A Puzzling Exit

Canadian doctor Samuel Bean created a curious tombstone for his first two wives, Henrietta and Susanna, who died in succession in the 1860s and are buried side by side in Rushes Cemetery near Crosshill, Wellesley Township, Ontario. The original stone weathered badly and was replaced with this durable granite replica in 1982. What does it say?

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Cemetery caretaker John Hammond decoded the inscription in 1947, some 80 years after the women were interred. The message begins at the I just west of the center; it starts in a counterclockwise spiral but then begins to zigzag: IN MEMORIAM HENRIETTA 1ST WIFE OF S BEAN M. D. WHO DIED 27TH SEP 1865 AGED 23 YEARS 2 MONTHS & 17 DAYS & SUSANNA HIS 2ND WIFE WHO DIED 27TH APRIL 1867 AGED 26 YEARS 10 MONTHS & 15 DAYS 2 BETTER WIVES 1 MAN NEVER HAD THEY WERE GIFTS FROM GOD BUT ARE NOW IN HEAVEN MAY GOD HELP ME S. B. TO MEET THEM THERE

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November 7, 2014November 23, 2014 | Death · Puzzles

Asking Directions

You’re a logician who wants to know which of two roads leads to a village. Standing nearby, inevitably, are three natives: one always lies, one always tells the truth, and one answers randomly. You don’t know which is which, and you can ask only two yes-or-no questions, each directed to a single native. How can you get the information you need?

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Ask native A, “Is B more likely than C to tell the truth?” Then go to the native he indicates is less trustworthy and ask, “If I were to ask you whether this road goes to the village, would you say yes?” This procedure ensures that you won’t put the second question to the random answerer. Once that’s sure, then you’re just facing the classic version of the puzzle, with one liar, one truth-teller, and one yes-no question. From mathematical physicists Vladas Sidoravicius and Senya Shlosman, via Dartmouth mathematician Peter Winkler’s “Seven Puzzles You Think You Must Not Have Heard Correctly,” from the seventh Gathering for Gardner, March 2006. Obligatory John Finnemore link:

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November 6, 2014February 20, 2017 | Puzzles

Black and White

loyd chess problem

By Sam Loyd. White to mate in two moves.

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1. Qh8! and the queen will mate. From Frank Leslie’s Illustrated Newspaper, 1858.

October 30, 2014November 9, 2014 | Puzzles

Hot and Cold

Suppose you have three identical Dewar flasks labeled A, B, and C. (A Thermos is a Dewar flask.) You also have an empty container labeled D, which has thermally perfect conducting walls and which fits inside a Dewar flask.

Pour 1 liter of 80°C water into flask A and 1 liter of 20°C water into flask B. Now, using all four containers, is it possible to use the hot water to heat the cold water so that the final temperature of the cold water is higher than the final temperature of the hot water? How? (You can’t actually mix the hot water with the cold.)

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Pour half the cold water into D and put D into the hot flask (A). The final temperature in both A and D will be about 60°C. Empty D into C and the repeat the procedure, pouring the remaining half of the cold water in B into D and immersing it in the hot water. Now the final temperature of the water in A and D will be about 47°C. Empty D again into C; the final temperature of the mixture there will be about 53°C. So in two operations we’ve warmed the cold water to 53°C and cooled the hot water to 47°C. From Christopher Jargodzki’s books Science Brain-Twisters (1976) and Mad About Physics (2002).

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October 24, 2014November 2, 2014 | Puzzles · Science & Math

A Dice Puzzle

Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colors on the second die?

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Throwing two six-sided dice produces 36 possible outcomes. If Timothy and Urban have equal chances of winning, then there must be 18 outcomes in which both dice display the same color. If x is the number of red faces on the second die, then: 18 = 5x + 1(6 – x) x = 3 The second die must have 3 red faces and 3 blue faces. (From Pierre Berloquin, 100 Numerical Games, 1994.)

October 23, 2014November 3, 2014 | Puzzles · Science & Math

Black and White

Herlin chess problem

By Theodore Adrien Louis Herlin. White to mate in two moves.

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1. Qe8! From Schachzeitung der Berliner Schachgesellschaft, 1852.

October 18, 2014October 26, 2014 | Puzzles

Weighty Matters

From the Second All Soviet Union Mathematical Competition, Leningrad 1968:

On a teacher’s desk sits a balance scale, on which are a set of weights. On each weight is the name of at least one student. As each student enters the classroom, she moves all the weights that bear her name to the other side of the scale.

Before any students enter, the scale is tipped to the right. Prove that there’s some set of students that you can let into the room that will will tip the scale to the left.

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Make a list of all possible subsets of the students, including the empty set and the full set, and suppose you let in all of these groups. Each weight would spend half its time in the left pan and half in the right, so the total weight for each pan is the same. Since the empty set produces a tip to the right, there must be some other set that produces a tip to the left.

October 16, 2014October 26, 2014 | Puzzles

Stretch Goals

stretch goals puzzle

Two circles intersect. A line AC is drawn through one of the intersection points, B. AC can pivot around point B — what position will maximize its length?

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Call the other intersection point D and draw AD and DC. All angles inscribed in a circle and subtended by the same chord are equal, so angle BAD retains the same measure as A travels around its circle. Similarly, angle BCD remains the same as C travels around its circle. This means that triangle ADC will always have the same shape: As line AC pivots around B, triangle ADC will vary in size but remain self-similar. So which position will maximize its size? AD and DC are chords of their respective circles, and the longest chord is a diameter. So turn the triangle until both of these lines are diameters (this will happen simultaneously); at that point triangle ADC will reach its maximum size and line AC its maximum length. From Mogens Larsen in Richard Guy and Robert Woodrow, eds., The Lighter Side of Mathematics, 1994.

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October 13, 2014October 19, 2014 | Puzzles · Science & Math

Black and White

morphy chess problem

A chess problem by Paul Morphy. White to mate in two moves.

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1. Ra6! From the New York Clipper, 1856.

October 9, 2014October 19, 2014 | Puzzles

Dirty Work

A puzzle by Pierre Berloquin:

Timothy, Urban, and Vincent are digging identical holes in a field.

When Timothy and Urban work together, they dig 1 hole in 4 days.
When Timothy and Vincent work together, they dig 1 hole in 3 days.
When Urban and Vincent work together, they dig 1 hole in 2 days.

Working alone, how long does it take Timothy to dig one hole?

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In one day: Timothy and Urban dig 1/4 hole. Timothy and Vincent dig 1/3 hole. Urban and Vincent dig 1/2 hole. If we made a clone of Timothy and set all four workers digging for one day, they’d dig 1/4 + 1/3 = 7/12 hole. Since Urban and Vincent working together dig 6/12 hole per day, Timothy and his clone dig 1/12. So Timothy alone digs 1/24 hole per day, and can dig one hole in 24 days.

October 1, 2014October 12, 2014 | Puzzles