1. Kf4 h8=N 2. Ke5 Ng6#.

Ingeniously, Abdurahmanovic now unveils a second puzzle: Play the first move of this solution, Kf4, and consider the resulting position as a new two-move helpmate.

And a third puzzle: Play the first move of that solution and again find a two-move helpmate.

And yet a fourth: Play the first move of that solution and find yet another two-move helpmate.

The solutions:

(after Kf4) 1. Kxg5 h8=B 2. Kh6 Be3#.

(after Kxg5) 1. Kxf6 h8=R 2. Kg7 Bd4#.

(after Kxf6) 1. Ke5 h8=Q# 2. Kxe4 Qd4#.

In all four solutions, White starts by promoting the pawn, and it promotes progressively to a knight, a bishop, a rook, and a queen!

White wins with the absurd-looking 1. Qa5!:

1. … bxa5 2. Bc5#
1. … Kf2 2. Ng4#
1. … Kd4 2. Qc3#
1. … f2 2. Qc3#

If Black moves the bishop, then the queen gives mate from d2. If he moves the a-pawn, then White captures the b-pawn with mate.

“[T]he key move — the queen unexpectedly leaving her strong position for one so odd and apparently weak — is one of the best points of the problem,” Laws writes, “and the general gossamery nature of the mates helps to keep the solver, though perhaps momentarily, off the scent; for when the squares immediately surrounding the Black king are attacked by slender means, the mind’s eye is liable to be confused or deceived, and the difficulty is thus increased.”

No. Any sequence of 12 consecutive integers is made up of six even and six odd numbers, so their sum must be even.

(James Tanton, “A Dozen Questions About a Dozen,” Math Horizons 14:4 [April 2007], 12-16.)

1. Rh8! and the bishop will mate.

From L’Alfiere di Re, 1921.

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.