## Arthur Christmas and the Least Common Multiple

I left Arthur Christmas and Grand-Santa in a hypothetical puzzle, inspired by the movie, with them stranded on a tiny island while their team of flying reindeer and sleigh carried on in a straight line without them. I am assuming for the sake of an interesting problem that this means the reindeer are carrying on the Great Circle route, favored by airplanes and satellites, and that the reindeer are in an orbit more like the satellite’s than the reindeers — that is, they keep to a circle in a plane which isn’t rotating while the Earth does, since otherwise, Arthur and Grand-Santa have to wait only for the reindeer to finish one lap around the planet and somehow get up to flying altitude to be picked up. If the reindeer aren’t rotating the with the Earth, then, when the reindeer finish one circuit our heroes are going to be … well, maybe east, maybe west, of the reindeer; the problem is, they’re going to be away.

This gets easier if we say how long it takes the reindeer to finish a loop. Since I’m just making this up — I’m inspired by but not following the exact plot points of the movie — we can say it takes them six hours to go all the way around. In that case, by the time the reindeer get back to where they lost Arthur and Grand-Santa, the pair have, with the rotation of the Earth, moved 90 degrees east. But that isn’t hopeless: in twelve hours, the reindeer will be back where they started and Arthur and Grand-Santa 180 degrees of longitude away, which seems to be making things more hopeless. Still, another six hours gets Arthur and Grand-Santa 270 degrees east, and on the next orbit that’s put 360 degrees of longitude between the reindeer and Arthur, which of course isn’t any difference at all. Fine for them, if they can hold out; not so good if they’re working under any deadlines.

The same holds if the reindeer need four hours to go around the world, or eight hours, or two hours, or so on. They’ll keep going around and around and after 24 hours, they and Arthur get back to the same spot. This works even if it takes longer than a day for the reindeer to finish a loop around the world: if they take 36 hours, well, Arthur and Grand-Santa need to hang around for 72 hours and right overhead will be, in principle, the ride home.

It might take a while: if the reindeer need five hours, they won’t meet up with Arthur and Grand-Santa for 120 hours, or five whole days. Worse, if the reindeer need 23 hours, 59 minutes, well, that’s 1,439 days, isn’t it, before they’re back overhead? I suppose if Arthur and Grand-Santa are lucky they fell on the equator, so it’ll be only half that time needed — right? Go ahead, see if I am — but still, that’s pretty rotten luck.

Could be worse still, though. Suppose it takes the reindeer $\sqrt{2}$ hours to loop around the world. In that case they’ll never be directly underneath the reindeer again, for reasons which are either obvious, or which I suggest you think about until they become obvious. This should come in a flash, and produce a grin when it hits you. Snapping of fingers is optional, but I recommend it because that flash of things becoming obvious is so much fun.

That Arthur and Grand-Santa have to wait forever for the reindeer to be overhead again looks desperate, but it’s not actually that bad.

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