Two pendulums are drawn aside and released at the same time. One has a time period of $$2\ \text{s}$$ and the other $$2.5\ \text{s}$$. What is the least time when they are at their extreme positions on opposite sides of equilibrium?

Both are released from the same side. The first pendulum is at an extreme every $$1\ \text{s}$$ (half its period), on the release side after an even number of half-periods and on the opposite side after an odd number. The second is at an extreme every $$1.25\ \text{s}$$, with the same rule. Both are at extremes together only at multiples of $$5\ \text{s}$$, and at $$t = 5\ \text{s}$$ the first has made 5 half-swings (opposite side) while the second has made 4 (release side), so they are on opposite sides. The least time is therefore $$5\ \text{s}$$.