Thus, the LCM of the periods is $ \frac124 $ minutes? No — correct interpretation: The time until alignment is the least $ t $ such that $ 48t $ and $ 72t $ are both integers and the angular positions coincide. Actually, the alignment occurs at $ t $ where $ 48t \equiv 0 \pmod360 $ and $ 72t \equiv 0 \pmod360 $ in degrees per rotation. Since each full rotation is 360°, we want smallest $ t $ such that $ 48t \cdot \frac360360 = 48t $ is multiple of 360 and same for 72? No — better: The number of rotations completed must be integer, and the alignment occurs when both complete a number of rotations differing by full cycles. The time until both complete whole rotations and are aligned again is $ \frac360\mathrmGCD(48, 72) $ minutes? No — correct formula: For two periodic events with periods $ T_1, T_2 $, time until alignment is $ \mathrmLCM(T_1, T_2) $, where $ T_1 = 1/48 $, $ T_2 = 1/72 $. But in terms of complete rotations: Let $ t $ be time. Then $ 48t $ rows per minute — better: Let angular speed be $ 48 \cdot \frac36060 = 288^\circ/\textsec $? No — $ 48 $ rpm means 48 full rotations per minute → period per rotation: $ \frac6048 = \frac54 = 1.25 $ seconds. Similarly, 72 rpm → period $ \frac512 $ minutes = 25 seconds. Find LCM of 1.25 and 25/12. Write as fractions: $ 1.25 = \frac54 $, $ \frac2512 $. LCM of fractions: $ \mathrmLCM(\fracab, \fraccd) = \frac\mathrmLCM(a, c)\mathrmGCD(b, d) $? No — standard: $ \mathrmLCM(\fracmn, \fracpq) = \frac\mathrmLCM(m, p)\mathrmGCD(n, q) $ only in specific cases. Better: time until alignment is $ \frac\mathrmLCM(48, 72)48 \cdot 72 / \mathrmGCD(48,72) $? No. - go-checkin.com