It's not really magic though. Here's a simpler variation of the problem: Two buses make different circular routes from the same depot. One bus takes 3 hours t

It's not really magic though. Here's a simpler variation of the problem: Two buses make different circular routes from the same depot. One bus takes 3 hours to do a loop, the other takes 5. At what times from the start of the day will both buses be at the depot at the same time? you could write a loop like this: for t in range(0, 10000): if t % 3 == 0 and t % 5 == 0: print("same at", t) which if you run, you'd see: same at 0 same at 15 same at 30 same at 45 same at 60 same at 75 same at 90 and even if you knew nothing about math, you could see that and think, oh, so they just meet every 15, which happens to be 3 times 5... You don't need to count up one at a time and check remainders, you can just increment by 15 and get the exact same result.. for t in range(0, 10000, 15): print("same at", t) That's more or less what I did when I solved it.. saw how the problem simplified to finding the common multiples of numbers, which when they are prime is just the product. apply this process iteratively and you have the solution.