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

Falling Currency

A problem from the October 1964 issue of Eureka, the journal of the Cambridge University Mathematical Society:

My friend tosses two coins and covers them with his hand. ‘Is there at least one “tail”?’ I ask. He affirms this (a).

Just then he accidentally knocks one of them to the floor (b). On finding the dropped coin under the table, we discover it to be a ‘tail’ (c).

‘That is all right,’ he says, ‘because it was a “tail” to start with.’ (d).

At each point (a), (b), (c) and (d) of this episode I calculated what, to the best of my knowledge, was the probability that both coins showed ‘tails’ at the time. What were these probabilities?

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(a) 1/3 (b) 1/3 (c) 2/3 (d) 1/2 01/14/2026 UPDATE: Reader Catalin Voinescu sent this illustration: (Thanks, Catalin.)

January 14, 2026January 14, 2026 | Puzzles

Black and White

wills chess problem

By W.F. Wills. White to mate in two moves.

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1. Qa8! If the knight on a7 moves, then 2. Nc2 is mate. If Black’s rook takes the knight, then 2. Qh8 is mate. If Black’s rook moves along the second rank, White can capture it, giving mate. And if the knight on d2 moves, then White’s rook on h2 captures Black’s on a2, giving mate. (If the knight captures the b1 bishop, then Rxb1 is mate.) From the Norwich Mercury Tournament, 1902.

January 10, 2026January 10, 2026 | Puzzles

Mixed Doubles

In a letter to Maud Standen dated Dec. 18, 1877, Lewis Carroll included a puzzle:

[M]y ‘Anagrammatic Sonnet’ will be new to you. Each line has 4 feet, and each foot is an anagram, i. e., the letters of it can be re-arranged so as to make one word. Thus there are 24 anagrams, which will occupy your leisure moments for some time, I hope. Remember, I don’t limit myself to substantives, as some do. I should consider ‘we dishwished’ a fair anagram.

As to the war, try elm. I tried.
The wig cast in, I went to ride.
‘Ring? Yes.’ We rang. ‘Let’s rap.’ We don’t.
‘O shew her wit!’ As yet she won’t.
Saw eel in Rome. Dry one: he’s wet.
I am dry. O forge! Th’rogue! Why a net?

For example, the first foot in the first line, “As to,” can be rearranged to spell OATS. Carroll left no solution, but he did add a parting riddle to which we have the answer:

“To these you may add ‘abcdefgi,’ which makes a compound word — as good a word as ‘summer-house.'” What is it?

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“Big-faced”. From Edward Wakeling’s 1995 book Rediscovered Lewis Carroll Puzzles, which gives some guesses as to the sonnet anagrams.

January 6, 2026 | Puzzles

Nonary

Take a whole number, reverse the order of its digits, and subtract one from the other. The difference will always be evenly divisible by 9.

Does this remain true if we just scramble the digits of the first number, rather than reversing them?

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Yes. Suppose we scramble 27910 to get 19702: 27910 = 2 × 10⁴ + 7 × 10³ + 9 × 10² + 1 × 10¹ + 0 × 10⁰ 19702 = 1 × 10⁴ + 9 × 10³ + 7 × 10² + 0 × 10¹ + 2 × 10⁰ Now collect the terms of 27910 – 19702 that have the same coefficients: 2 × (10⁴ – 10⁰) + 7 × (10³ – 10²) + 9 × (10² – 10³) + 1 × (10¹ – 10⁴) + 0 × (10⁰ – 10¹) Each term has a factor that’s either 10^m(10ⁿ – 1) or -10^m(10ⁿ – 1) for some integers m, n. So each term, and therefore the sum, is divisible by 9. From Keith Kendig, Sink or Float?: Thought Problems in Math and Physics, 2008.

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January 2, 2026January 1, 2026 | Puzzles · Science & Math

A Harrowing Puzzle

A gigantic tire, with a radius of 100 miles, is rolling down Broadway at 60 mph. One driver fails to notice the tire’s approach until its descending surface is just touching the roof of her car, 6 feet above the road. If she leaves the car immediately and can shrink to within 2 feet of the road’s surface, how long does she have to crawl out of the tire’s path?

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Surprisingly, she has more than 12 seconds. In this diagram, which is not remotely to scale, r = 100 miles, u = 2 feet, and h = 6 feet. The tire moves at v = 88 feet per second. By Pythagoras, a² = r² – (r – u)² = 2ru – u² and, similarly, b² = 2rh – h². The time we seek is x/v, or (b – a)/v, which works out to 12.089356 seconds. From Dick Hess’ All-Star Mathlete Puzzles, 2009.

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December 26, 2025December 25, 2025 | Puzzles