Lee writes, “That such a thing could exist tends to defy belief.” I’ve walked through it and the two squares do indeed inventory one another’s contents. Well done!

Any two words in the set have exactly one “crash” — one letter that’s in the same position in each word.

(A. Ross Eckler, “Every Word-Pair in This Set Has One Crash,” Word Ways 32:1 [February 1999], 3-7.)

Roy Sinclair posed this question in the November-December 2000 issue of MIT Technology Review. Bruce Layton offered a solution in the May 2001 issue: The values are 1, 2, 3, and 4. For n = 1 there are no distances to be unequal, and for n = 2 there’s only one distance, with nothing to be unequal to. The cases of n = 3 and n = 4 can be shown by choosing two points on the sphere’s surface and centering on each of them a sphere whose radius is the distance between the two points (so that now each of the two points is on the surface of one constructed sphere and at the center of the other). Any point that falls at the intersection of these two spheres and on the surface of the original sphere satisfies the requirement. There can be one or two such points, meaning n can be 3 or 4, but not higher. If it’s 4 then the four points are the vertices of a regular tetrahedron; if it’s 3 then the original two chosen points and one of the intersection points form an equilateral triangle (roughly, on Earth, reader Eugene Sard suggested that three such points fall at the South Pole, on Wake Island, and in southern Saudi Arabia).

Each of the 52 integers can be expressed as 100a + b, where b can range from -49 to 50. That means that the absolute value of b can take only 51 possible values (0 to 50), so at least two of our 52 integers must share one such value. If these two integers have opposite signs associated with b, then the sum of those integers is divisible by 100; if they have the same sign, then their difference is divisible by 100.

I found this in “The Olympiad Corner” in the September 2003 issue of Crux Mathematicorum; reportedly it appeared originally in the Russian Mathematical Olympiad, but the date isn’t known.

After 1. Nd2! one of the black knights must relax its guard of b3 or c4, and White can mate.

1. a8=N d3 2. Nb6 cxb6 3. c7 b5 4. c8=N b4 5. Nd6 exd6 6. e7 d5 7. e8=N d4 8. Nf6 gxf6 9. g7 f5 10. g8=N#

Four knight underpromotions, and three knight sacrifices!

You can see this played out on the excellent ChessNetwork YouTube channel here.

This appeared originally in the Journal of Recreational Mathematics in July 1969, posed by David L. Silverman. Benjamin L. Schwartz of McLean, Va., sent in three different arguments, the simplest of which is purely heuristic. Imagine that the flea is on a dodecahedron (20 vertices), and that he hops around it for a long period of time, say 10,000 hops. He’ll occupy each vertex for approximately the same time, in this case 500 visits. “Then obviously, his average time between calls at that vertex is 20, the number of vertices.”

More formal proofs are possible; see R. Robinson Rowe, “Random Hops on Polyhedral Edges and Other Networks,” Journal of Recreational Mathematics 4:2 [April 1971], 124-130, as well as “Solutions to Problems” in that issue.

Each number in the table is just the sum of its row and column headers, so choosing six numbers in this way is equivalent to just adding up all the headers: (7 + 2 + 3 + 6 + 7 + 5) + (2 + 9 + 8 + 3 + 4 + 1) = 57.

From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

Any pair of squares that are a knight’s move apart are of different colors — one is white, one black. But any “knight’s tour” on our m × n board will visit mn squares, and mn is odd (since it’s the product of two odd numbers). That means that the knight’s starting and ending squares will be of the same color, and hence can’t be a knight’s move apart.