We have two gears. By connecting them, the ratio of the rotational speeds of the driving and driven gears (the transmission ratio) is $7: 9$ or $9: 7$. By replacing one of the gears with a gear that has 3 fewer teeth and the other with a gear that has 3 more teeth, the transmission ratio becomes $1: 3$ or $3: 1$. - Which gear was replaced with one that has fewer teeth? How many teeth does each of the 4 gears have?
The ratio of the number of revolutions of the two wheels (over the same time period) being $7:9$ means that there is a positive integer $k$ such that the number of teeth on the first wheel is $9k$, and on the second wheel, it is $7k$. Indeed, with $63k$ teeth passing, the first wheel makes 7 complete revolutions, and the second wheel makes 9 complete revolutions. Therefore, the number of teeth on the wheels after the replacement is either $9k+3$ and $7k-3$, or $9k-3$ and $7k+3$.

In the first case, it is certain that $9k+3 > 7k-3$, so only

$$
(9k+3):(7k-3) = 3:1 \text{ is possible, from which } k=1
$$

In the second case, we must consider both the subcases $9k-3 > 7k+3$ and $9k-3 < 7k+3$, but neither yields a positive integer $k$:

$$
\begin{aligned}
& (9k-3):(7k+3) = 3:1 \text{ yields } k=-1 \\
& (9k-3):(7k+3) = 1:3 \text{ yields } k=3/5
\end{aligned}
$$

Therefore, the number of teeth is $k=1$, so 9 and 7, and the 9-toothed wheel was replaced with a 12-toothed wheel, and the 7-toothed wheel with a 4-toothed wheel.

Kócza Éva (Fót-Gyermekváros, specialized secondary school I. o. t.)

Note. Several people naturally assumed that "since the ratio after the replacement, 3:1 or 1:3, deviates more strongly from the value 1 than the original ratios 9:7 or 7:9, the only possible solution is to replace the wheels with fewer teeth with even fewer teeth." This needed to be proven. The above solution shows that indeed only this case is possible. If the increase or decrease in the number of teeth had not been the same, the mentioned "solvers" would have likely shown more caution.