The whip.

John lives in Pensacola, Fla., which is in the Central time zone. Mary lives in Ontario, Ore., which is in the Mountain time zone.

At 2 a.m. on the first Sunday in November, John sets his clock back 1 hour to standard time. One hour will pass before Mary too sets her clock back. So for one hour each year, they observe the same time.

From Frank Morgan, The Math Chat Book, 2000.

See also Four-Dimensional Basketball, Pop Fly, and A Freak of Navigation.

1. … dxc5
2. (any) cxb4
3. (any) bxa3
4. (any) axb2#

The answer is not 45 mph (with the idea that 15 + 45 / 2 = 30). In that event the man would cover 2 miles in 1/15 + 1/45 of an hour, for an average speed of only 22.5 mph.

In fact he can’t possibly drive fast enough. To average 30 miles per hour, he would have to cover the 2 miles in 1/15 hour, or 4 minutes. But he’s already consumed 4 minutes in driving to the top of the hill (1 mile / 15 miles per hour = 1/15 hour, or 4 minutes). So it can’t be done.

It’s twice as long. You had to climb three flights to get from the first floor to the fourth; now you must climb six flights to get from the first to the seventh.

That was the solution intended by Russian mathematician Boris Kordemsky, who invented the puzzle. It’s not valid in Britain, interestingly, which counts the ground-level story as the “ground floor,” the one above that as the first, etc. Evidently Russians and Americans have more in common than we thought.

Each door has two sides, so the total number of door sides is even.

Since the number of door sides inside the house is even, the number on the exterior must be even as well.

From L.A. Graham, Ingenious Mathematical Problems and Methods, 1959.

The trains are traveling at 100 mph relative to one another, so one hour before they meet they’ll be 100 miles apart.

I don’t have Bennett’s solution, but here’s mine:

Take the simplest case, with three people, A, B, and C. Each can see smudges on his friends’ foreheads, so all are laughing. A reasons: Suppose my own forehead is clean. Then B can see only one smudge, the one on C’s forehead. But B sees that C is laughing. This should prove to B that his own forehead must have a smudge, and he should wipe it off. He isn’t doing so. This must mean that this whole supposition — that my own forehead is clean — is false.

With that realization, A wipes his own forehead. And at that point his supposition comes true, and B cleans his own forehead, following the reasoning given above. Now C sees that A and B are laughing, and that they both have clean foreheads. This tells C that his own forehead must be smudged, and he cleans it.

So that’s the case with three people. Now start over and add D. D can think: If my own forehead were clean, all of the above reasoning would hold and A would clean his forehead. He isn’t doing so. That must mean that my own forehead is dirty. D cleans his forehead and everything else plays out as above.

Then similarly you can add E, and F, and as many additional people as you like, following the same idea.