Artur wants to draw a "spiral" 4 meters long, formed of straight segments. He has already traced seven segments, as shown in the figure. How many segments are still missing?
(a) 28
(b) 30
(c) 24
(d) 32
(e) 36
The correct option is (d).

The figure shows that the "spiral" is made up of segments whose lengths form a finite sequence of the form $1,1,2,2,3,3,4,4, \ldots, n, n$ (if the last two segments of the spiral have the same length) or, alternatively, of the form $1,1,2,2,3,3,4,4, \ldots$, $n, n, n+1$ (if the last two segments of the spiral have different lengths).
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-055.jpg?height=514&width=1310&top_left_y=1947&top_left_x=452)

The sum of the first $k$ natural numbers is given by

$$
1+2+3+\cdots+k=\frac{k(k+1)}{2}
$$

and the total length of the spiral is $4 \mathrm{~m}=400 \mathrm{~cm}$. Therefore,

$$
2 \times \frac{n(n+1)}{2}=1+1+2+2+3+3+\cdots+n+n=400
$$

or, alternatively,

$$
2 \times \frac{n(n+1)}{2}+n+1=1+1+2+2+\cdots+n+n+n+1=400
$$

so that $n(n+1)=400$ or $(n+1)^{2}=400$.

However, there are no two consecutive natural numbers whose product is 400, that is, the equation $n(n+1)=400$ has no solution. Thus, $(n+1)^{2}=400$, so $n+1=20$. Therefore, the last segment of the spiral is $20 \mathrm{~cm}$ and the second-to-last is $19 \mathrm{~cm}$. The lengths of the segments of the spiral form the sequence of numbers $1,1,2,2,3,3,4,4, \ldots, 19,19,20$. Thus, there are $19 \times 2+1=39$ segments. Since seven have already been drawn, 32 remain to be drawn.