A measuring tape has a length of $l$ meters and a thickness of $d^{\mathrm{m}} / \mathrm{m}$. If we roll it up in the usual way, how many turns will it make, given that the circumference of the first turn is $k \mathrm{~cm}$? What is the length of the last turn?

For example, $l=1.5, d=1, k=1$.
If the circumference of the first lap is $k_{1}=k \mathrm{~cm}$, then the second one, whose radius is $d^{\mathrm{m}} / \mathrm{m}=0.1 d \mathrm{~cm}-\mathrm{rel}$ larger,

$$
k_{2}=\left(\frac{k_{1}}{2 \pi}+0.1 d\right) 2 \pi=k_{1}+2 \pi \cdot 0.1 d=k+2 \pi \cdot 0.1 d
$$

Similarly, $k_{3}=k_{2}+2 \pi \cdot 0.1 d$ and so on.

Accordingly, the lengths of the individual laps form an arithmetic progression, and thus their sum $l$ meters, i.e., $100 l \mathrm{~cm}$, so

$$
100 l=\frac{k_{1}+k_{1}+(n-1) 2 \pi \cdot 0.1 d}{2} n
$$

or

$$
2 \pi \cdot 0.1 d n^{2}+\left(2 k_{1}-2 \pi \cdot 0.1 d\right) n-200 l=0
$$

With numerical values: $0.628 n^{2}+1.372 n-300=0$.

This equation has real roots of opposite signs, of which only the positive one makes sense in the context of our problem, and it is $n \sim 20.8$ laps.

Accordingly, the number of laps is $20<n<21$. The last lap is not complete after the 20th lap. To find its length, let's calculate the length of 20 laps. The length of the 20th lap is

$$
\begin{gathered}
k_{20}=k_{1}+2 \pi \cdot 19 d=1+6.28 \cdot 19 \cdot 0.1=12.932 \mathrm{~cm} \\
S_{20}=10\left(k_{1}+k_{20}\right)=10 \cdot 13.932=139.32 \mathrm{~cm}
\end{gathered}
$$

The length of the last lap: $150-139.32=10.68 \mathrm{~cm}$.

Sommer György (Áll. Dobó István r. VI. o. Eger).