The Interplanetary Alignment Celebration Day

There are many ways to solve this, I'm just posting one.

Let's call the time it takes for each planet to make a full revolution $T$ . Thus we have

$T_{in}=1.5$ , $T_{out}=2.5$

Let's now define the angular velocities $\omega_{in}$ and $\omega_{out}$ which can be derived from the circular motion formulas $\omega=\frac{2 \pi}{T}$ .

While the inner planet makes a full revolution, the outer only makes a fraction. If we watch the difference of angular velocities, we are watching at which speed the angular distance between planets is increasing. $\Delta \omega = \omega_{in}-\omega_{out}$

By multiplying this $\Delta \omega$ for the elapsed time $t$ we have the angular phase between the planets $\Delta \theta = \Delta \omega t$

We now just have to require this $\Delta \theta$ to be $2 \pi$ , so that the angular distance between the two planets has now become a 360° (or $2 \pi$ in radiants) turn. By doing this we now get

$\Delta \omega t= 2 \pi$ And replacing the espression for $\Delta \omega$ with the previous definitions we get $(\frac{2 \pi}{T_{in}}-\frac{2 \pi}{T_{out}})t=2 \pi$ where now $2 \pi$ cancels out and we get the general synodic period formula

$t=\frac{T_{in} T_{out}}{T_{out}-T_{in}}$

By putting given numbers in this formula we get $t_{celebration}=3.75 years$

Note that you can't just find the least common multiple, because by doing that you're requiring that the planets are aligned at exactly the same angle as they were before, but in general that is not the least amount of time it occurs between two alignments, which can take place at different angular positions.