B5. Three wheels are pressed against each other so that they do not slip when rotating. The circumferences of the wheels are 14, 10, and \(6 \, \text{cm}\) respectively. Each wheel has an arrow pointing straight down. The largest wheel is set in motion, and the other wheels rotate along with it. They stop at the first moment when all three arrows point down again. A whistle sounds whenever one of the arrows points straight up.
![](https://cdn.mathpix.com/cropped/2024_04_17_88f9acd303931df7c94dg-2.jpg?height=668&width=622&top_left_y=201&top_left_x=1228)
If two or three arrows point up at the same time, only one whistle sounds. How many whistles sound in total?

## C-Problems

For the C-Problems, the answer is not the only thing that matters; a reasoning must also be provided that shows your answer is correct. Each correctly worked-out C-Problem is worth 10 points. Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper as well.

NOTE: Solve each C-Problem on a separate sheet of paper and submit the corresponding scratch paper per problem.
B5. 57

Consider the situation where we add a fourth wheel to the row with a circumference of $2 \mathrm{~cm}$. The smallest wheel rotates the fastest. When the wheel with a circumference of $14 \mathrm{~cm}$ has made one round, the fourth wheel has made 7 rounds. One round of the wheel with a circumference of $10 \mathrm{~cm}$ corresponds to 5 rounds of the fourth wheel, and one round of the wheel with a circumference of $6 \mathrm{~cm}$ corresponds to 3 rounds of the smallest wheel.

The total number of rounds the smallest wheel must make until we return to the initial situation must be a multiple of 3, 5, and 7. The smallest number that is a multiple of 3, 5, and 7 is 105. In total, the wheels from largest to smallest then make 15, 21, and 35 rounds. This means there are at most $15 + 21 + 35 = 71$ whistles, but there might be fewer, as it is possible that two arrows point upwards simultaneously.

Exactly halfway through the 105 rounds of the fourth wheel, the other wheels have made $7 \frac{1}{2}$, $10 \frac{1}{2}$, and $17 \frac{1}{2}$ rounds. This means they all point upwards. So, there are at least 2 whistles that we have counted too many.

Now we will examine how often the two largest wheels point upwards simultaneously. From the situation we had exactly halfway, the fourth wheel must then rotate a multiple of 5 and a multiple of 7 rounds (or rotate back). This situation occurs after $52 \frac{1}{2} - 35$ and $52 \frac{1}{2} + 35$ rounds of the fourth wheel, counted from the initial situation. In these situations, the arrow of the third wheel does not point upwards, since 35 is not a multiple of 3.

Similarly, we find that the first and third wheels point upwards simultaneously after $52 \frac{1}{2} - 42$, $52 \frac{1}{2} - 21$, $52 \frac{1}{2} + 21$, and $52 \frac{1}{2} + 42$ rounds of the fourth wheel. In these situations, the second wheel does not point upwards, since 21 and 42 are not multiples of 5.

Finally, the second and third wheels point upwards simultaneously after $52 \frac{1}{2} - 45$, $52 \frac{1}{2} - 30$, $52 \frac{1}{2} - 15$, $52 \frac{1}{2} + 15$, $52 \frac{1}{2} + 30$, and $52 \frac{1}{2} + 45$ rounds of the fourth wheel. In these situations, the first wheel does not point upwards, since 15, 30, and 45 are not multiples of 7.

In total, there are $2 + 4 + 6 = 12$ situations where exactly two of the arrows point upwards. This means we have counted 12 whistles too many. In total, there are $71 - 2 - 12 = 57$ whistles to be heard.

## C-problems