21. There are four gears, A, B, C, and D, with the number of teeth being 84, 36, 60, and 48, respectively. How many revolutions will each gear make before the teeth that are currently engaged mesh together again?
21. Solution: When the teeth that are engaged simultaneously come back to engage again, each gear has turned an integer number of revolutions. Therefore, the number of teeth turned is the least common multiple of the number of teeth on each gear. Since
$$\begin{array}{l}
84=2^{2} \times 3 \times 7 \\
36=2^{2} \times 3^{2} \\
60=2^{2} \times 3 \times 5 \\
48=2^{4} \times 3
\end{array}$$

Thus, $\{84,36,60,48\}=2^{4} \times 3^{2} \times 5 \times 7=5040$. Also,
$$\frac{5040}{84}=60, \quad \frac{5040}{36}=140, \frac{5040}{60}=84, \quad \frac{5040}{48}=105$$

Answer: The number of revolutions each gear turns is: Gear A 60 revolutions, Gear B 140 revolutions, Gear C 84 revolutions, Gear D 105 revolutions.