Piecework

In his Canterbury Puzzles of 1907, Henry Dudeney posed a now-famous challenge: How can you cut an equilateral triangle into four pieces that can be reassembled to form a perfect square?

Dudeney’s beautiful solution was accompanied by a rather involved geometric derivation. It seems unlikely that he worked this out laboriously in approaching an answer to the problem, but how then did he reach it?

Here’s one possibility: If a strip of squares is draped adroitly over a strip of triangles, their intersection forms a wordless proof of the task’s feasibility:

https://commons.wikimedia.org/wiki/File:Haberdasher_strips.svg — Image: Wikimedia Commons

Whether that was Dudeney’s path to the solution is not known, but it appears at least plausible.

August 8, 2025August 7, 2025 | Puzzles · Science & Math

“Locker Lotto”

A puzzle by Ying Zhou, Daniel Irving, and Walter Gall of Rhode Island College, from the February 2012 issue of Math Horizons:

A sports team is divided into “red” and “blue” groups of 10 players each. Each player puts his belongings into a bag of his team’s color and puts it into one of 20 lockers, choosing at random. All the players leave the room. Presently one of the red team returns and can’t remember which locker is his. He and the janitor make a bet: The player can keep opening lockers so long as each bag he discovers is red. If he finds his bag, the janitor will give him $7. If not, he’ll owe the janitor $1. Should the player take the bet?

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No. This solution is by Adilet Otemissov and the Ashland University Problem Solving Group. If there are m red and 20 – m blue players at the start, then the red player will always be 20 – m times more likely to find a blue bag than to find his own bag on opening the kth locker, no matter the value of k. So the probability that he finds his own bag is $\displaystyle p\left ( m \right )=\frac{1}{\left ( 20-m \right ) + 1} = \frac{1}{21-m}.$ In our case, p(10) = 1/11, so the red player shouldn’t accept — his expected winnings are $\displaystyle \left ( 1/11 \right ) \times \$ 7 + \left ( 10/11 \right ) \times \left ( -\$ 1 \right ) = -\$ 3/11,$ which is negative. (Derek Smith, “The Playground!”, Math Horizons 19:3 [February 2012], 30-33.)

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August 4, 2025August 3, 2025 | Puzzles

Nowhere

This is charming somehow: a detailed portrait of a place that doesn’t exist. During the Cold War, U.S Army cryptologist Lambros D. Callimahos devised a “Republic of Zendia” to use in a wargame for codebreakers simulating the invasion of Cuba. (Callimahos’ maps of the Zendian province of Loreno are below; click to enlarge.)

The Zendia map now hangs on the wall of the library at the National Cryptologic Museum. The “Zendian problem,” in which cryptanalysts students were asked to interpret intercepted Zendian radio messages, formed part of an advanced course that Callimahos taught to NSA cryptanalysts in the 1950s. Graduates of the course were admitted to the “Dundee Society,” named for an empty marmalade jar in which Callimahos kept his pencils.

08/02/2025 UPDATE: Apparently they speak Esperanto in Zendia, or at least their cartographers do. “Respubliko” is Esperanto for “Republic,” “Bovinsulo” and “Kaprinsulo” are “Cow-Island” and “Goat-Island”, and so on. (Thanks, Ed and David.)

August 1, 2025August 2, 2025 | Art · Puzzles

Black and White

andrade chess problem

By J. de C. Andrade. White to mate in two moves.

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At first this looks absurdly easy — surely promoting either white pawn leads directly to mate? The trouble is that once either pawn advances, the other is pinned and can’t move. Instead, the key is the crazy-looking 1. Qh7! Black’s rook has no safe refuge on the eighth rank, and if it takes the queen then a pawn promotion is fatal. From The Problemist, 1941. 07/29/2025 My description isn’t quite accurate — a few readers have pointed out that Black’s rook is safe on b8. In that case White can mate with 2. Nc7, a smothered checkmate. Thanks to everyone who wrote in about this.

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July 28, 2025July 29, 2025 | Puzzles

Hand Holding

From an old copy of Games magazine:

Four players describe the cards they held in their last game of five-card poker:

Harry: “I had the two, nine, and jack of hearts, and the two and nine of diamonds.”

Edna: “I held four fives and the queen of spades.”

Dave: “I had a black straight flush.”

Frankie: “I held a club flush.”

Georgiana: “I had a straight. Two of my cards were spades; the others weren’t.”

Cleo: “I just had one lousy pair.” (to Frankie:) “But my lowest-ranking card was higher than any of yours.”

An onlooker asks how many of the undealt cards were black, and Dave answers, “No more than three.” What were the contents of Dave’s, Frankie’s, Georgiana’s, and Cleo’s hands?

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Dave: ♠6 ♠7 ♠8 ♠9 ♠10 Frankie: ♣2 ♣3 ♣4 ♣6 ♣7 Georgiana: ♣9 ♣10 ♠J ♣Q ♠K Cleo: ♣8 ♣J ♣K ♣A ♠A (By Bob Stanton.)

July 19, 2025July 18, 2025 | Puzzles

Jumping Kangaroos

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

Joey, the baby kangaroo has been kidnapped and placed at 2¹⁰⁰ on a number line.

His mother, Kandice the Kangaroo, is at 0 on the number line, and will try to save him. Kandice normally jumps forward 6 units at a time. Guards have been placed at n³ on the number line, for every integer n≥1. If Kandice lands on a number with a guard on it, she will be caught and her mission will fail. Otherwise, she will safely sneak past the guard. Whenever she successfully sneaks past a guard, she gets an adrenaline rush that causes her next jump (the first jump after passing the guard) to take her 1 unit farther than it normally would (7 units instead of 6). (After a single 7-unit jump, she resumes jumping 6 units at a time, until the next time she sneaks past a guard.)

Will Kandice the Kangaroo reach (or pass) her son Joey safely?

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Yes. First, note that two consecutive cubes are always separated by one more than a multiple of 6. To see this, note that (n+1)³ = n³ + 3n² + 3n + 1, so (n+1)³ – n³ = 3n(n+1) + 1. n must be either even or odd; if it is even, then n(n + 1) is even, and if it is odd, then n+1 is even, so again, n(n+1) is even. Therefore, 3n(n+1) is divisible by both 2 and 3, so it must be a multiple of 6, so the difference (n+1)³ – n³ is one more than a multiple of 6. Now, it’s easy to check that Kandice will safely jump over the first two guards (at 1 and 8). Suppose Kandice later gets caught and Gary the Guard is the first guard who catches her. Since passing the guard previous to Gary, Kandice would have finished one jump of size 7 and several jumps of size 6. Because the distance between the guards is one more than a multiple of 6 (as shown in this illustration), counting backwards we see this pattern would have forced Kandice to land on the previous guard also. But this is impossible, since Gary was the first guard to catch Kandice. So Gary does not exist and Kandice will never be caught! Key: Cube numbers are the positions of the guards. Evens and odds alternate, so if some number n isn’t even, then the next number (n+1) must be even.

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July 18, 2025July 17, 2025 | Puzzles

Current Affairs

A problem by G. Galperin, from the May-June 1995 issue of Quantum:

A raft and a motorboat depart simultaneously from Point A on a riverbank and begin drifting and speeding downstream, respectively, toward Point B. At the same moment, a second motorboat, of the same type as the first, sets out from Point B heading upstream. When the first motorboat reaches B, will the floating raft be closer to Point A or to the second motorboat?

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This intuitive solution is by V. Dubrovsky. Imagine a third point, C, that’s as far upstream from A as A is from B (so CA = AB). And imagine a third motorboat that parallels the movements of the first, setting out from point C and heading downstream at the same time. The third and second motorboats are equidistant from the raft when they begin their journeys, and both move at the same speed relative to the raft (and to the surface of the water). So they’ll remain equidistant from the raft as they progress toward it. When the first motorboat reaches B, the third will reach A. At that moment the raft will be the same distance from the second motorboat and from Point A (now occupied by the third motorboat). That’s the answer. 07/20/2025 UPDATE: From reader Catalin D. Voinescu: Even simpler, from reader Joe McVeety: “Since you are not given the speed of the river, it must not matter. Therefore, set it to 0. The raft never moves. Boats 1 and 2, traveling at the same speed, will cover the same distance in the same time. When Boat 1 reaches B, Boat 2 is reaching A. The raft never moved, so it is equidistant to Boat 2 and Point A. 0, in this case.”

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July 17, 2025July 20, 2025 | Puzzles

Odd Job

A problem by Russian mathematician Viktor Prasolov:

On a piece of graph paper, is it possible to paint 25 cells so that each of them has an odd number of painted neighbors? (“Neighboring” cells have a common side.)

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Let n_k be the number of painted cells with exactly k painted neighbors, and let N be the number of common sides of painted cells. Each common side belongs to exactly two painted cells, so $\displaystyle N = \frac{n_{1} + 2n_{2} + 3n_{3} + 4n_{4}}{2} = \frac{n_{1} + n_{3}}{2} + n_{2} + n_{3} + 2n_{4}.$ Since N is an integer, n₁ + n₃ is even and thus can’t be 25. (From Prasolov’s Problems in Plane and Solid Geometry.)

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July 10, 2025July 10, 2025 | Puzzles

The Sure Thing

In a 1905 short story by Jacob Elson, Mr. Brown laments that he cannot solve chess problems.

Mr. Pincus wagers $10 that “I can show you a two-move problem with three different lines of play which you would have to solve whether you wanted to or not.”

Brown accepts. After studying the board for 10 minutes, he says, “It’s a humbug, a confounded silly swindling humbug, but I am beat.” Here’s the position:

sure thing chess position

July 9, 2025July 8, 2025 | Puzzles

The Studious Bather

A puzzle from Chris Maslanka’s The Pyrgic Puzzler, 1987:

A bathtub will fill in 3 minutes if the plug is in and the cold tap only is turned on full. It will fill in 4 minutes if the plug is in and the hot tap only is turned on full. With the plug out and both taps off, a full tub will drain in 2 minutes. How long will it take to fill the empty tub if the plug is out and both taps are turned on full?

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If the bath’s volume is V, then the cold tap will fill it at a rate of V/3 and the hot tap at a rate of V/4. It will drain at a rate of V/2. Let t be the time it will take to fill it with the plug out and both taps on. Then (V/3 + V/4 – V/2)t = V (1/3 + 1/4 – 1/2)t = 1 and t = 12. (Via Barry R. Clarke’s Mathematical Conundrums, 2023.)

June 26, 2025June 24, 2025 | Puzzles