Bishop Samuel Wilberforce was fond of riddles. After his death in 1873, this one was found among his literary papers:
I’m the sweetest of sounds in Orchestra heard,
Yet in Orchestra never was seen.
I’m a bird of gay plumage, yet less like a bird,
Nothing ever in Nature was seen.
Touch the earth I expire, in water I die,
In air I lose breath, yet can swim and can fly;
Darkness destroys me, and light is my death,
And I only keep going by holding my breath.
If my name can’t be guessed by a boy or a man,
By a woman or girl it certainly can.
No one knows the answer.
A pilot is about to depart in his plane when he meets a young woman on the airport concourse. She has missed her flight.
“I can give you a lift if you like,” he says.
“But you don’t know where I’m going,” she says.
“It doesn’t matter. I can drop you off wherever you like and continue to my destination without going out of my way.”
This seems preposterous until he explains where he’s going. Where is it?
William West noticed this inscription in an alehouse near Brighton. What does it mean?
Sort the numbers 0, 1, 2, …, 123456 into two sets. In one set put all the numbers who digits add to an even sum; in the other put those whose digits produce an odd sum. Which set is larger?
A logic exercise by Lewis Carroll — what conclusion can be drawn from these premises?
- I despise anything that cannot be used as a bridge.
- Nothing that is worth writing an ode to would be an unwelcome gift to me.
- A rainbow will not bear the weight of a wheelbarrow.
- Whatever can be used as a bridge will bear the weight of a wheelbarrow.
- I would not take, as a gift, a thing that I despise.
Prove that the product of four consecutive positive integers cannot be a perfect square.
On Christmas Day 1877, assailed by two young ladies with “nothing to do,” Lewis Carroll invented a new “form of verbal torture”: Presented with two words of the same length, the solver must convert one to the other by changing a single letter at a time, with each step producing a valid English word. For example, HEAD can be converted to TAIL in five steps:
Carroll called the new pastime Doublets and published it in Vanity Fair, which hailed it as “so entirely novel and withal so interesting, that … the Doublets may be expected to become an occupation to the full as amusing as the guessing of the Double Acrostics has already proved.”
In some puzzles the number of steps is specified. In Nabokov’s Pale Fire, the narrator describes a friend who was addicted to “word golf.” “He would interrupt the flow of a prismatic conversation to indulge in this particular pastime, and naturally it would have been boorish of me to refuse playing with him. Some of my records are: HATE-LOVE in three, LASS-MALE in four, and LIVE-DEAD in five (with LEND in the middle).” I’ve been able to solve the first two of these fairly easily, but not the last.
But even without such a constraint, some transformations require a surprising number of steps. Carroll found that 10 were required to turn BLUE into PINK, and in 1968 wordplay expert Dmitri Borgmann declared himself unable to convert ABOVE into BELOW at all.
In a computer study of 5,757 five-letter English words, Donald Knuth found that most could be connected to one another, but 671 could not. One of these, fittingly, was ALOOF. In the wider English language, what proportion of words are “aloof,” words that cannot be connected to any of their fellows? Is ALOOF itself one of these?
In 1917 Sam Loyd and Thomas Edison made this short, which plays with similar ideas. The goat at the end was animated by Willis O’Brien, who would bring King Kong to life 16 years later:
By Carl Kockelkorn and Johannes Kohtz, 1876. White to mate in two moves.
Prove that if each point in the plane is colored red, yellow, or blue, a unit segment must exist whose endpoints are the same color.
From The Black-Out Book, a book of entertainments to amuse Londoners during the air-raid blackouts of 1939:
When a certain commodity was rationed, a woman wrote to her tradesman asking his advice.
If the B mt, put :
If the B . putting :
What was the commodity, and what did he mean?
Using a 7-quart and a 3-quart jug, how can you obtain exactly 5 quarts of water from a well?
That’s a water-fetching puzzle, a familiar task in puzzle books. Most such problems can be solved fairly easily using intuition or trial and error, but in Scripta Mathematica, March 1948, H.D. Grossman describes an ingenious way to generate a solution geometrically.
Let a and b be the sizes of the jugs, in quarts, and c be the number of quarts that we’re seeking. Here, a = 7, b = 3, and c = 5. (a and b must be positive integers, relatively prime, where a is greater than b and their sum is greater than c; otherwise the problem is unsolvable, trivial, or can be reduced to smaller integers.)
Using a field of lattice points (or an actual pegboard), let O be the point (0, 0) and P be the point (b, a) (here, 3, 7). Connect these with OP. Then draw a zigzag line Z to the right of OP, connecting lattice points and staying as close as possible to OP. Now “It may be proved that the horizontal distances from OP to the lattice-points on Z (except O and P) are in some order without repetition 1, 2, 3, …, a + b – 1, if we count each horizontal lattice-unit as the distance a.” In this example, if we take the distance between any two neighboring lattice points as 7, then each of the points on the zigzag line Z will be some unique integer distance horizontally from the diagonal line OP. Find the one whose distance is c (here, 5), the number of quarts that we want to retrieve.
Now we have a map showing how to conduct our pourings. Starting from O and following the zigzag line to C:
- Each horizontal unit means “Pour the contents of the a-quart jug, if any, into the b-quart jug; then fill the a-quart jug from the well.”
- Each vertical unit means “Fill the b-quart jug from the a-quart jug; then empty the b-quart jug.”
So, in our example, the map instructs us to:
- Fill the 7-quart jug.
- Fill the 3-quart jug twice from the 7-quart jug, each time emptying its contents into the well. This leaves 1 quart in the 7-quart jug.
- Pour this 1 quart into the 3-quart jug and fill the 7-quart jug again from the well.
- Fill the remainder of the 3-quart jug (2 quarts) from the 7-quart jug and empty the 3-quart jug. This leaves 5 quarts in the 7-quart jug, which was our goal.
You can find an alternate solution by drawing a second zigzag line to the left of OP. In reading this solution, we swap the roles of a and b given above, so the map tells us to fill the 3-quart jug three times successively and empty it each time into the 7-quart jug (leaving 2 quarts in the 3-quart jug the final time), then empty the 7-quart jug, transfer the remaining 2 quarts to it, and add a final 3 quarts. “There are always exactly two solutions which are in a sense complementary to each other.”
Grossman gives a rigorous algebraic solution in “A Generalization of the Water-Fetching Puzzle,” American Mathematical Monthly 47:6 (June-July 1940), pp. 374-375.
A puzzle by Lewis Carroll:
Two travelers, starting at the same time, went opposite ways round a circular railway. Trains start each way every 15 minutes, the easterly ones going round in 3 hours, the westerly in 2. How many trains did each meet on the way, not counting trains met at the terminus itself?
The digits 123456789 can be arranged to form 362,880 distinct 9-digit numbers.
How many of these are prime?
Alan and Bob are playing a game of marbles. Alan has two marbles, Bob has one, and each rolls to try to come nearest to a fixed point. If the two have equal skill, what is the chance that Alan will win?
There seem to be two contradictory arguments. On the one hand, each of the three marbles has an equal chance of winning, and two of them belong to Alan, so it seems that there’s a 2/3 chance that Alan will win.
On the other hand, there are four possible outcomes: (a) both of Alan’s rolls are better than Bob’s, (b) Alan’s first roll is better than Bob’s, but his second is worse, (c) Alan’s first roll is worse than Bob’s, but his second is better, and (d) both of Alan’s rolls are worse than Bob’s. In 3 of the 4 cases, Alan wins, so it appears that his overall chance of winning is 3/4.
Which argument is correct?