dr/ds = {1 \over ds/dr} = R = constant$, which is true as far as it goes...

  • Note: $dr/ds$ is not the no. of rounds covers by a sheet! That is Option #2 below.

• Anyway, integrating we have $r = sR + C$. Again we find the integration constant via the boundary conditions, giving: $r = sR - R = R(s-1)$

  • Sanity check: this is the same as $s = 1 + r/R$.

The next step is the big question. What is the integral? Since $s$ is number of sheets in a specific round, all we have is $180 = \int_0^R s(r) dr$. In particular both of the following integrals are wrong:

$$\int_0^{180} r(s) ds$$

is wrong because the variable $s$ does not go from $0$ to $180$, and

$$\int_1^2 r(s) ds$$

is also wrong because it's measuring a different area in the $(s,r)$ plane! If you draw the $(s,r)$ plane, the relation between $s$ and $r$ is the line segment connecting $(1, 0)$ to $(2, R)$. The "correct" integral $\int_0^R s(r) dr$ measures the area from the line segment to the $r$-axis, which is a trapezoid, whose area we relate to $180$. This last "wrong" integral measures the area from the line segment to the $s$-axis, which is a triangle, whose area cannot be related to $180$, nor to $R$. In fact, since it is a triangle of base $2-1 = 1$ and height $R$, its area is indeed $R/2$:

$$\int_1^2 r(s) ds = R[s^2/2 - s]_1^2 = R/2$$

The point, however, is that it cannot be related to $180$.

Option #2 is to consider the number of rounds covered by a sheet. However, this option requires totally different variables than the $s, r$ defined above! In this kind of model, the "independent" variable is $\hat{s}$ which goes from $0$ (or $1$) to $180$, and the "dependent" variable is $\hat{r}(\hat{s})$ which is the number of rounds covered by sheet $\hat{s}$, and which goes from $1$ to $1/2$. However, these are very different variables from $s, r$ and in particular

$$d \hat{r} / d \hat{s} \neq {1 \over ds/ dr}$$

and indeed we have no reason to believe $d\hat{r} / d \hat{s}$ is a constant. And if it isn't a constant, then $\hat{r}$'s unweighted average (also median) value of $3/4$ hardly matters.

Answer 6

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.