The key is that the number of fingers can tell us uniquely the number of Martians. That’s a tall order. It eliminates, say, 246 fingers, because that’s too ambiguous: There might be 82 Martians with 3 fingers each, or 3 Martians with 82 fingers each, and so on. The only possibility that avoids this uncertainty is that the quantity of Martians and the quantity of fingers per Martian are expressed by the same number, and that this number is not composite. That means we’re looking for the square of a prime number, and the only such number in the range 200-300 is 289, or 17². So there are 17 Martians, each of which has 17 fingers. This is the only number between 200 and 300 that permits such a definite assertion.

1. Rd6! and White will mate with the knight or the bishop.

From Der Schachfreund, 1898.

Invert the smaller triangle and it becomes clear that the ratio is 4 to 1.

The Astronomer Royal. Six months ago we were all on the other side of the sun.

After 1. Qh8! Black must move his knight, allowing 2. e6#.

(If 1. … Nd4 or 1. … Ng7, then 2. Qa8#.)

From the Ottawa Citizen, 1902.

Wrap the whole group in an imaginary film. The film takes the shape of parts of cylinders, parts of planes, and parts of planets. The parts of the planets that contact the film are precisely those that are invisible to the other planets; a planet floating inside the group is entirely visible.

The invisible regions on the planets are bordered by spherical arcs, parallel two by two. If we translate all these arcs onto a single planet, they’ll partition it into several parts, each corresponding to the invisible region of some planet. Together these parts cover the planet completely.

A Soviet problem from the 22nd International Mathematical Olympiad in 1981. Here’s the same idea in two dimensions.

I create a gap between us by slowing down, then break the speed limit in catching up.

With the waiting move 1. Kh2! White forces Black to abandon some vital defense:

If the Nb3 moves, then 2. b4#.
If 1. … Nb5, then 2. Qa6#.
If the Nd6 moves elsewhere, then 2. Nb7#.
If 1. … Bb5, then 2. Qc7#.
If the black bishop moves elsewhere along the a6-f1 diagonal, then 2. Nxb3#.
If the black bishop moves along the a2-g8 diagonal, then 2. Qa6#.

Note that the white king’s move must be to h2 — any other move leaves him exposed to a check by Black.

Yes. The top of this trapezoid represents the unstretched snake, and the bottom represents the stretched snake. Every point on the top corresponds to one on the bottom, as denoted by the green lines. Since at least one of these lines must be perpendicular to both top and bottom, there must be some point on the snake’s body that does not move.