Well, whose turn is it? If White has the move he can mate immediately with the pawn. But if it’s White’s turn, then Black must just have moved, and there’s no legal move he can just have made. So it must be Black’s turn to move, and after 1. … axb6 the black pawn can scurry down the board to the queening square, just beyond the reach of the pursuing king. Black wins.

The Germans call this the Vielväter (“many fathers”) position, because the idea has inspired literally thousands of problems.

(From Fairy Chess Review.)

Now try the same challenge with a horse.

First, three observations: (1) The number of dollars that the men received for the herd of cattle was a perfect square. (2) The number of sheep they bought was odd, because it was necessary to throw in the dog in order to divide the animals evenly. (3) Since they bought an odd number of sheep at $10.00 per head, the next-to-last digit in the square payment from observation (1) must be odd.

Now, as it happens, in any perfect square, if the next-to-last digit is odd, the last figure must be a 6. The simplest way to see this is to consider that any number can be expressed as 10a + b, where a comprises all the digits except the last and b is the last digit (0 to 9). The square of this number is 100a² + 20ab + b². But 100a² has zeros in the last two places, and 20ab has a zero in the last place and is preceded by an even number. So, in order for the square to have an odd digit in the next-to-last position, b² must have an odd first digit, and the only possibilities are 4² = 16 and 6² = 36.

In either case, the dog cost $6.00, and the rancher owed his partner $2.00 for accepting a $6.00 dog in place of a $10.00 sheep.

The next number is 9. This is Golomb’s sequence, a non-decreasing sequence of integers in which a_n is the number of times that n occurs in the sequence. We’re given a₁ = 1 to start with, and this indicates that the number 1 appears only once in the whole sequence. That means that a₂ is greater than 1. Setting a₂ to 2 immediately tells us that 2 appears twice in the sequence, so a₃ also must be 2. That tells us that 3 occurs twice in the sequence, and so on. In this way each number in the sequence is uniquely determined.

Yes — here’s one possibility:

(From E.G. Kozlova, Skazki i podskazki: Zadachi dlya matematicheskogo kruzhka [Fairy Tales and Hints: Problems for a Mathematical Circle], 2004 [in Russian], via Jennifer Beineke and Jason Rosenhouse, The Mathematics of Various Entertaining Subjects, 2017.)

1. e4 e6 2. Bb5 Ke7! 3. Bxd7 c6 4. Be8 Kxe8.

(Thanks, Jacques.)

No, it’s not possible. Imagine that the corner sub-cubes are black and that any two sub-cubes that share a face are of different colors:

Because the mouse always travels orthogonally, it will visit black and white squares alternately. And because the large cube contains an odd number of sub-cubes, any such route that starts in a black sub-cube and tours the whole cube must end in a black sub-cube. But the center sub-cube is white. So the task is impossible.

(From “Mathematical Mayhem,” Crux Mathematicorum 28:1 [February 2002], 34.)

For special arc PQ, construct the smallest isosceles right triangle ABC with hypotenuse AB on line PQ that contains the arc. Let arc PQ touch the legs of this triangle at R and S. Now reflect arc PR in AC and arc SQ in BC, obtaining arcs RP’ and SQ’. Now arc PRSQ equals arc P’RSQ’ in length and lies between lines AP’ and BQ’, which are distance AB apart. The arc length is 1, and it’s not less than AB, which proves the theorem. “Wetzel concludes, ‘Pause and reflect!'”

(From Clayton W. Dodge, “Reflections of a Problems Editor,” in Joby Milo Anthony and Howard Whitley Eves, In Eves’ Circles, 1994.)