a) In the figure below, two gears are meshed, one gear $A$ with 6 teeth and another gear $B$ with 8 teeth. All the teeth are the same size. If gear $A$ makes 12 revolutions, how many revolutions will gear $B$ make?

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-04.jpg?height=433&width=589&top_left_y=1695&top_left_x=615)

b) Consider 5 meshed gears. The first has 10 teeth and is meshed with a second gear with 20 teeth, which in turn is meshed with a third gear with 40 teeth, which is meshed with a fourth gear with 80 teeth, which is meshed with a fifth gear with 160 teeth. When the largest gear makes one revolution, what is the sum of the revolutions made by all the gears?

c) Consider three gears $C, D$, and $E$, where gear $C$ is meshed with gear $D$ and gear $D$ is meshed with gear $E$. Each of them has a certain number of teeth, all of the same size. It is known that when gear $C$ made 160 revolutions, gear $D$ made 1007 revolutions, and gear $E$ made 38 revolutions. What is the smallest total number of teeth on the three gears combined for this to be possible?
Solution

a) Let $N$ be the number of turns made by gear $B$. The reader can verify that in the contact region between the two gears, each tooth of gear $A$ is touched above by a tooth of gear $B$. The number of such contacts is equal to $12 \times 6$, since gear $A$ makes a total of 12 turns and has 6 teeth. On the other hand, since gear $B$ has 8 teeth and makes $N$ turns, this number is also equal to $N \times 8$. Thus, we obtain that $8 N=12 \times 6$, which gives us $N=9$.

b) Since the fifth gear has twice as many teeth as the fourth, for each turn it makes, the fourth gear makes 2 turns. Since the fifth gear makes 1 turn, the fourth makes 2.

Following the same reasoning, we see that for each turn made by one of the gears, the smaller gear connected to it makes 2 turns. Since the fourth gear makes 2 turns, the third makes 4 turns. Then the second makes 8 turns and the first makes 16 turns.

The total number of turns made by all the gears is then equal to $1+2+4+8+16=31$.

c) Let $N_{C}, N_{D}$, and $N_{E}$ be the number of teeth of gears $C, D$, and $E$ respectively. Since gear $C$ makes 160 turns while gear $D$ makes 1007 turns and gear $E$ makes 38 turns, a similar reasoning to that used in item a) shows that:

$$
160 N_{C}=1007 N_{D}=38 N_{E}
$$

Thus, the following equations hold:

$$
\begin{aligned}
& 160 N_{C}=1007 N_{D} \\
& 160 N_{C}=38 N_{E}
\end{aligned}
$$

Decomposing into prime factors, we find that $38=2 \times 19, 160=2^{5} \times 5$, and $1007=19 \times 53$. Thus, the first equation above gives us that:

$$
2^{5} \cdot 5 \cdot N_{C}=19 \cdot 53 \cdot N_{D}
$$

From this first equation, we see that $N_{C}$ must be a multiple of $19 \cdot 53$ because these prime factors appear on the right side of the equation but not in the term $2^{5} \cdot 5$. Similarly, $N_{D}$ must be a multiple of $2^{5} \cdot 5$. The smallest numbers that satisfy these conditions are $N_{C}=19 \cdot 53$ and $N_{D}=2^{5} \cdot 5$.

Substituting these values into the second equation, we get:

$$
2^{5} \cdot 5 \cdot 19 \cdot 53=2 \cdot 19 \cdot N_{E}
$$

which gives us $N_{E}=2^{4} \cdot 5 \cdot 53$.

Thus, the smallest number of teeth of gears $C, D$, and $E$ are 1007, 160, and 4240, respectively, with the smallest total sum being 5407.