The Liars

A problem by British puzzlist Hubert Phillips:

In writing home about an examination, five schoolgirls each made one true statement and one untrue one. The relevant passages:

Betty: Kitty was second in the examination. I was only third.
Ethel: You’ll be glad to hear that I was top. Joan was second.
Joan: I was third, and poor old Ethel was bottom.
Kitty: I came out second. Mary was only fourth.
Mary: I was fourth. Top place was taken by Betty.

In what order did they place?

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Kitty Joan Betty Mary Ethel It’s possible to begin with any of the girls’ statements, but here’s one solution: Two girls say that Mary placed fourth. If this is not true then Kitty was second (per Kitty’s statement) and Betty first (per Mary’s statement). But if Betty is first then Joan is second (per Ethel’s statement), which is a contradiction. That means it’s true that Mary placed fourth, and it follows that Kitty is not second (per Kitty’s statement). Hence Betty placed third (per Betty’s statement), Ethel placed fifth (per Joan’s statement), and Joan placed second (per Ethel’s statement). That leaves Kitty, who by process of elimination must have placed first. From Problem Omnibus, 1960.

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March 18, 2025March 17, 2025 | Puzzles

One Way

https://commons.wikimedia.org/wiki/File:Maze_Type_Arrow.png — Image: Wikimedia Commons

A maze by Eric Fisk. Following the arrows, find a path that leads away from the star and back to it.

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Image: Wikimedia Commons

March 12, 2025 | Puzzles

A Number Maze

https://commons.wikimedia.org/wiki/File:Maze_Type_Number.png — Image: Wikimedia Commons

By Wikimedia user Efbrazil. Begin at the star. The number at your current position tells you the number of blocks that your next jump must span. All jumps must be orthogonal. So, for example, your first jump must take you to the 1 in the lower left corner or the 2 in the upper right. What sequence of jumps will return you to the star?

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Image: Wikimedia Commons 02/21/2025 UPDATE: There are at least three additional solutions. Label the columns A-D and the rows 1-4, like a miniature chessboard. Now: A4-A1-B1-D1-D4-D2-C2-C4-C1-C3-A3-A4 (from reader Justin Hilyard) A4–D4–B4–B2–C2–C4–C1–C3–A3–A4 (from reader Heinrich Hemme) A4-A1-A2-D2-C2-C4-C1-C3-A3-A4 (from reader Paul Wagner)

February 20, 2025February 23, 2025 | Puzzles

Wine and Roses

This portrait of Bacchus contains the images of two lovers. Where are they?

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They’re surprisingly plain once you see them — their joined lips are just to the left of Bacchus’ nose. From “Old Puzzle Cards,” Strand, November 1899.

February 16, 2025February 15, 2025 | Puzzles

Shipshape

https://commons.wikimedia.org/wiki/File:Cmglee_Cambridge_Trinity_College_punt_names.jpg — Image: Wikimedia Commons

These are the punts of Trinity College, Cambridge, moored on the River Cam. What is the significance of their names?

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Inspired by the college’s name, each punt evokes the number three: In the game NOUGHTS & CROSSES, players compete to place three marks in a row. A CODON is a sequence of three nucleotides in a DNA or RNA molecule. In Greek mythology, beauty was deified in the three GRACEs Euphrosyne, Aglaia, and Thalia. A HAIKU has three lines. Three WISE MONKEYs illustrate the maxim “See no evil, hear no evil, speak no evil.” NENYA is one of the three Rings of Power in J.R.R. Tolkien’s fantasy legendarium. FLUFFY is a three-headed dog in J.K. Rowling’s Harry Potter universe. Traditionally after two failures one hopes for a third LUCKY TIME. CHARLES III is king of the United Kingdom. Not shown: LITHIUM has atomic number 3. A BARYON is a subatomic particle made up of three quarks. 02/11/2025 UPDATE: Wow, yet more! The additional names include LITTLE KITTEN, CHEER, BRONZE, WISE MAN, PEACE SWEET, GRUFF, MAD GEORGE, DIABOLUS IN MUSICA, HARRY LIME, and POINT TURN. (Thanks, Dmitry.)

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February 11, 2025February 11, 2025 | Puzzles · Trivia

Freight

A problem from Cambridge mathematician J.E. Littlewood’s Miscellany (1953):

Is it possible to pack a cube with a finite number of smaller cubes, no two of which are the same size?

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No. Think of the large cube as a square well with a flat bottom. If the task is possible, then this floor is tiled perfectly with cubes of different sizes. Consider the smallest of these cubes. This can’t adjoin an edge of the floor, because in that position it would be impossible to surround it with cubes larger than itself. So it lies somewhere on the expanse of the floor, surrounded by larger cubes. But now we’re back where we started. The top of this small cube forms the floor of a square well, and all the reasoning above applies again. And so on forever: At each stage we’ll be forced to supply smaller cubes, and our task will never be finished.

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February 3, 2025February 2, 2025 | Puzzles

Oh

A puzzle by Lewis Carroll:

Dreaming of apples on a wall,
And dreaming often, dear,
I dreamed that, if I counted all,
How many would appear?

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Ten.

January 22, 2025January 20, 2025 | Puzzles

Self-Destruction

pauly selfmate

This chess problem, from Wolfgang Pauly’s 1912 Theory of Pawn Promotion, is a selfmate in two: How can White force Black to deliver checkmate within two moves, despite Black’s best effort to avoid doing so? White moves first.

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The bishops are nicely aligned with White’s king; if Black had no other option, he’d have to play … Bxg2, giving mate. But White can’t simply withdraw his bishop along the long diagonal, as then Black could shift his own bishop mildly to g2, avoiding mate. What other options does White have? He mustn’t move his knight, as that would allow the Black king to move. 1. e6 allows Black to take the f6 pawn. Moving the f6 pawn allows Black to take the g7 pawn. Promoting the g pawn to a queen or a rook counterproductively protects the g2 bishop. Promoting it to a knight checkmates Black (!). Promoting it to a bishop almost works, but after 1. g8=B exf6 2. exf6 Bxg2+ White now has 3. Bd5, which again prevents checkmate. The only resource that works is to promote the c7 pawn to a knight. Now Black has only two ways to avoid giving mate, and both are only temporary: 1. … exf6 2. exf6 Bxg2# 2. … e6 2. g8=B Bxg2# (as now the promoted bishop can’t reach d5) Note that the c7 pawn must promote to a knight — any other piece could block the final check.

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January 21, 2025January 18, 2025 | Puzzles

Gun Control

Marksman A hits a certain small target 75 percent of the time. Marksman B hits it 25 percent of the time. The two of them aim at that target and fire simultaneously. One bullet hits it. What’s the probability that it came from A?

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At first it might seem that the answer is 75 percent, since A is three times as good a shot as B. “But there are two things happening here: hitting and missing,” writes Dartmouth mathematician Peter Winkler. The probability that A hits and B misses is 3/4 × 3/4 = 9/16, and the probability that B hits and A misses is 1/4 × 1/4 = 1/16. So there’s fully a 9/10 chance that the successful bullet was A’s. From Winkler’s excellent Mathematical Puzzles, 2021.

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January 15, 2025January 14, 2025 | Puzzles

Branch Manager

A puzzle from The Book of the Thousand Nights and One Night:

“Solve the following problem if you can: a flock of pigeons alighted upon a tree, some perching upon the upper branches and some upon the lower; those upon the upper branches said to those upon the lower: ‘If one of you flies up to us our number will be double yours; if one of us flies down to you, our numbers will be equal.'”

How large was the flock?

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“There were twelve pigeons in all, seven upon the upper branches and five upon the lower branches. If one of the lower ones joined the higher, there would be eight above, which would be the double of four; but if one of the seven flew down to the lower ones, there would be six in each position. But Allah is all-knowing.” Via Aplusclick‘s excellent collection of math and logic problems. (Thanks, Igor.)

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January 9, 2025January 8, 2025 | Puzzles