Fanciful Creatures

In What Is the Name of This Book? (1986), Raymond Smullyan describes two curious denizens of the Forest of Forgetfulness. The Lion lies on Mondays, Tuesdays, and Wednesdays, and the Unicorn lies on Thursdays, Fridays, and Saturdays. Each tells the truth on the days it doesn’t lie.

One day Alice encounters the two of them resting under a tree. They tell her:

Lion: Yesterday was one of my lying days.

Unicorn: Yesterday was one of my lying days too.

“From these two statements, Alice (who was a very bright girl) was able to deduce the day of the week. What day was it?”

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The Lion can say “I lied yesterday” only on Mondays and Thursdays. The Unicorn can say it only on Thursdays and Sundays. So today must be Thursday. Obligatory John Finnemore sketch:

March 19, 2026March 18, 2026 | Puzzles

The Number Checks Puzzle

From Henry Dudeney’s Amusements in Mathematics, 1917. Without removing these checks from their ring, divide them into three groups so that the first group multiplied by the second makes the third. For example, one valid try might be 28, 907, 15463, except that 28 × 907 doesn’t equal 15463.

“Of course, you may have as many of the checks as you like in any group. The puzzle calls for some ingenuity, unless you have the luck to hit on the answer by chance.”

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715 × 46 = 32890.

March 18, 2026March 17, 2026 | Puzzles

Charlie’s Birthday

A puzzle by National Security Agency mathematician Stephen C., from the agency’s July 2015 Puzzle Periodical:

Charlie presents a list of 14 possible dates for his birthday to Albert, Bernard, and Cheryl.

Apr 14, 1999
Feb 19, 2000
Mar 14, 2000
Mar 15, 2000
Apr 16, 2000
Apr 15, 2000
Feb 15, 2001
Mar 15, 2001
Apr 14, 2001
Apr 16, 2001
May 14, 2001
May 16, 2001
May 17, 2001
Feb 17, 2002

He then announces that he is going to tell Albert the month, Bernard the day, and Cheryl the year.

After he tells them, Albert says, “I don’t know Charlie’s birthday, but neither does Bernard.”

Bernard then says, “That is true, but Cheryl also does not know Charlie’s birthday.”

Cheryl says, “Yes, and Albert still has not figured out Charlie’s birthday.”

Bernard then replies, “Well, now I know his birthday.”

At this point, Albert says, “Yes, we all know it now.”

What is Charlie’s birthday?

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When Albert claims that Bernard does not know Charlie’s birthday, he is saying that he knows that the correct day occurs more than once in the list. In other words, he is saying that Charlie’s birthday is not Feb 19, 2000, and the only way he could know that is that he knows that the month is not February. So by making this claim Albert has reduced the list for everyone to: Apr 14, 1999 Mar 14, 2000 Mar 15, 2000 Apr 16, 2000 Apr 15, 2000 Mar 15, 2001 Apr 14, 2001 Apr 16, 2001 May 14, 2001 May 16, 2001 May 17, 2001 When Bernard says that it is true that he does not know Charlie’s birthday, it tells everyone that even with the restricted list the correct day occurs more than once in the list. So everyone can thus eliminate May 17, 2001. Furthermore, the claim that Cheryl also does not know Charlie’s birthday is a claim that Bernard knows that the year occurs more than once on the remaining list. This rules out Apr 14, 1999, and since Bernard could only rule this out by knowing that the day was not 14, everyone can further reduce the list to: Mar 15, 2000 Apr 16, 2000 Apr 15, 2000 Mar 15, 2001 Apr 16, 2001 May 16, 2001 When Cheryl says that Albert still has not figured out Charlie’s birthday, she is telling everyone that given the new list the month occurs more than once and thus rules out May 16, 2001. This tells everyone that Cheryl knows that the year is not 2001 and the list can be reduced to: Mar 15, 2000 Apr 16, 2000 Apr 15, 2000 When Bernard then claims to know the date he is saying that the day occurs only once among the 3 remaining choices, thus telling everyone that Charlie’s birthday is Apr 16, 2000.

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March 11, 2026March 10, 2026 | Puzzles

Sliding Dominoes

https://commons.wikimedia.org/wiki/File:100_grid.svg — Image: Wikimedia Commons

The squares of a 9×9 board are colored as shown, and then its surface is covered with 40 dominoes. Each domino covers two orthogonally adjacent squares, and the uncovered square is a black square on the boundary.

A move shifts a domino along its length by one square, so that it covers one empty square and exposes another. Prove that, for each of the black squares on the board, there’s a sequence of moves that will uncover it.

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This solution is by Li Zhou of Polk Community College in Winter Haven, Florida. Draw an arrow on each domino that’s covering a black square, starting on the black square and continuing through the white square. Now every black square on the board (except for the one that was exposed at the start) is covered by the tail of an arrow, and each arrow points toward a neighboring black square. This provides an answer. For any black square that’s covered at the start, the arrows now indicate a route leading to the uncovered black square. Shifting the dominoes along this route will uncover the chosen black square. Will every such route find its way to the uncovered square? Yes. The alternative would be some “loop” of arrows within the diagram. Any such loop would enclose an odd number of squares, and such an interior could not be tiled by dominoes. So we know it doesn’t exist. From the 1997 Russian Olympiad for high school students, via American Mathematical Monthly, April 2004 (Problem 10960). A solution using a graph model appears in the 1999 pamphlet Mathematical Olympiads 1997-1998: Problems and Solutions From Around the World, by T. Andreescu and K. Kedlaya.

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March 7, 2026March 6, 2026 | Puzzles

Words and Numbers

Andrzej Bartz offered these “doubly true” alphametics in the May 2017 issue of Word Ways. If the letters in each equation encode digits, what mathematical facts do these expressions represent?

CCCLVI + CCCI + CCLI = CMVIII

ONE + THIRTYNINE + NINETYONE = THREE + NINE + THIRTY + EIGHTYNINE

TWO × TWO + TEN × FIVE = SIX × NINE

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555870 + 5550 + 5580 = 567000 974 + 3861307674 + 767430974 = 38144 + 7674 + 386130 + 4628307674 250 × 250 + 286 × 9378 = 431 × 6368

March 5, 2026March 4, 2026 | Puzzles

In a Word

anfractuous
adj. having many windings and turnings

loof
n. the palm of the hand

penetralia
n. the innermost recesses of a building

swither
n. a state of perplexity

It’s commonly said that you can defeat a hedge maze by placing one hand on a wall and carefully maintaining that contact as you advance. If the hedges are all connected, this method will reliably lead you to the center of the maze (and, indeed, to every other part of it before you return to the entrance).

The Chevening maze, in Kent, was designed deliberately to thwart this technique. Its center is concealed in an “island” of hedges distinct from the outer wall, so following either a left- or a right-hand rule will return you to the entrance without ever passing the goal.

February 23, 2026February 22, 2026 | Puzzles · Science & Math

Busywork

What’s the sum of all the digits used in writing out all the numbers from one to a billion?

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Add 0 to the start of the list, and pad out each entry to 10 digits by prepending zeroes: 0000000000 0000000001 0000000002 0000000003 0000000004 ... 0999999996 0999999997 0999999998 0999999999 1000000000 Omit the final entry for the moment and pair the smallest entry in the remaining list with the largest, the second smallest with the second largest, and so on: 0000000000 + 0999999999 = 999999999 0000000001 + 0999999998 = 999999999 0000000002 + 0999999997 = 999999999 0000000003 + 0999999996 = 999999999 ... Each of these pairs produces the same sum, 999999999, the total of whose digits is 9 × 9 = 81. There are 500 million such pairs, so the sum of all those digits is 500,000,000 × 81, plus 1 from the last entry, 1,000,000,000, which we’d omitted above, giving a final total of 40,500,000,001. From Ross Honsberger’s More Mathematical Morsels, 1991. Allegedly the 10-year-old Carl Friedrich Gauss had used the same trick as a schoolboy.

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Add 0 to the start of the list, and pad out each entry to 10 digits by prepending zeroes:

Omit the final entry for the moment and pair the smallest entry in the remaining list with the largest, the second smallest with the second largest, and so on:

0000000000 + 0999999999 = 999999999
0000000001 + 0999999998 = 999999999
0000000002 + 0999999997 = 999999999
0000000003 + 0999999996 = 999999999
...

Each of these pairs produces the same sum, 999999999, the total of whose digits is 9 × 9 = 81. There are 500 million such pairs, so the sum of all those digits is 500,000,000 × 81, plus 1 from the last entry, 1,000,000,000, which we’d omitted above, giving a final total of 40,500,000,001.

From Ross Honsberger’s More Mathematical Morsels, 1991. Allegedly the 10-year-old Carl Friedrich Gauss had used the same trick as a schoolboy.

February 2, 2026February 1, 2026 | Puzzles

Dead Center

In a shooting match, Andryusha, Volodya, and Borya each fired 6 shots, and each totaled 71 points.

Andryusha’s first 2 shots earned him 22 points, and Volodya’s first shot earned 3 points.

Who hit the bullseye?

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The only way to partition the 18 shots into 3 equal sets of 6 is: 25, 20, 20, 3, 2, 1 25, 20, 10, 10, 5, 1 50, 10, 5, 3, 2, 1 The first row must be Andryusha’s, since it’s the only one that contains two numbers totaling 22. The score 3 appears only in the first and third rows, so the third must be Volodya’s. Thus he hit the bullseye. From Boris Kordemsky’s The Moscow Puzzles, 1956.

January 23, 2026January 23, 2026 | Puzzles

A Roman New Year

George Sicherman, of Sicherman dice fame, has designed a puzzle to welcome the new year: Print out the PDF, cut out the letters MMXXVI (the Roman numerals for 2026), and fit them into the tray provided so that no pieces with the same shape touch one another.

Sicherman’s website has a whole range of puzzles, most of them geometric.

(Thanks, William.)

January 20, 2026 | Puzzles

Poser

Does the sequence of squares contain an infinite arithmetic subsequence?

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Consider three squares in arithmetic progression, $a_{1}^{2}$ , $a_{2}^{2}$ , and $a_{3}^{2}$ . Then $a_{3}^{2} - a_{2}^{2} = a_{2}^{2} - a_{1}^{2}$ , or (a₃ – a₂)(a₃ + a₂) = (a₂ – a₁)(a₂ + a₁). We know that (a₂ + a₁) < (a₃ + a₂), so (a₂ – a₁) must be greater than (a₃ – a₂). Now suppose that $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, ...$ is an infinite arithmetic progression. Then (a₂ – a₁) > (a₃ – a₂) > (a₄ – a₃) > … . That’s impossible, since there’s no infinite decreasing sequence of positive integers. So the answer is no. From Arthur Engel’s excellent Problem-Solving Strategies, 2008.

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January 18, 2026 | Puzzles