Mass Transit

A problem from the October 1964 issue of Eureka, the journal of the Cambridge University Mathematical Society:

The planet Kophikkup is in the shape of a torus or ring-doughnut. There is a direct mono-rail line from each of the four space-ports to each of the major cities. No lines join or cross. What is the greatest possible number of major cities? Draw a diagram for this case.

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Four. The map above shows that each of four spaceports can communicate directly with each of four cities on the surface of the torus, but there’s nowhere to “put” a fifth city that wouldn’t require some of the monorail lines to cross. (From the Eureka Digital Archive, published under a Creative Commons license.) 03/26/2024 Reader Drake Thomas sent a better diagram: (Thanks, Drake.)

March 22, 2024March 26, 2024 | Puzzles

“The Twenty-Four Monks”

During the middle ages there existed a monastery, in which lived twenty-four monks, presided over by a blind abbot. The cells of the monastery were planned as shown in the accompanying figure, passages being arranged along two sides of each of the outer cells and all round the inner cell, in which the abbot took up his quarters. Three monks were allotted to each cell, making, of course, nine monks in each row of cells. The abbot, being lazy as well as blind, was very remiss in making his rounds, but provided he could count nine heads on each side of the monastery he retired into his own cloister, contented and satisfied that the monks were all within the building, and that no outsiders were keeping them company. The monks, however, taking advantage of their abbot’s blindness and remissness, conspired to deceive him, a portion of their number sometimes going out and at other times receiving friends in their cells. They accomplished their deception, and it never happened that strangers were admitted when monks were out, yet there never were more nor less than nine persons upon each side of the building. Their first deception consisted in four of their number going out, upon which four monks took possession of each of the cells numbered 1, 3, 6, and 8, one monk only being left in each of the other cells; nine monks being thus on each side of the building. Upon returning, the four monks brought in four friends, when it was necessary to arrange the twenty-eight persons, two in each of the cells 1, 3, 6, and 8, and five in each of the others; still nine heads only were to be counted in either row. Emboldened by success, eight outsiders were introduced, and the thirty-two persons now were arranged, one only in each of the cells 1, 3, 6, and 8, but seven in each of the other cells; again, according to the abbot’s system of counting, all was well. In the next endeavour, the strangers all went away and took six monks with them, leaving but eighteen at home to represent twenty-four; these eighteen placed themselves five in each of the cells 1 and 8 and four in each of the cells 3 and 6; the remaining cells were empty, but the cells on each side of the building still contained nine monks. On returning, the six truants each brought two friends to pass the night, and the thirty-six retired to rest, nine in each of the cells 2, 4, 5, and 7; the remainder were empty, and the abbot was quite satisfied that the monks were alone in the monastery.

— Cassell’s Book of In-Door Amusements, Card Games and Fireside Fun, 1882

March 6, 2024March 5, 2024 | Puzzles

Target Practice

An arrow has a 1/4 chance of hitting its target. If four arrows are shot at one target, what’s the chance that the target will be hit?

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Each arrow has 3/4 chance of missing, so the chance that all four arrows miss is 3/4 × 3/4 × 3/4 × 3/4 = 81/256. The chance that at least one arrow hits the target is 1 – 81/256 = 175/256, or about 0.684.

March 3, 2024March 2, 2024 | Puzzles

Migration

chessboard

Here’s a checkerboard. Suppose we put a checker on each of the nine squares in the lower left corner. And suppose that any checker can move in any direction by jumping over an adjacent checker, provided that the square beyond it is vacant. Is there some combination of moves by which we can transfer the nine checkers to the nine squares in the upper left corner of the board?

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No. Each move carries a checker from one square to another of the same color. But in their starting array the nine checkers stand on five black and four white squares, and the destination contains five white and four black squares. So the task is impossible.

March 2, 2024March 1, 2024 | Puzzles

The Four Oaks Puzzle

https://commons.wikimedia.org/wiki/File:4_hrasta_-_1.jpg — Image: Wikimedia Commons

One other notable problem from Sam Loyd’s Cyclopedia of 5000 Puzzles: A father left to his four sons this square field, with the instruction that they divide it into four pieces, each of the same shape and size, so that each piece of land contained one of the trees. How did they manage it?

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Image: Wikimedia Commons

February 8, 2024February 7, 2024 | Puzzles

Three Hats

Donald Aucamp offered this problem in the Puzzle Corner department of MIT Technology Review in October 2003. Three logicians, A, B, and C, are wearing hats. Each of them knows that a positive integer has been painted on each of the hats, and each of them can see her companions’ integers but not her own. They also know that one of the integers is the sum of the other two. Now they engage in a contest to see which can be the first to determine her own number. A goes first, then B, then C, and so on in a circle until someone correctly names her number. In the first round, all three of them pass, but in the second round A correctly announces that her number is 50. How did she know this, and what were the other numbers?

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Howard Haber gave this solution in the March-April 2004 issue. The key observation to bear in mind is that if one of the logicians sees that both of her opponents’ hats bear the same number, she can conclude immediately that her own number is twice this value. Each of the logicians knows that her own number is either the sum or the positive difference of the other two. A, going first, reasons that the three numbers A, B, C must stand in one of two ratios, 1:2:3 or 5:2:3. She can’t decide between these possibilities, so she remains silent. But in the second round A has the benefit of knowing that each of the other logicians also remained silent in the first round. Now, suppose that the ratio had been 1:2:3. Then, when B took her turn in the first round, she would have concluded that the ratio must be either 1:2:3 or 1:4:3. Like A, she would have remained silent, unable to decide between these possibilities. But C, going next, would have determined that the ratio must be either 1:2:3 or 1:2:1 — and she could immediately have excluded the latter because in that case B would have been able to win the game by employing the key observation above. Noting that B didn’t do this would have allowed C to win the game herself by announcing her only remaining possibility, 1:2:3. The fact that C didn’t do this tells A that the ratio can’t be 1:2:3, and she concludes that it must be the other ratio she’d been considering, 5:2:3. “For the problem as stated, just multiply all numbers above by 10.”

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February 5, 2024February 4, 2024 | Puzzles

“Pythagoras’ Classical Problem”

https://commons.wikimedia.org/wiki/File:2_kvadrata_-_1.jpg — Image: Wikimedia Commons

A pretty dissection puzzle by Sam Loyd. Cut this figure into three pieces that will fit together to make a square.

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Image: Wikimedia Commons From Loyd’s Cyclopedia of 5000 Puzzles, Tricks, and Conundrums With Answers (1914).

February 4, 2024 | Puzzles

Card Algebra

From Henry Dudeney:

Take an ordinary pack of playing cards and regard all the court cards as tens. Now, look at the top card — say it is a seven — place it on the table face downwards and play more cards on top of it, counting up to twelve. Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile. Then look again at the top card of the pack — say it is a queen — then count 10, 11, 12 (three cards in all), and complete the second pile. Continue this, always counting up to twelve, and if at last you have not sufficient cards to complete a pile, put these apart. Now, if I am told how many piles have been made and how many unused cards remain over, I can at once tell you the sum of all the bottom cards in the piles. I simply multiply by 13 the number of piles less 4, and add the number of cards left over. Thus, if there were 6 piles and 5 cards over, then 13 times 2 (i.e. 6 less 4) added to 5 equals 31, the sum of the bottom cards. Why is this?

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Every pile must contain thirteen cards, less the value of the bottom card. Therefore, thirteen times the number of piles less the sum of the bottom cards, and plus the number of cards left over, must equal fifty-two, the number in the pack. Thus thirteen times the number of piles plus number of cards left over, less fifty-two, must equal sum of bottom cards. Or, which is the same thing, the number of piles less four, multiplied by thirteen, and plus the cards left over gives the answer as stated. The algebraically inclined reader can easily express this in terms of his familiar symbols. From Modern Puzzles and How to Solve Them, 1926.

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February 1, 2024January 31, 2024 | Puzzles

Self-Reference

A problem from the October 1959 issue of Eureka, the journal of the Cambridge University Mathematical Society:

A. The total number of true statements in this problem is 0 or 1 or 3.
B. The total number of true statements in this problem is 1 or 2 or 3.
C. The total number of true statements in this problem (excluding this one) is 0 or 1 or 3.
D. The total number of true statements in this problem (excluding this one) is 1 or 2 or 3.

Which of these statements are true?

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B and D are true, A and C false.

January 26, 2024January 25, 2024 | Puzzles

Coming and Going

Two runners start from the same point on a circular track and run at different constant speeds. If they run in opposite directions on the track, they meet after a minute. If they run in the same direction, they meet after an hour. What’s the ratio of their speeds?

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Dartmouth mathematician Peter Winkler writes, “This is simultaneously confusing and easy.” Let L be the length of the track. Now, when the two runners are going in opposite directions, they’ll meet when they’ve covered a combined distance L. When they’re going in the same direction, they’ll meet when the difference between the distances they’ve covered is L. So $\displaystyle \frac{(s+t)L}{(s-t)L} = 60,$ and we find that s/t = 61/59. From Dick Hess, Golf on the Moon, 2014, via Winkler’s excellent collection Mathematical Puzzles, 2021.

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January 21, 2024January 20, 2024 | Puzzles