Black and White

gold chess problem

Henry Dudeney in Strand, June 1911: “It would be difficult to find a prettier little chess problem in three moves, produced from such limited material as a rook and a pawn, than the one given this month, by Dr. S. Gold. The novice will probably find the task of discovering the key move quite perplexing. White plays and checkmates in three moves.”

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Solution: 1. Rc2 b5 2. Kc7 (any) 3. Ra2# 1. Rc2 Ka8 2. Kc7 (any) 3. Ra2# 1. Rc2 Kb8 2. Kxb6 Ka8 3. Rc8#

September 27, 2021September 25, 2021 | Puzzles

The Switchback Puzzle

“Some years ago there was a craze for rolling pellet puzzles,” wrote Henry Dudeney in 1909, “though they are really more trials of patience than puzzles.”

One exception was this undulated glass tube, which contained three shots or pellets. The task was to get them into the three depressions at A, B, and C, which are unfortunately positioned at high points in the tube.

This “could be solved by a puzzle trick which I was surprised to notice how few people discovered,” Dudeney wrote. What was it?

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“The trick was to first reverse the tube so that the three depressions, D, E, and F, were at the bottom. It was quite easy to get the shots into these hollows, and when you had them in position you had merely to twist the tube with a quick turn of the fingers, holding it at the ends, when the pellets would fall into the required positions. You could hardly fail once in a hundred attempts, yet I have seen people try the puzzle for hours without success, while this simple trick never occurred to them.” (Henry E. Dudeney, “The World’s Best Puzzles,” Strand 36:216 [January 1909], 692-700.)

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September 20, 2021 | Puzzles

Made to Order

Arrange the digits 0-9 into a 10-digit number such that the leftmost n digits comprise a number divisible by n. For example, if the number is ABCDEFGHIJ, the number ABC must be divisible by 3, ABCDE must be divisible by 5, and so on.

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3816547290. (By Nob Yoshigahara.) 09/14/2021 UPDATE: I came across this puzzle in Yoshigahara’s 2004 book Puzzles 101: A Puzzlemaster’s Challenge. In an article in Quanta last year, Pradeep Mutalik credits it to John Horton Conway. I don’t know who came up with it first; possible the two devised it independently. (Thanks, Dharmesh.) 09/17/2021 UPDATE: Reader Ben Hamula has worked out programmatically that this solution is unique. 09/17/2021 UPDATE: This is getting interesting. Reader Hans Havermann finds that neither Yoshigahara nor Conway originated the problem. The earliest known mention is in the December 1984 issue of the New Zealand Mathematical Society newsletter (Problem 15 on page 17). David B. Gauld, mentioned in that problem, is sometimes cited as its author; Gauld contributed OEIS sequence A181736 in 2010. Immediately after the problem appeared in the New Zealand newsletter, Kishor N. Gordhandas posed it in the January 1985 issue of Games (page 2) and it generally escaped into the wild. Hans contacted Gauld, who recalls that he encountered 3816547290 in François Le Lionnais and Jean Brette’s 1983 book Les nombres remarquables and passed it on to Teddy Zulauf, sub-editor of the NZMS newsletter. Many thanks to Hans for all this yeoman sleuthing; I’ll post a further update if Le Lionnais turns out to cite a source. 09/30/2021 UPDATE: Le Lionnais cites no source but calls the fact “Trés remarquable.” See Hans’ blog for more on the search. 10/04/2021 UPDATE: It appears Hans has identified the originator of the puzzle: Lea Gorodisky, in the Spanish puzzle magazine The Snark. See the link immediately above. (Thanks again, Hans.)

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September 13, 2021October 4, 2021 | Puzzles

Black and White

ensor chess puzzle

From Henry Dudeney’s Perplexities column, Strand, March 1911:

“Here is a pretty little chess puzzle, made some years ago by Mr. F. S. Ensor. White has to checkmate the Black king without ever moving a queen off the bottom row, on which they at present stand. It is not difficult. As the White king is not needed in this puzzle, His Majesty’s attendance is dispensed with. His three wives can dispose of the enemy without assistance — in seven moves.”

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“Play the three queens as follows. There is no necessity to give the moves of the Black king, as they are all forced.” 1. Qe1 (“just a waste move”) 2. Qh1 3. Qhg1 4. Qgf1 5. Qce1 6. Qa1+ 7. Qeb1#

September 10, 2021September 9, 2021 | Puzzles

“A Match Puzzle”

This puzzle, by T.E. Maw of the Luton Public Library, appeared in the Strand in April 1911:

“Take fifteen matches and place them as shown in the diagram, then take away three, change the position of one, and the result will be the word showing what matches are made of.”

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September 9, 2021September 8, 2021 | Puzzles

Exam Week

A problem submitted by the United States and shortlisted for the 16th International Mathematical Olympiad, Erfurt-Berlin, July 1974:

Alice, Betty, and Carol took the same series of examinations. There was one grade of A, one grade of B, and one grade of C for each examination, where A, B, C are different positive integers. The final test scores were

Alice: 20
Betty: 10
Carol: 9

If Betty placed first in the arithmetic examination, who placed second in the spelling examination?

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Call the number of exams n. Then the total number of points distributed is n(A + B + C) = 20 + 10 + 9 = 39. Since A, B, and C are distinct, their sum must be at least 1 + 2 + 3 = 6, and the only combination of factors that can give us 39 is n = 3 and A + B + C = 13. Assume without loss of generality that A > B > C. We’re told that Betty placed first in arithmetic, so she received A points on that exam. But we also know that she received 10 points overall, which is fewer than A + B + C = 13, so she must have received C points in both remaining exams (one of which is spelling). Carol received 9 points overall, which is also less than A + B + C = 13. But she must have received at least B points on two of the exams (since Betty accounts for two Cs), one of which is spelling. If she’d received A points in the spelling exam, though, she’d have a total of at least A + B + C = 13. That’s a contradiction. So Carol must have scored B points in spelling, placing second. (It can be worked out that Alice, Betty, and Carol scored B + A + A, A + C + C, and C + B + B, respectively, and that A = 8, B = 4, and C = 1.) (From Dušan Djukic et al., The IMO Compendium: A Collection of Problems Suggested for the International Mathematical Olympiads: 1959-2004, 2006.)

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September 8, 2021September 7, 2021 | Puzzles

To Town

A problem from the 1949 problems drive of the Archimedeans, the mathematical society of Cambridge University:

A farmer lives in a cottage 4/17 of a mile from a main road. There is a lane leading from his farm to the nearest point Q on the road. The road is straight running north and south, and there is a village two miles south of Q at which he keeps a bicycle. He wishes to go to a town on the road four miles north of Q. He can walk across the fields surrounding the roads at 1 1/2 miles per hour, but along the roads he can walk at 3 1/2 miles per hour. He can cycle at 14 miles per hour. Should he collect his bicycle in order to get to the town from his farm as quickly as possible?

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He should collect the bicycle. By symmetry he can reach point P two miles north of Q in the same time as it takes to reach the village. He can cycle from the village to the town quicker than he can walk there from P. (From the October 1949 issue of the society’s journal, Eureka. The Eureka archive is published under a Creative Commons license.)

September 7, 2021September 6, 2021 | Puzzles

$50 for All

A puzzle by Ben H., a systems engineer at the National Security Agency, from the agency’s August 2016 Puzzle Periodical:

At a work picnic, Todd announces a challenge to his coworkers. Bruce and Ava are selected to play first. Todd places $100 on a table and explains the game. Bruce and Ava will each draw a random card from a standard 52-card deck. Each will hold that card to his/her forehead for the other person to see, but neither can see his/her own card. The players may not communicate in any way. Bruce and Ava will each write down a guess for the color of his/her own card, i.e. red or black. If either one of them guesses correctly, they both win $50. If they are both incorrect, they lose. He gives Bruce and Ava five minutes to devise a strategy beforehand by which they can guarantee that they each walk away with the $50.

Bruce and Ava complete their game and Todd announces the second level of the game. He places $200 on the table. He tells four of his coworkers — Emily, Charles, Doug and Fran — that they will play the same game, except this time guessing the suit of their own card, i.e. clubs, hearts, diamonds or spades. Again, Todd has the four players draw cards and place them on their foreheads so that each player can see the other three players’ cards, but not his/her own. Each player writes down a guess for the suit of his/her own card. If at least one of them guesses correctly, they each win $50. There is no communication while the game is in progress, but they have five minutes to devise a strategy beforehand by which they can be guaranteed to walk away with $50 each.

For each level of play — 2 players or 4 players — how can the players ensure that someone in the group always guesses correctly?

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The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game. Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards. The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies: Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8). Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5). Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6). Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7). In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

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The 2-person case is simpler. Let r denote a red card and b denote a black card. Let’s list Bruce’s card’s color first. Then the 4 distinct possibilities are rr, bb, rb, and br. The following strategy is foolproof. Bruce guesses that his card is the same color as Ava’s (covering the cases rr and bb). Ava takes the opposite approach and assumes that her card’s color is different from Bruce’s (covering the cases rb and br). One of them will necessarily be correct. If Bruce and Ava properly employ these strategies, they cannot lose the game.

Before diving into the 4-person case, let’s reexamine the 2-person case. We could interpret a red card as a 0 and a black card as a 1. Then Bruce’s strategy, as described above, handles the cases 00 and 11. His strategy could be interpreted as “the sum of our cards is even.” (0+0=0, 1+1=2) In this language, Ava’s strategy would be “the sum of our cards is odd.” Again, one of them must be right, no matter the cards. Note that Bruce and Ava have divided the sample space of all possible outcomes into 2 cases, based on the sum of the cards.

The 4-person case is more general and depends upon the notion of modular arithmetic, and in particular remainders when dividing by 4. Rather than purely guessing “the suits are the same,” or “the suits are different,” the players must be more clever. Let’s map each suit to a number. Clubs are 0, diamonds are 1, hearts are 2, and spades are 3. So, if Emily looks out and sees hearts (Charles), hearts (Doug), and spades (Fran), she interprets this as 2+2+3=7. Suppose that Emily’s card’s suit is clubs, which she obviously doesn’t know. Consider the following strategies:

Emily guesses her suit so that the total sum is a multiple of 4. She sees hearts, hearts, spades and guesses diamonds (2+2+3+1=8).
Charles guesses his suit so that the total sum is 1 more than a multiple of 4. He sees clubs, hearts, spades and guesses clubs (0+2+3+0=5).
Doug guesses his suit so that the total sum is 2 more than a multiple of 4. He sees clubs, hearts, spades and guesses diamonds (0+2+3+1=6).
Fran guesses her suit so that the total sum is 3 more than a multiple of 4. She sees clubs, hearts, hearts and guesses spades (0+2+2+3=7).

In this example, Fran guesses correctly because the total sum is 7, which is 3 more than a multiple of 4. For any combination of suits, one of them must be right. After all, the total sum must be a whole number. And all whole numbers have remainder 0, 1, 2, or 3 when divided by 4. As long as each of them sticks to his/her strategy, the group can ensure they won’t lose the game. Interestingly, it doesn’t matter which numbers they choose to use, as long as they all have different remainders when divided by 4. So, choosing any four consecutive whole numbers is a sufficient mapping.

August 27, 2021August 26, 2021 | Puzzles

Paint Job

chessboard

Given a standard chessboard, you can choose any rank or file and repaint each of its squares to the opposite color (white squares turn black, and black squares turn white). By doing this repeatedly, is it possible to produce a board with 63 white squares and one black square?

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No. Repainting a rank or file in this way transforms its k black and (8 – k) white squares into (8 – k) black and k white squares. The number of black squares changes by (8 – k) – k = 8 – 2k. This is an even number, which means that the parity of the number of black squares is preserved. Since we start with an even number (32) of black squares, there is no way to reach the odd outcome 1. (From Viktor Prasolov’s Problems in Plane and Solid Geometry, 1988.)

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