Black and White

rottigni chess problem

By G.E. Rottigni. White to mate in two moves.

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1. Rh8! and the bishop will mate. From L’Alfiere di Re, 1921.

October 1, 2019 | Puzzles

Podcast Episode 266: Lateral Thinking Puzzles

Here are seven new lateral thinking puzzles — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

September 30, 2019October 7, 2019 | Podcast · Puzzles

The Two Squares Puzzle

Lee Sallows just sent me this — the puzzle is difficult, but the solution is stunning:

sallows two squares puzzle

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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!

September 28, 2019 | Puzzles

Query

What’s remarkable about this set of words?

BIOLOGY
DEATHLY
SLOSHED
BASTARD
SELVAGE
FISSILE
DALLIES
FLAVORS

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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.)

September 27, 2019September 27, 2019 | Puzzles

Keeping Distance

For which values of n can n points be placed on a sphere so that all of them are equidistant from each other?

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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).

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September 20, 2019September 19, 2019 | Puzzles

Some Two

Given 52 integers, prove that it’s always possible to find some two of them whose sum or difference is evenly divisible by 100.

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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.

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September 17, 2019September 16, 2019 | Puzzles

Black and White

friedländer chess problem

From Louis Friedländer’s 1919 collection Chess Poetry. White to mate in two moves.

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

September 11, 2019 | Puzzles

Black and White

van dehn chess puzzle

A remarkable thematic chess puzzle by Bodo Van Dehn, 1951. White to move and win.

The solution is 10 moves long, but all Black’s moves are forced. (That’s a very valuable hint.)

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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.

August 21, 2019August 20, 2019 | Puzzles

A Flea’s Journey

https://commons.wikimedia.org/wiki/File:Euclid_Tetrahedron_4.svg — Image: Wikimedia Commons

A flea sits on one vertex of a regular tetrahedron. He hops continually from one vertex to another, resting for a minute between hops and choosing vertices without bias. Prove that, counting the first hop, we’d expect him to return to his starting point after four hops.

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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.

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August 17, 2019 | Puzzles

The Magic Total

magic total

Each of the 36 numbers in this table is the sum of the numbers at the head of its column and at the left of its row. For example, 3 = 2 + 1 and 13 = 5 + 8. The six bold numbers have been chosen so that each of them falls in a different row and a different column. The underlined numbers were chosen in the same way. But each of these two sextets produces the same total: 16 + 6 + 5 + 14 + 8 + 8 = 8 + 10 + 7 + 8 + 10 + 14 = 57. In fact, any six numbers chosen in this way will produce the total 57. Why is this?

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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.

August 7, 2019August 6, 2019 | Puzzles