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

A Plate of 1,000 Cookies

https://commons.wikimedia.org/wiki/File:Chocolate-chip-cookie-crumbs-on-plate.png — Image: Wikimedia Commons

A puzzle by David B., a mathematician at the National Security Agency, from the agency’s May 2017 Puzzle Periodical:

Steve, Tony, and Bruce have a plate of 1,000 cookies to share. They decide to share them in the following way: beginning with Steve, each of them in turn takes as many cookies as he likes (they must take an integer amount, greater than or equal to 1), and then passes the plate clockwise (with Tony sitting to Steve’s left, and Bruce sitting to Tony’s left). Nobody wants to feel like he hogged too many cookies, so they all want to avoid being the player at the end who has taken the most cookies. Additionally, nobody wants to feel cheated by finishing with the fewest cookies. Finally, given that the previous two conditions are definitely met, or definitely cannot be met, each player would like to maximize the number of cookies he eats. The players’ objectives can be summarized as follows:

Objectives:

Have one player who has eaten more cookies than you, and one player who has eaten fewer cookies than you.
Eat as many cookies as possible.

Objective #1 takes infinite priority over Objective #2. Assuming that all players are perfectly rational, that they are all aware of each other’s rationality and objectives, and that they cannot communicate with each other in any way, how many cookies should Steve take to ensure he meets both objectives and how many cookies will Tony and Bruce take if Steve takes the winning amount?

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First, suppose Steve takes 335 cookies. We will argue that Tony can then win by taking 334 cookies. If Tony does take 334, Bruce will be left with a plate of 331 cookies to take from. No matter how Bruce plays from here, he cannot meet objective #1, because he will need at least 335 cookies to do so, and has only 331 to take from. Therefore, Bruce will maximize the number of cookies he eats, and take the remaining 331. Tony finishes with 334, the middle amount, so he meets objective #1, justifying his decision to take 334 cookies. (Note that if Tony had taken more than 334, Bruce would still have taken all the remaining cookies, but Tony would not have met objective #1, so taking 334 is in fact Tony’s best play.) Therefore, Steve will fail objective #1 if he takes 335 cookies. Furthermore, we can repeat the argument above if Steve takes more than 335 cookies, by always having Tony take 1 fewer cookie than Steve (or all remaining cookies, if this amount is less than the number Steve took). Now suppose instead that Steve takes 334 cookies. We will show that Tony cannot take any amount of cookies resulting in him (Tony) meeting objective #1. First, note that if Tony takes 333 or more cookies, Bruce will be unable to meet objective #1 (he will need at least 334, but there are at most 333 remaining), and so will take all remaining cookies, resulting in a “loss” for Tony. Now suppose instead that Tony takes x cookies, where x ≤ 332. Then there will be 666 – x cookies remaining. Suppose Bruce takes 333 cookies. This leaves 333 – x cookies on the plate. Even if Steve takes only 1 cookie each turn from now on, Bruce can ensure that he will have fewer cookies than Steve by also taking 1. Suppose Steve takes s cookies. Then on Tony’s next turn, he has 333 – x – s cookies to take from. Even if Tony takes all remaining cookies, total over his two turns, he will have taken 333 – x – s + x = 333 – s, which is at most 332 (since s ≥ 1), so Tony will definitely finish with fewer cookies than Bruce. Thus, if Tony takes 332 or fewer cookies, Bruce can win by taking 333 cookies, since this guarantees he will finish with less than Steve and more than Tony. Therefore, if Steve takes 334 cookies, Tony cannot meet objective #1, and so will take as many cookies as possible, leaving none for Bruce. This leaves Steve with the middle amount of cookies, so taking 334 cookies is in fact a winning strategy for Steve. As shown above, Steve cannot do any better than this. Steve: 334 Tony: 666 Bruce: 0

Click for Answer

First, suppose Steve takes 335 cookies. We will argue that Tony can then win by taking 334 cookies. If Tony does take 334, Bruce will be left with a plate of 331 cookies to take from. No matter how Bruce plays from here, he cannot meet objective #1, because he will need at least 335 cookies to do so, and has only 331 to take from. Therefore, Bruce will maximize the number of cookies he eats, and take the remaining 331. Tony finishes with 334, the middle amount, so he meets objective #1, justifying his decision to take 334 cookies. (Note that if Tony had taken more than 334, Bruce would still have taken all the remaining cookies, but Tony would not have met objective #1, so taking 334 is in fact Tony’s best play.)

Therefore, Steve will fail objective #1 if he takes 335 cookies. Furthermore, we can repeat the argument above if Steve takes more than 335 cookies, by always having Tony take 1 fewer cookie than Steve (or all remaining cookies, if this amount is less than the number Steve took).

Now suppose instead that Steve takes 334 cookies. We will show that Tony cannot take any amount of cookies resulting in him (Tony) meeting objective #1. First, note that if Tony takes 333 or more cookies, Bruce will be unable to meet objective #1 (he will need at least 334, but there are at most 333 remaining), and so will take all remaining cookies, resulting in a “loss” for Tony.

Now suppose instead that Tony takes x cookies, where x ≤ 332. Then there will be 666 – x cookies remaining. Suppose Bruce takes 333 cookies. This leaves 333 – x cookies on the plate. Even if Steve takes only 1 cookie each turn from now on, Bruce can ensure that he will have fewer cookies than Steve by also taking 1. Suppose Steve takes s cookies. Then on Tony’s next turn, he has 333 – x – s cookies to take from. Even if Tony takes all remaining cookies, total over his two turns, he will have taken 333 – x – s + x = 333 – s, which is at most 332 (since s ≥ 1), so Tony will definitely finish with fewer cookies than Bruce. Thus, if Tony takes 332 or fewer cookies, Bruce can win by taking 333 cookies, since this guarantees he will finish with less than Steve and more than Tony.

Therefore, if Steve takes 334 cookies, Tony cannot meet objective #1, and so will take as many cookies as possible, leaving none for Bruce. This leaves Steve with the middle amount of cookies, so taking 334 cookies is in fact a winning strategy for Steve. As shown above, Steve cannot do any better than this.

Steve: 334

Tony: 666

Bruce: 0

June 27, 2021June 26, 2021 | Puzzles

Evolution

I just ran across this anecdote by Jason Rosenhouse in Notices of the American Mathematical Society. In a middle-school algebra class Rosenhouse’s brother was given this problem:

There are some horses and chickens in a barn, fifty animals in all. Horses have four legs while chickens have two. If there are 130 legs in the barn, then how many horses and how many chickens are there?

The normal solution is straightforward, but Rosenhouse’s brother found an alternative that’s even easier: “You just tell the horses to stand on their hind legs. Now there are fifty animals each with two legs on the ground, accounting for one hundred legs. That means there are thirty legs in the air. Since every horse has two legs in the air, we find that there are fifteen horses, and therefore thirty-five chickens.”

(Jason Rosenhouse, “Book Review: Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles,” Notices of the American Mathematical Society, 67:9 [October 2020], 1382-1385.)

June 23, 2021 | Puzzles · Science & Math