Black and White

baird chess problem

By Edith Baird. White to mate in two moves.

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1. Bf4! 1. … exf4 2. Qe8# 1. … dxe3 2. Qa4# 1. … f2 2. Kxc7# 1. … d2 2. f6# From Field, July 16, 1898.

August 6, 2021August 5, 2021 | Puzzles

Tank Hunt

A puzzle from Daniel J. Velleman and Stan Wagon’s excellent 2020 problem collection Bicycle or Unicycle?:

Before you is a field of 225 squares arranged in a 15×15 grid. One of the squares contains a perfectly camouflaged tank that you’re trying to destroy. You have a weapon that will destroy one square of the grid with each shot, but it takes two shots to destroy the tank, and you know that when the tank has been hit the first time (and only then) it will flee invisibly to an adjacent square (horizontally or vertically). What’s the minimum number of shots you’ll need to be sure of destroying it?

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Imagine the field as a checkerboard with white corners. Shoot the 112 black squares, then the 113 white squares, then the black squares again. Two of these shots must hit the tank; in the worst case this will take 337 shots. To see that this is best, imagine covering the field with 112 dominoes, leaving one square uncovered, and suppose that the tank is restricted to its domino. We can study each domino independently: Each will require three shots, and the uncovered square requires a shot as well. So the lower bound is 3 × 112 + 1 = 337.

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August 4, 2021August 3, 2021 | Puzzles

The Wine-Bin

A puzzle by Claude Gaspar Bachet de Méziriac, from 1612, via Henry Dudeney:

A gentleman had a wine-bin of eight compartments, as in the illustration, containing 60 bottles, arranged as shown. His dishonest servant stole 4 bottles and rearranged the remainder. The gentleman noticed that the bottles had been redistributed, but as there were still 21 bottles on every side he innocently concluded that all the 60 were there. The servant, emboldened by his success, again stole 4 bottles and rearranged the remainder without discovery. In fact, on two more occasions he repeated his theft of 4 bottles, always leaving the remainder so arranged symmetrically that there were 21 on every side. How did he arrange them on the four occasions so as to steal the 16 bottles?

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(Henry E. Dudeney, “The World’s Best Puzzles,” Strand 36:216 [January 1909], 692-700; Henry E. Dudeney, “The World’s Best Puzzles — Solutions,” Strand 36:217 [February 1909], 113-116.)

August 2, 2021August 1, 2021 | Puzzles

Implementia

A puzzle by Yoshinao Katagiri: A boy and a girl played rock paper scissors 10 times. Altogether the boy played rock three times, scissors six times, and paper once, and the girl played rock twice, scissors four times, and paper four times (though, in each case, the order of these plays is unknown). There were no ties. Who won?

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Since there were no ties, each time the boy played scissors the girl must have played rock or paper. The girl’s record shows us that he lost two of those six rounds and won four. Likewise, in the four rounds in which the girl played scissors, the boy played rock three times and paper once. She lost three times and won once. Altogether the boy won seven rounds and lost three. (Via Nobuyuki Yoshigahara’s Puzzles 101: A Puzzlemaster’s Challenge, 2004.)

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July 28, 2021July 27, 2021 | Puzzles

Fixing a Point

A problem proposed by Richard Hoshino and Sarah McCurdy for Crux Mathematicorum, September 2008:

Five points lie on a line. Here are the 10 distances between pairs of points, listed from smallest to largest:

2, 4, 5, 7, 8, k, 13, 15, 17, 19

What’s k?

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This solution is by Gustavo Krimker of Universidad CAECE, Buenos Aires. The distance between the first and fifth points is 19. That means that 19 is also the sum of each of three pairs of intermediate distances: (the distance from the first point to the second point) + (the distance from the second point to the fifth point) = 19 (the distance from the first point to the third point) + (the distance from the third point to the fifth point) = 19 (the distance from the first point to the fourth point) + (the distance from the fourth point to the fifth point) = 19 Looking over the list of distances we’ve been given, the only pairs that add to 19 are (2, 17) and (4, 15). So those pairs correspond to two of the bulleted sums above. What’s the third pair? It must be (7, k) or (8, k). (It can’t be (5, k) or (k, 13) because we know that k falls between 8 and 13 and so those pairs can’t be made to total 19.) And if the third case is (7, k) or (8, k), then k must be 11 or 12. Now, 17 is the second-greatest distance on the list; that means it’s the distance between one of the end points in the row of five and the penultimate point on the opposite side. Either way, that means that 17 is the sum of two pairs of intermediate distances, using the same reasoning as above. One of those pairs is (4, 13), and the other must be (5, k), (7, k), or (8, k). That means that k must be 12, 10, or 9. The only one of these that matches a candidate above is 12. (Ian VanderBurgh, “Mathematical Mayhem,” Crux Mathematicorum 34:5 [September 2008], 275-286.)

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July 20, 2021July 19, 2021 | Puzzles

Black and White

carney chess problem

By E. Carney Jr. White to mate in two moves.

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1. Rb3! and the queen will mate. From the Southern Trade Gazette, via Charles F. Wadworth’s Unique Chess Problems, 1890. 07/20/2021 UPDATE: This is cooked. 1. Qd6+ works just as well and is simpler. (Thanks, Travis.) 07/20/2021 UPDATE: Blimey, doubly cooked! 1. Rc3 works too. (Thanks, Chris.) And the pawn seems to have no role in any of this, not even in Carney’s intended solution. Not sure how Wadworth can have chosen this without noticing the errors — but I guess I just did the same thing!

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July 19, 2021July 20, 2021 | Puzzles

Enlightenment

You’re in a dark room. The only light comes from an old LED digital alarm clock with four seven-segment displays. The time is displayed in 24-hour format, HH:MM (no seconds), and the leading digit is blank if not used. How much time passes between the room’s darkest state and its lightest?

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18 hours 57 minutes (1:11 to 20:08). (Thanks, Nick.) 07/16/2021 UPDATE: Properly speaking the interval is one minute less than that, because it extends from the last moment of 1:11 to the first of 20:08. Also, given my wording, the answer might equally be 5 hours 3 minutes, as that’s the interval “between” 20:08 and 1:11. (Thanks, Michael and Joe.)

July 15, 2021July 16, 2021 | Puzzles

Black and White

From La Nuova Rivista degli Scacchi. White to mate in two moves.

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1. Qa7! 1. … Kg5 2. Qe3#. 1. … Kxg3 2. Ne2#. 1. … Ke5 2. Nd3#. 1. … Ne5 2. Qe3#. On any other knight move, 2. Qc7 is mate. Via Arthur Ford Mackenzie, Chess: Its Poetry and Its Prose, 1887.

July 2, 2021July 2, 2021 | Puzzles

Harmony

A problem from the January 1990 issue of Quantum: Forty-one rooks are placed on a 10 × 10 chessboard. Prove that some five of them don’t attack one another. (Two rooks attack one another if they occupy the same row or column.)

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Roll the board into a cylinder and paint each of its diagonals a different color. These 10 regions contain 41 = 4 × 10 + 1 rooks, so at least one region must contain 5 rooks. And rooks on the same diagonal don’t attack one another. (Alexander Soifer and Edward Lozansky, “Pigeons in Every Pigeonhole,” Quantum 1:1 [January 1990], 25-28, 32.)

June 29, 2021June 28, 2021 | Puzzles

Quickie

One other quick item from Eureka, the journal of the Cambridge University Mathematical Society:

In its 1947 problem drive, the society proposed the following problem:

To find unequal positive integers x, y, z such that

x³ + y³ = z⁴.

“Although there were some research students in Theory of Numbers among those who tried, not one person succeeded in solving it within the time, yet the solution is extremely simple.” What is it?

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Take the equation 2³ + 3³ = 35 and multiply by 35³, getting 70³ + 105³ = 35⁴. (“Diogenes,” “Note on a Diophantine Equation,” Eureka 10 [March 1948], 5.)

June 28, 2021June 27, 2021 | Puzzles