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Dudeney's Amusements @dudeney_puzzles@mathstodon.xyz

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re-entrant knight's tour may be made on each part. Cuts along the dotted lines will not do, as the four central squares of the board would be either detached or hanging on by a mere thread. (2/2)

339. THE FOUR KNIGHTS' TOURS. I will repeat that if a chessboard be cut into four equal parts, as indicated by the dark lines in the illustration, it is not possible to perform a knight's tour, either re-entrant or not, on one of the parts. The best re-entrant attempt is shown, in which each knight has to trespass twice on other parts. The puzzle is to cut the board differently into four parts, each of the same size and shape, so that a (1/2)

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (6/6) Show more

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (5/6) Show more

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (4/6) Show more

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (3/6) Show more

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (2/6) Show more

SOLUTION TO 375. FIVE JEALOUS HUSBANDS. (1/6) Show more

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Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. To go over and return counts as two crossings. No tricks such as ropes,
swimming, currents, etc., are permitted. (2/2)

375. FIVE JEALOUS HUSBANDS. During certain local floods five married couples found themselves surrounded by water, and had to escape from their unpleasant position in a boat that would only hold three persons at a time. Every husband was so jealous that he would not allow his wife to be in the boat or on either bank with another man (or with other men) unless he was himself present. Show the quickest way of getting these five men and their wives across into safety.
(1/2)

266. A TENNIS TOURNAMENT. Four married couples played a "mixed double" tennis tournament, a man and a lady always playing against a man and a lady. But no person ever played with or against any other person more than once. Can you show how they all could have played together in the two courts on three successive days? This is a little puzzle of a quite practical kind, and it is just perplexing enough to be interesting.

SOLUTION TO 269. THREE MEN IN A BOAT. (7/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (6/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (5/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (4/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (3/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (2/7) Show more

SOLUTION TO 269. THREE MEN IN A BOAT. (1/7) Show more

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The men within each pair of brackets are here seen to be in the same boat, and therefore A can never go out with B or with C again, and C can never go out again with B. The same applies to the other four boats. The figures show the number on the boat, so that A, B, or C, for example,
can never go out in boat No. 1 again. (5/5)

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readers to discover how the men should have been placed in the boats. As their names happen to have been Andrews, Baker, Carter, Danby, Edwards,
Frith, Gay, Hart, Isaacs, Jackson, Kent, Lang, Mason, Napper, and Onslow, we can call them by their initials and write out the five groups for each of the seven days in the following simple way:

1 2 3 4 5
First Day: (ABC) (DEF) (GHI) (JKL) (MNO).
(4/5)