Twelve:

From Loyd’s Cyclopedia of Puzzles, 1914.

Call the bugs A, B, C, and D. In the initial position, when D looks at the other three bugs, she sees triangle ABC labeled clockwise, but in the final position she sees it labeled counterclockwise (or vice versa). Since the bugs move continuously, some position must arise during the walk in which either D is on the plane ABC or ABC do not determine a plane. In the latter case they’re collinear, which means they share a plane with any other point, including D.

“The frogs that jump are George in the third horizontal row; Chang, the artful-looking batrachian at the end of the fourth row; and Wilhelmina, the fair creature in the seventh row. George jumps downwards to the second tumbler in the seventh row; Chang, who can only leap short distances in consequence of chronic rheumatism, removes somewhat unwillingly to the glass just above him — the eighth in the third row; while Wilhelmina, with all the sprightliness of her youth and sex, performs the very creditable saltatory feat of leaping to the fourth tumbler in the fourth row. In their new positions it will be found that of the eight frogs no two are in a line vertically, horizontally, or diagonally.”

(Henry E. Dudeney, “The Professor’s Puzzles,” Strand 12:12 [December 1896], 720-726; “The Professor’s Puzzles: Solutions,” Strand 13:1 [January 1897], 106-109.)

1. Bb5 (there’s a metaphor in here somewhere) and the queen will mate.

From the Central New Jersey Herald, via Charles F. Wadworth, Unique Chess Problems, 1890.

Keep the goat in the boat with you for the first two trips, taking first the dog and then the cabbage across and leaving them on the opposite bank. Take the goat back with you and leave it on the riverbank by itself as you take the two wolves across. Leave them with the cabbage and return with the dog. Collect the goat and bring it and the dog to the farther shore.

(By E. Chernyshov, via Quantum, May 1990.)

If we work backward, we can see that at each step there’s only one way in which the money can have been distributed:

After three rounds, they have $24, $24, and $24.

So after two rounds the amounts must have been $12, $12, and $48.

Hence after one round the amounts must have been $6, $42, and $24.

And thus originally the amounts were $39, $21, and $12.

Under the rules as described, with each man losing once, these must be the amounts that they started with. Note that the total amount held by the group is always $72.

From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

Solution:

1. Rc2 b5 2. Kc7 (any) 3. Ra2#

1. Rc2 Ka8 2. Kc7 (any) 3. Ra2#

1. Rc2 Kb8 2. Kxb6 Ka8 3. Rc8#

“The trick was to first reverse the tube so that the three depressions, D, E, and F, were at the bottom. It was quite easy to get the shots into these hollows, and when you had them in position you had merely to twist the tube with a quick turn of the fingers, holding it at the ends, when the pellets would fall into the required positions. You could hardly fail once in a hundred attempts, yet I have seen people try the puzzle for hours without success, while this simple trick never occurred to them.”

(Henry E. Dudeney, “The World’s Best Puzzles,” Strand 36:216 [January 1909], 692-700.)

3816547290.

(By Nob Yoshigahara.)

09/14/2021 UPDATE: I came across this puzzle in Yoshigahara’s 2004 book Puzzles 101: A Puzzlemaster’s Challenge. In an article in Quanta last year, Pradeep Mutalik credits it to John Horton Conway. I don’t know who came up with it first; possible the two devised it independently. (Thanks, Dharmesh.)

09/17/2021 UPDATE: Reader Ben Hamula has worked out programmatically that this solution is unique.

09/17/2021 UPDATE: This is getting interesting. Reader Hans Havermann finds that neither Yoshigahara nor Conway originated the problem. The earliest known mention is in the December 1984 issue of the New Zealand Mathematical Society newsletter (Problem 15 on page 17). David B. Gauld, mentioned in that problem, is sometimes cited as its author; Gauld contributed OEIS sequence A181736 in 2010. Immediately after the problem appeared in the New Zealand newsletter, Kishor N. Gordhandas posed it in the January 1985 issue of Games (page 2) and it generally escaped into the wild.

Hans contacted Gauld, who recalls that he encountered 3816547290 in François Le Lionnais and Jean Brette’s 1983 book Les nombres remarquables and passed it on to Teddy Zulauf, sub-editor of the NZMS newsletter. Many thanks to Hans for all this yeoman sleuthing; I’ll post a further update if Le Lionnais turns out to cite a source.

09/30/2021 UPDATE: Le Lionnais cites no source but calls the fact “Trés remarquable.” See Hans’ blog for more on the search.

10/04/2021 UPDATE: It appears Hans has identified the originator of the puzzle: Lea Gorodisky, in the Spanish puzzle magazine The Snark. See the link immediately above. (Thanks again, Hans.)

“Play the three queens as follows. There is no necessity to give the moves of the Black king, as they are all forced.”

1. Qe1 (“just a waste move”) 2. Qh1 3. Qhg1 4. Qgf1 5. Qce1 6. Qa1+ 7. Qeb1#