Edward Elgar loved hidden meanings. The English composer filled his letters with wordplay and musical riddles, and he named one of his family homes Craeg Lea, an anagram of the names (C)arice (A)lice and (E)dward ELGAR.
Two of his puzzles have never been solved. One lies at the heart of the so-called Enigma variations, a set of 14 variations on a theme that Elgar said is “not played.” What does this mean? “The Enigma I will not explain,” he wrote in a program note for the first performance in 1899. “Its ‘dark saying’ must be left unguessed, and I warn you that the connection between the Variations and the Theme is often of the slightest texture; further, through and over the whole set another and larger theme ‘goes’, but is not played. … So the principal Theme never appears, even as in some late dramas … the chief character is never on the stage.” He took the solution to his grave, and music lovers have been searching for the hidden theme for more than 100 years.
The second mystery lies in a letter that Elgar wrote two years before Enigma premiered. On July 14, 1897, he sent the message above to 22-year-old Dora Penny, the daughter of his close friend Alfred Penny, rector of St. Peter’s, Wolverhampton. “It is well known that Elgar was always interested in puzzles, ciphers, cryptograms, and the like,” she wrote in her book Edward Elgar: Memories of a Variation. “The cipher here reproduced — the third letter I had from him, if indeed it is one — came to me enclosed in a letter from [Elgar's wife] to my stepmother. On the back of it is written ‘Miss Penny.’ It followed upon their visit to us at Wolverhampton in July 1897. I have never had the slightest idea what message it conveys; he never explained it and all attempts to solve it have failed. Should any reader of this book succeed in arriving at a solution it would interest me very much to hear of it.”
No satisfactory solution has ever been found. Elgar named Dora herself as the inspiration for the 10th variation of the Enigma, so some wonder whether this was a clue. When she asked him in later life about the musical puzzle, he said, “I thought you, of all people, would guess it.” But no one knows what this means, and if Dora ever found the answer she told no one before her death in 1964. Perhaps the solution is now beyond our reach.
You’re taking care of a friend’s house while he’s on vacation. One hot day you pull the chain on a ceiling fan, and when it doesn’t respond you realize the house has temporarily lost power. Unfortunately, you have to leave now, and you’ll be away for several days.
You know that the fan was in the “off” position before you pulled the chain, and that pulling the chain successively will cycle it through its remaining settings (“off,” “high,” “medium,” etc.). You don’t know how many settings there are, but you’re sure there aren’t more than 4.
How can you ensure that the fan will be in the “off” position when power is restored? (Assume that you can’t simply cut the fan’s power.)
One hundred people stand in a line, all facing in the same direction. Each is wearing a red or a blue hat, assigned at random. Each person can see all the hats before him in the line, but not his own or those of the people behind him. Starting at the back of the line, each person in turn must guess the color of his own hat. Each person can hear all the prior guesses. If the group are allowed to discuss strategy beforehand, how many can be sure of guessing correctly?
Young telegraph operator Joseph Orton Kerbey was enlisted as a spy for the federal forces during the Civil War. In 1861, laid up in a sick bed in Richmond, he needed a way to communicate his latest discoveries to his friends in the north. The message would have to appear innocent and contain the key to its own decipherment. Here’s what he sent:
He directed it, not to his father’s name and address, but to a friend in the telegraph office at Annapolis. What was the hidden message?
Prove that, if we choose nine vertices of a regular icosagon, some three of these will form an isosceles triangle.
Speaking of Lewis Carroll — and further to Wednesday’s logic exercise — here’s the king of all Carroll’s logic problems. What’s the strongest conclusion that can be drawn from these premises?
- When the day is fine, I tell Froggy “You’re quite the dandy, old chap!”
- Whenever I let Froggy forget that 10 pounds he owes me, and he begins to strut about like a peacock, his mother declares “He shall not go out a-wooing!”
- Now that Froggy’s hair is out of curl, he has put away his gorgeous waistcoat.
- Whenever I go out on the roof to enjoy a quiet cigar, I’m sure to discover that my purse is empty.
- When my tailor calls with his little bill, and I remind Froggy of that 10 pounds he owes me, he does not grin like a hyena.
- When it is very hot, the thermometer is high.
- When the day is fine, and I’m not in the humor for a cigar, and Froggy is grinning like a hyena, I never venture to hint that he’s quite the dandy.
- When my tailor calls with his little bill and finds me with an empty pocket, I remind Froggy of that 10 pounds he owes me.
- My railway shares are going up like anything!
- When my purse is empty, and when, noticing that Froggy has got his gorgeous waistcoat on, I venture to remind him of that 10 pounds he owes me, things are apt to get rather warm.
- Now that it looks like rain, and Froggy is grinning like a hyena, I can do without my cigar.
- When the thermometer is high, you need not trouble yourself to take an umbrella.
- When Froggy has his gorgeous waistcoat on, but is not strutting about like a peacock, I betake myself to a quiet cigar.
- When I tell Froggy that he’s quite a dandy, he grins like a hyena.
- When my purse is tolerably full, and Froggy’s hair is one mass of curls, and when he is not strutting about like a peacock, I go out on the roof.
- When my railways shares are going up, and when it’s chilly and looks like rain, I have a quiet cigar.
- When Froggy’s mother lets him go a-wooing, he seems nearly mad with joy, and puts on a waistcoat that is gorgeous beyond words.
- When it is going to rain, and I am having a quiet cigar, and Froggy is not intending to go a-wooing, you had better take an umbrella.
- When my railway shares are going up, and Froggy seems nearly mad with joy, that is the time my tailor always chooses for calling with his little bill.
- When the day is cool and the thermometer low, and I say nothing to Froggy about his being quite the dandy, and there’s not the ghost of a grin on his face, I haven’t the heart for my cigar!
Unfortunately, Carroll died before he was able to publish the solution — but he warned that it contains “a beautiful ‘trap.'”
1. Find an expression for the number 1 that uses each of the digits 0-9 once.
2. Do the same for the number 100.
3. Write 31 using only the digit 3 five times.
4. Express 11 with three 2s.
5. Express 10 with two 2s.
6. Express 1 with three 8s.
7. Express 5 with two 2s.
A logic exercise by Lewis Carroll. What conclusion can be drawn from these premises?
- Animals are always mortally offended if I fail to notice them.
- The only animals that belong to me are in that field.
- No animal can guess a conundrum unless it has been properly trained in a Board-School.
- None of the animals in that field are badgers.
- When an animal is mortally offended, it rushes about wildly and howls.
- I never notice any animal unless it belongs to me.
- No animal that has been properly trained in a Board-School ever rushes about wildly and howls.
This puzzle was devised by W.T. Williams. The goal is to discover the age of Father Dunk’s mother-in-law (2 down), but the clues contain so much cross-reference and the grid so many interlocking solutions that practically the whole puzzle must be completed to find it.
The year is 1939. There are 20 shillings in a pound; 4840 square yards in an acre; a quarter of an acre in a rood; and 1760 yards in a mile.
1. Area in square yards of Dog’s Mead
5. Age of Martha, Father Dunk’s aunt
6. Difference in yards between length and breadth of Dog’s Mead
7. Number of roods in Dog’s Mead times 8 down
8. The year the Dunks acquired Dog’s Mead
10. Father Dunk’s age
11. Year of Mary’s birth
14. Perimeter in yards of Dog’s Mead
15. Cube of Father Dunk’s walking speed in miles per hour
16. 15 across minus 9 down
1. Value in shillings per rood of Dog’s Mead
2. Square of the age of father Dunk’s mother-in-law
3. Age of Mary, father Dunk’s other daughter
4. Value in pounds of Dog’s Mead
6. Age of Ted, father Dunk’s son, who will be twice the age of his sister Mary in 1945
7. Square of the breadth of Dog’s Mead
8. Time in minutes it takes Father Dunk to walk 4/3 times round Dog’s Mead
9. The number which, when multiplied by 10 across, gives 10 down
10. See 9 down
12. Addition of the digits of 10 down plus 1
13. Number of years Dog’s Mead has been in the Dunk family
Hint: Start with 15 across, and keep your wits about you — most of the clues require some sort of insight or intelligent narrowing of the possible solutions; there’s very little mechanical plugging of numbers. With enough careful, dogged reasoning, it’s possible to complete the entire grid, but it’s stupendously hard.
Suppose we have a clock whose hour and minute hands are identical. How times times per day will we find it impossible to tell the time, provided we always know whether it’s a.m. or p.m.?
A boat floats in a swimming pool. In the boat is a block of ice. If the block is dumped into the water and melts, does the water level rise or fall?
When eight certain cards are removed from a standard poker deck, it becomes impossible to deal a straight flush. What are the cards? (Assume the deck contains no jokers.)
On a 12 × 12 grid, some squares are infected and some are healthy. On each turn, a healthy square becomes infected if it has two or more infected orthogonal neighbors. (In the example above, the black squares are infected, the white squares are healthy, and the gray squares will be infected next turn.) What’s the smallest number of initially infected squares that can spread an infection over the whole board?
A more challenging version of the Counterfeit Coin puzzle from 2011:
You have 12 coins, one of which has been replaced with a counterfeit. The false coin differs in weight from the true ones, but you don’t know whether it’s heavier or lighter. How can find it using three weighings in a pan balance?
This, now, is straightforward and business-like: A. applied to B. for a loan of $100. B. replied, ‘My dear A., nothing would please me more than to oblige you, and I’ll do it. I haven’t $100 by me, but make a note and I’ll indorse it, and you can get the money from the bank.’ A. proceeded to write the note. ‘Stay,’ said B., ‘make it $200. I want $100 myself.’ A. did so, B. indorsed the paper, the bank discounted it, and the money was divided. When the note became due, B. was in California, and A. had to meet the payment. What he is unable to cipher out is whether he borrowed $100 of B., or B. borrowed $100 of him.
— Henry C. Percy, Our Cashier’s Scrap-Book, 1879