Figures

tintin

I think Jacques Jouet was the first to notice this. In the first few pages of the Tintin adventure The Secret of the Unicorn, as Tintin visits the Vossenplein antique market in Brussels, Snowy the dog keeps scratching himself. Why?

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It’s a flea market. I can’t find any record that author Hergé ever acknowledged this explicitly, but both he and his hero were Belgian, and apparently he included it as a quiet inside joke. Locals know the market as the marché aux puces.

January 7, 2018January 9, 2018 | Literature · Puzzles

Black and White

köhnlein chess problem

By Friedrich Köhnlein. White to mate in two moves.

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1. Qa8! and Black can’t evade the rook. From Tägliche Rundschau, 1904.

January 3, 2018January 2, 2018 | Puzzles

Podcast Episode 184: Lateral Thinking Puzzles

Here are six new lateral thinking puzzles — play along with us as we try to untangle some perplexing situations using yes-or-no questions.

See full show notes …

January 1, 2018January 15, 2018 | Podcast · Puzzles

Eban Numbers

What number comes next in this sequence?

2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, __

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50 comes next. It’s a list of natural numbers whose English names lack the letter E (“e-ban numbers”). (Proposed by N.J.A. Sloane.) From the excellent puzzle page at the Online Encyclopedia of Integer Sequences.

December 27, 2017December 27, 2017 | Puzzles

The Last Digit

A problem from the 1996 Georg Mohr mathematics competition in Denmark:

n is a positive integer. The next-to-last digit in the decimal expression of n² is 7. What’s the last digit?

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This solution is by Michel Bataille of Rouen, France: Write n as 10a + b, where a is a nonnegative integer and b is a single nonnegative digit. Then n² = 100a² + 20ab + b², and the last two digits of n² are the same as those of 20ab + b². Now, 20ab is 10 × 2ab — that is, it’s an even number multiplied by 10. So it must end with 00, 20, 40, 60, or 80. We know that the next-to-last digit of 20ab + b² is 7, so the next-to-last digit of b² must be odd. That can occur only when b is 4 or 6, giving b² = 16 or 36. In both cases the last digit is 6. (“The Olympiad Corner,” Crux Mathematicorum 27:4 [May 2001], 239.)

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December 23, 2017December 24, 2017 | Puzzles · Science & Math

RSS Quiz

Another holiday challenge: The Royal Statistical Society’s 2017 Christmas quiz presents 13 problems that require general knowledge, logic, and lateral thinking but no particular math skills. For example:

4. CAN YOU DIG IT? [11 points]

Identify the following from the clues. What do all ten answers have in common?

A cockerel of human dimensions, performed by a prolific informer
William’s book (1956), Charles’s film (1957), or Mike & Al’s song (1971)
Challenger first appeared here, over a century ago
A cricketer, a rugby player, or a commentator
2001 boy-band album – a remix of “NOW FOR LOUD ROW”
The official title of Guinness
One who reigned for almost 999 million seconds
King’s collection, ordered by cards
The Bassett country residence, according to Plum
ATR co-founder, name-checked by “The Tiger” between The Slits and Dickens

You can use any tools or resources you like, including books, search engines, and computer programs. Anyone can enter, and you stand to win £150 if you’re an RSS member. The deadline for entries is January 7. The full quiz is here.

(Thanks, Dave.)

December 21, 2017 | Puzzles

A Thoughtful Gift

University College London mathematician Matthew Scroggs has created a mathematical Christmas card for Chalkdust magazine.

Solve 10 puzzles, convert the answers to base 3, write them in the grid, and color them accordingly to reveal a Christmassy picture.

He’s provided both a PDF and an online version that will color the squares for you.

December 19, 2017 | Puzzles

Black and White

kling chess problem

By Josef Kling. White to mate in two moves.

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1. f8=B! Kxf6 2. Rf4# From The Chess Euclid, 1849.

December 17, 2017December 16, 2017 | Puzzles

Two Puzzles

In 2005 Keele University computer scientist Gordon Rugg published two ciphers to the web.

The first is called the Penitentia Manuscript. The image above is only one panel; you can view and download the whole thing here. Rugg’s website provides one clue: “Most modern codes are based on a shared set of underlying assumptions. He wondered what would happen if you deliberately ignored those assumptions. What sorts of code might that produce?” There’s some more info here.

The second cipher, called the Ricardus Manuscript, was inspired by Rugg’s work on another famous puzzle: “When Gordon was working on the Voynich Manuscript, he started wondering what a real code based on the components of the Voynich Manuscript would look like. This code is the result.” Again, the image below is only a sample; you can find the whole thing here. More info here.

Both of these ciphers have been freely available on the web for more than 10 years, and both remain unsolved. Any takers?

December 16, 2017December 15, 2017 | Puzzles

‘Tis the Season

Charles Trigg proposed this festive cryptarithm in the American Mathematical Monthly in 1956:

MERRY XMAS TO ALL

If each letter is a unique representation of a digit, and each word is a square integer, what are these four numbers?

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L.A. Ringenberg of Eastern Illinois State College showed that there are two solutions: A table of squares shows that ALL must be 100, 144, 400, or 900. If it’s either of the first two, then XMAS has to be 2916 or 9216, and TO can’t be a square. If ALL is 400, then XMAS has to be 1849, 3249, 6241, or 8649. Taking these four possibilities in turn: If ALL = 400 and XMAS = 1849, then M = 8, MERRY = 81225, and E and X now both represent 1, so that can’t be right. If ALL = 400 and XMAS = 3249, then M = 2, MERRY = 27556, and TO = 81, so produces a valid solution: 27556 3249 81 400. If ALL = 400 and XMAS = 6241, then TO isn’t a square. If ALL = 400 and XMAS = 8649, then M = 6 and L = 0 and there’s no solution for MERRY. Finally, going back to the possibility that ALL = 900, in this case XMAS = 1296 or 7396: If ALL = 900 and XMAS = 1296, then TO isn’t a square. If ALL = 900 and XMAS = 7396, then M = 3, MERRY = 34225, and TO = 81, giving a second solution: 34225 7396 81 900. (“Problems for Solution,” American Mathematical Monthly 63:10 [December 1956], 723.) 12/08/2017 UPDATE: A variant with a unique solution, from reader Jean-Noël Monette: MERRY XMAS TO YA ALL 27556 3249 81 64 400

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December 7, 2017December 8, 2017 | Puzzles