“King Charles the First walked and talked; half an hour after, his head was cut off.”

From Peter Puzzlewell, A Choice Collection of Riddles, Charades, and Rebuses, 1794.

Add a comma after cannot.

Anton was the last passenger on and the second one off, so no one entered the elevator after him and only one person exited before him. He went up at least two floors. Edsel got off before Gerda got on, so Edsel must be the passenger who exited the elevator before Anton did. This means that Filbert entered the car sometime before Anton and left it after him, which means he went up at least four floors. Edsel traveled exactly twice as many floors as Filbert did, so Edsel went up at least eight floors. If we assume that Filbert went up only four floors, then at a minimum we have this:

Floor 1: Edsel gets on
Floor 9: Edsel gets off
Floor 10: Gerda gets on
Floor 11: Filbert gets on
Floor 12: Anton gets on
Floor 13: No one gets on or off
Floor 14: Anton gets off
Floor 15: Filbert gets off
Floor 17: No one gets on or off

This scheme leaves exactly four floors (16, 18, 19, 20) for four passengers (Bessie, Clara, Dagwood, and Gerda) to get off, so this minimum scheme must be correct, and Filbert cannot have risen more than four floors.

So that gives us 11 known floor stops (1, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20). Five of these are odd and six even. We’re told that the elevator’s doors opened at more odd-numbered floors than even-numbered floors, so there were eight stops at odd floors, and the three in addition to the ones we’ve named must be 3, 5, and 7. These must be the floors on which Dagwood, Bessie, and Clara (in that order) got on. Bessie went up more floors than Dagwood, so he must have got off at 16 (after rising 13 floors) and then Bessie got off at 19 or 20 (after rising 14 or 15 floors). If Dagwood had gotten off at floor 18 or higher then Bessie would have had to get off above floor 20. We’re told that no two passengers traveled the same number of floors, so Gerda, who got on at 10, couldn’t have exited at 18. We also know Gerda wasn’t the last one to exit, so she must have got off at 19, which means Bessie got off at 20 and Clara got off at 18:

Anton: on at 12, off at 14, rising 2
Bessie: on at 5, off at 20, rising 15
Clara: on at 7, off at 18, rising 11
Dagwood: on at 3, off at 16, rising 13
Edsel: on at 1, off at 9, rising 8
Filbert: on at 11, off at 15, rising 4
Gerda: on at 10, off at 19, rising 9

From Games magazine, April 1993.

The flag measures 4 feet by 3 feet, so its diagonal is 5 feet. “All you need to do is to deduct half the length of this diagonal (2 1/2 feet) from a quarter of the distance all round the edge of the flag (3 1/2 feet) — a quarter of 14 feet. The difference (1 ft.) is the required width of the arm of the red cross.”

Proof:

If the flag measures 4 feet by 3 feet, the red area is given by

4x + 3x – x² = 6,

so

x² – 7x + 6 = 0.

Solving the quadratic equation gives x – 3.5 = ± 2.5, and as x can’t be 6, the answer must be 1 foot.

From Dudeney’s Amusements in Mathematics, via Erwin Brecher’s Ultimate Book of Puzzles, Mathematical Diversions, and Brainteasers (1996).

04/18/2025 UPDATE: Reader Catalin Voinescu has a simpler solution:

“Consider: the banner has the same areas of red and white as a banner of the same size with a red L of the same thickness as the cross along two of the edges. (Cut the original banner in four and rearrange the fragments.) In this case, the white area is a rectangle. The problem states that it must be half of the area of the flag, that is, 6 sq ft. Its sides are shorter than the sides of the banner by the same amount each, namely, the thickness of the red L. Without any equation, one immediately notices that 3 by 2 gives 6, and that’s 1 less than 4 by 3. So the red part is 1 ft wide.”

(Thanks, Catalin.)

In place of each decagon, consider a circle whose circumference is 99 units long and divided into segments whose length corresponds to the vertex integers, in order. So if the first few vertices around one decagon bear the integers 14, 9, 30, then the corresponding segments of that circle will have those lengths.

Now superimpose the two circles and rotate one of them. If at any point two pairs of segment endpoints coincide simultaneously, then we have our solution — these endpoints identify two sets of successive vertices on the two decagons that produce the same sum.

What if we never get two simultaneous coincidences? Well, that’s impossible: During a full rotation, each endpoint on the first circle must encounter 10 endpoints on the other circle, so there are 100 coincidences altogether. These must occur over the course of 99 discrete opportunities (the lengths of the two circles’ circumferences), so some two coincidences must occur together.

From the January-February 1995 issue of Quantum.

OILILY and BAILIWICK.

See Pillow Talk.

Insert a C before each capitalized word (except those that begin sentences):

“The Coptic creed has a chaste charm. It chose, one assumes, not to be clever; it neither cheats the cripple, nor chides the crazed sinner for his crude crime; it is not cold. Then cleave to it; ere thy grave closes, thou mayst find here the chart to the crest thou cravest to climb.”

From The Game of Words, 1987.

No. Call the current six numbers a₁ … a₆. Now I = a₁ – a₂ + a₃ – a₄ + a₅ – a₆ is initially 2, and I will remain unchanged no matter what we do. So there’s no way to reach I = 0.

Via Arthur Engel, Problem-Solving Strategies, 1999.

Yes.

(From Quantum, November-December 1994.)