My Aunt Maria asked me to read the life of Dr. Chalmers, which, however, I did not promise to do. Yesterday, Sunday, she was heard through the partition shouting to my Aunt Jane, who is deaf, ‘Think of it! He stood half an hour today to hear the frogs croak, and he wouldn’t read the life of Chalmers.’

Donald Aucamp offered this problem in the Puzzle Corner department of MIT Technology Review in October 2003. Three logicians, A, B, and C, are wearing hats. Each of them knows that a positive integer has been painted on each of the hats, and each of them can see her companions’ integers but not her own. They also know that one of the integers is the sum of the other two. Now they engage in a contest to see which can be the first to determine her own number. A goes first, then B, then C, and so on in a circle until someone correctly names her number. In the first round, all three of them pass, but in the second round A correctly announces that her number is 50. How did she know this, and what were the other numbers?

Howard Haber gave this solution in the March-April 2004 issue. The key observation to bear in mind is that if one of the logicians sees that both of her opponents’ hats bear the same number, she can conclude immediately that her own number is twice this value.

Each of the logicians knows that her own number is either the sum or the positive difference of the other two. A, going first, reasons that the three numbers A, B, C must stand in one of two ratios, 1:2:3 or 5:2:3. She can’t decide between these possibilities, so she remains silent. But in the second round A has the benefit of knowing that each of the other logicians also remained silent in the first round. Now, suppose that the ratio had been 1:2:3. Then, when B took her turn in the first round, she would have concluded that the ratio must be either 1:2:3 or 1:4:3. Like A, she would have remained silent, unable to decide between these possibilities. But C, going next, would have determined that the ratio must be either 1:2:3 or 1:2:1 — and she could immediately have excluded the latter because in that case B would have been able to win the game by employing the key observation above. Noting that B didn’t do this would have allowed C to win the game herself by announcing her only remaining possibility, 1:2:3. The fact that C didn’t do this tells A that the ratio can’t be 1:2:3, and she concludes that it must be the other ratio she’d been considering, 5:2:3.

“For the problem as stated, just multiply all numbers above by 10.”

“Garçon!” the diner was chargin’,
“My butter has been writ large in!”
“But I had to write there,”
Exclaimed waiter Pierre,
“I didn’t have room in the margarine.”

Speaking of swans: By royal prerogative, all mute swans in open water in Britain are the property of the British Crown. Historically the Crown shares ownership with two livery companies, the Worshipful Company of Vintners and the Worshipful Company of Dyers, and so, accordingly, each year in the third week of July three skiffs make their way up the Thames from Sunbury to Abingdon, catching, tagging, and releasing the swans they encounter. Nominally they’re apportioning the birds among themselves; in practice they’re counting them and checking their health.

Magnificently, the Crown’s swans are recorded by the Marker of the Swans, a recognized official in the Royal Household since this tradition began in the 12th century. Queen Elizabeth II attended the Swan Upping ceremony in 2009, as “Seigneur of the Swans,” the first time a reigning monarch had done so. The entire operation was shut down for the first time in 2020, due to COVID-19, but it commenced again the following year.

While we’re at it: All whales and sturgeons caught in Britain become the personal property of the monarch — they are “royal fish.” Plan accordingly.

Some say “If God sees everything before
It happens — and deceived He cannot be —
Then everything must happen, though you swore
The contrary, for He has seen it, He.”
And so I say, if from eternity
God has foreknowledge of our thought and deed,
We’ve no free choice, whatever books we read.

Fletcher Christian’s first son was named Thursday October Christian.

SLICES OF BREAD = DESCRIBES LOAF (Dean Mayer)

16384 = 16^{3} × (8 – 4)

Of the 46 U.S. presidents to date, 16 have had no middle name.

“It is ill arguing against the use of anything from its abuse.” — Elizabeth I, in Walter Scott’s Kenilworth

Star Trek costume designer William Ware Theiss offered the Theiss Theory of Titillation: “The degree to which a costume is considered sexy is directly proportional to how accident-prone it appears to be.”

Take an ordinary pack of playing cards and regard all the court cards as tens. Now, look at the top card — say it is a seven — place it on the table face downwards and play more cards on top of it, counting up to twelve. Thus, the bottom card being seven, the next will be eight, the next nine, and so on, making six cards in that pile. Then look again at the top card of the pack — say it is a queen — then count 10, 11, 12 (three cards in all), and complete the second pile. Continue this, always counting up to twelve, and if at last you have not sufficient cards to complete a pile, put these apart. Now, if I am told how many piles have been made and how many unused cards remain over, I can at once tell you the sum of all the bottom cards in the piles. I simply multiply by 13 the number of piles less 4, and add the number of cards left over. Thus, if there were 6 piles and 5 cards over, then 13 times 2 (i.e. 6 less 4) added to 5 equals 31, the sum of the bottom cards. Why is this?

Every pile must contain thirteen cards, less the value of the bottom card. Therefore, thirteen times the number of piles less the sum of the bottom cards, and plus the number of cards left over, must equal fifty-two, the number in the pack. Thus thirteen times the number of piles plus number of cards left over, less fifty-two, must equal sum of bottom cards. Or, which is the same thing, the number of piles less four, multiplied by thirteen, and plus the cards left over gives the answer as stated. The algebraically inclined reader can easily express this in terms of his familiar symbols.

The 1672 painting Easel With Still Life of Fruit, by the Flemish painter Cornelius Gijsbrechts, is a sort of apotheosis of trompe-l’œil: The whole thing — not just the still life itself but the easel, all the tools, the other pictures, and the letter — have been painted on a wooden cutout; it’s all an illusion.