Spin Doctor

I have 13 magnetic tapes on reels; one empty reel; and a machine that can unwind a tape from a full reel onto an empty one, reversing the tape’s direction. I need to reverse the direction of each tape while leaving it on its original reel. Can this be done?

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No. If each of the 13 reels must end with the tape it started with, then each of these reels must go through an even number of tape transfers. The empty reel, too, must go through an even number of transfers to remain empty. But each of the 13 tapes must be wound or rewound an odd number of times to end up reversed. These facts can’t be reconciled, so the task can’t be done. From Dick Hess’ excellent All-Star Mathlete Puzzles, 2009.

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September 18, 2026 | Puzzles

Kakuro

This logic puzzle is the numerical equivalent of a crossword. Each white cell in the diagram is to be filled with a digit from the range 1-9. The sum of each string of digits must match the “clue” given in the diagram. For example, the clue 29 above means that the four digits to its right must sum to that total. No digit may be duplicated within any entry. Can you complete the diagram?

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Image: Wikimedia Commons

September 12, 2026September 11, 2026 | Puzzles

Self-Study

A puzzle from the October 1962 issue of Eureka, the journal of the Cambridge University Mathematical Society:

(a) At least two of these statements, apart from this one, are true.
(b) At least two of these statements, apart from this one, are false.
(c) At least one of these statements is false.
(d) x of these statements are true.

Given that, if you knew the value of x, you could determine uniquely which statements are true and which are false, determine the value of x.

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The journal states only that x = 3, but here’s an attempt. We’re told that the value of x in (d) is vital, so (d) must be true; otherwise its claim about x would be useless. It’s impossible for all four statements to be true, because then (a) and (b) would contradict one another. This means that at least one statement is false, which means that (c) is true. And if (c) and (d) are both true, then (a) is true, which means that (b) is false: (a) true (b) false (c) true (d) true

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September 11, 2026 | Puzzles

Orb Weaver

An elegant alphametic by Jonathan Joseph, from the October 1991 issue of Games magazine:

   VENUS
    MARS
  SATURN
+ URANUS
--------
 NEPTUNE

If each letter consistently represents a particular digit, what equation does this expression encipher?

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20163 4593 358691 + 695163 -------- 1078610

September 8, 2026September 7, 2026 | Puzzles

Flight From Moscow

Russian chess master Alexander Petrov composed this puzzle to commemorate Napoleon’s defeat in 1812:

petrov chess puzzle

Currently the emperor stands in Moscow, on h7. The goal is to chase him all the way back to Paris and to checkmate him there. The white knights are Cossack cavalry, and the white bishops are Russian partisans. Given these clues, which square represents Paris?

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Paris is a3. The white knights will check from g5, h6, h7, g8, f8, e7, d7, c8, b8, a7, a6, and b5. All of Black’s replies are forced. Via the Batsford Chess Puzzle Book, 2002.

September 2, 2026September 1, 2026 | History · Puzzles

For Short

In 2007 I posted this cryptic verse by William Whewell:

U 0 a 0 but I 0 U,
O 0 no 0 but O 0 me;
O let not my 0 a 0 go,
But give 0 0 I 0 U so.

You sigh for a cipher but I sigh for you,
O sigh for no cipher, but O sigh for me;
O let not my sigh for a cipher go,
But give sigh for sigh, for I sigh for you so.

In An Almanac of Words at Play (1975), Willard R. Espy offered this addition. What does it say?

Oy OO thO he O
2 On an OO. OO bO
An OO but he never thO
He’d O the OO dealer O
For the OO OO bO.
Say, has OO’s OO brO
OO O that OO sO?
OO’s OO’s good for nO.

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Naughty Otto thought he ought To own an auto. Otto bought An auto but he never thought He’d owe the auto dealer aught For the auto Otto bought. Say, has Otto’s auto brought Otto aught that Otto sought? Otto’s auto’s good for naught.

August 29, 2026August 28, 2026 | Language · Puzzles

Checkmate

A puzzle by A. Savin, from the July-August 1996 issue of Quantum:

A mother, her brother, her daughter, and her son take part in a family chess tournament. Two of the four are twins. At the end of the tournament, the winner and the “loser” (the player in last place) are found to be the same age, and the winner is the opposite gender to the loser’s twin. Who won the tournament?

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The twins must be either the mother and her brother or the son and daughter. Either way, they are of opposite genders. This means that the winner and the loser are of the same gender. These two can’t be the mother and daughter, because they must be the same age. That means they’re the son and brother. The son can’t be the winner, because that would require that the brother’s twin (the mother) be the same age as the son. So the only possibility is that the brother won the tournament.

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August 21, 2026August 20, 2026 | Puzzles

Around the World

A brainteaser from the British intelligence agency GCHQ:

Explain the following: falkun, home de guerra, kick, lavoro, moed, mosca, raccordement, schafer, stobhach, tre, zevk.

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Falkun is the Maltese word for falcon — it’s the Maltese falcon. In the same sense, the remaining items present a Portuguese man o’ war, the Swedish chef, the Italian job, Dutch courage, Spanish fly, the French connection, a German shepherd, Irish stew, Norwegian wood, and Turkish delight.

August 15, 2026August 15, 2026 | Language · Puzzles

CyberChance

A puzzle by National Security Agency applied research mathematician Nicholas R., from the agency’s October 2016 Puzzle Periodical:

Eddie, Layne, Kurt, and Chris are locked in a heated game of CyberChance which only one of them will win. Their fans are on the edge of their seats: everyone knows that in CyberChance, things can change at any moment. One fan, who prefers Kurt or Chris, is feeling worried since there is only a 4-in-10 chance that one of them will win. An Eddie fan brags that Eddie is twice as likely as Layne to come out on top. Another Eddie fan concurs, and adds the following observation: Chris is Eddie’s main competition. He figures that, given that Chris doesn’t win, Eddie has a 4-in-7 chance of winning. A Kurt fan has been quietly standing in the corner, listening to all the other fans, and wonders: What chance does Kurt have of winning CyberChance?

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Let E, L, K, and C be the chances that Eddie, Layne, Kurt, and Chris win the game, respectively. There is a 100% chance that one of them will win, so E + L + K + C = 100%. We know that K + C, the probability that Kurt or Chris will win, is 40%. This tells us that E + L = 100% – 40% = 60%. Now, we also know that the probability that Eddie will win is twice the probability that Layne will win. In numbers, this means that E = 2L, i.e. E – 2L = 0%. Knowing these two facts lets us find out what Eddie’s chance of winning is: add the equation E + L = 60% twice to the equation E – 2L = 0%. This tells us that 3E = 120%. Therefore E = 40%. We are almost there. We know that the conditional probability that Eddie wins, given that Chris loses, is 40%/70%. This probability is equal to the probability that Eddie wins and Chris loses, divided by the probability that Chris loses. But notice: CyberChance has only one winner, so the probability that Eddie wins and Chris loses is the same as the probability that Eddie wins. Also notice that the probability that Chris loses equals the probability that anybody else wins. In numbers, the probability that Chris loses is 100% – C. So we have the following equation: 40%/70% = E / (100% – C) = 40% / (100% – C) Solving for C tells us that C = 30%. Recall that K + C = 40%. Therefore K = 10%. So Kurt has a measly 10% chance of winning CyberChance. No wonder the Kurt fan was sulking in the corner!

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August 13, 2026August 12, 2026 | Puzzles

Tile Swap

You’re tiling a floor with 4×1 and 2×2 tiles when you accidentally break one. A tile of the other shape is available. Show that it’s not possible to cover the floor by rearranging the tiles.

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Color the tiles as shown. Now every 4×1 tile must cover 0 or 2 black squares, and every 2×2 tile must cover one black square. So we can’t successfully exchange one for the other. From Arthur Engel’s excellent Problem-Solving Strategies, 1998.

August 12, 2026August 11, 2026 | Puzzles