65. THE RAILWAY STATION CLOCK. A clock hangs on the wall of a railway station, 71 ft. 9 in. long and 10 ft. 4 in. high. Those are the dimensions of the wall, not of the clock! While waiting for a train we noticed that the hands of the clock were pointing in opposite directions, and were parallel to one of the diagonals of the wall. What was the exact time?
SOLUTION TO 265. A PUZZLE FOR CARD-PLAYERS. (4/4) Show more
It will be seen that no man ever plays with or against his own wife--an ideal arrangement. If the reader wants a hard puzzle, let him try to arrange eight married couples (in four courts on seven days) under exactly similar conditions. It can be done, but I leave the reader in this case the pleasure of seeking the answer and the general solution.
SOLUTION TO 265. A PUZZLE FOR CARD-PLAYERS. (3/4) Show more
twice as his opponent.
266.--A TENNIS TOURNAMENT.
Call the men A, B, D, E, and their wives a, b, d, e. Then they may play as follows without any person ever playing twice with or against any other person:--
First Court. Second Court.
1st Day | A d against B e | D a against E b
2nd Day | A e " D b | E a " B d
3rd Day | A b " E d | B a " D e
SOLUTION TO 265. A PUZZLE FOR CARD-PLAYERS. (2/4) Show more
A H -- D G | K E -- B F | L C -- I J
A I -- E H | L F -- C G | B D -- J K
A J -- F I | B G -- D H | C E -- K L
A K -- G J | C H -- E I | D F -- L B
A L -- H K | D I -- F J | E G -- B C
It will be seen that the letters B, C, D ...L descend cyclically. The solution given above is absolutely perfect in all respects. It will be found that every player has every other player once as his partner and
SOLUTION TO 265. A PUZZLE FOR CARD-PLAYERS. (1/4) Show more
In the following solution each of the eleven lines represents a sitting,
each column a table, and each pair of letters a pair of partners.
A B -- I L | E J -- G K | F H -- C D
A C -- J B | F K -- H L | G I -- D E
A D -- K C | G L -- I B | H J -- E F
A E -- L D | H B -- J C | I K -- F G
A F -- B E | I C -- K D | J L -- G H
A G -- C F | J D -- L E | K B -- H I
265. A PUZZLE FOR CARD-PLAYERS. Twelve members of a club arranged to play bridge together on eleven evenings, but no player was ever to have the same partner more than once, or the same opponent more than twice. Can you draw up a scheme showing how they may all sit down at three tables every evening? Call the twelve players by the first twelve letters of the alphabet and try to group them.
SOLUTION TO 341. THE FOUR FROGS. (7/7) Show more
last moved _from_." This is not, of course, the only way of placing the counters, but it is the simplest solution to carry in the mind.
There are several puzzles in this book that the reader will find lend themselves readily to this method.
SOLUTION TO 341. THE FOUR FROGS. (6/7) Show more
to the next vacant point (in either direction), where you deposit the counter. You proceed in the same way until all the counters are placed. Remember you always touch a vacant place and slide the counter from it to the next place, which must be also vacant. Now, by the "buttons and string" method of simplification we can transform the diagram into E. Then the solution becomes obvious. "Always move _to_ the point that you
SOLUTION TO 341. THE FOUR FROGS. (5/7) Show more
moves given above, and you will find that every little difficulty has disappeared.
In Diagram D I give another familiar puzzle that first appeared in a book published in Brussels in 1789, _Les Petites Aventures de Jerome Sharp_. Place seven counters on seven of the eight points in the following manner. You must always touch a point that is vacant with a counter, and then move it along a straight line leading from that point
SOLUTION TO 341. THE FOUR FROGS. (4/7) Show more
the diagram in the form shown in Diagram C, where the relationship between the buttons is precisely the same as in B. Any solution on C will be applicable to B, and to A. Place your white knights on 1 and 3 and your black knights on 6 and 8 in the C diagram, and the simplicity of the solution will be very evident. You have simply to move the knights round the circle in one direction or the other. Play over the
SOLUTION TO 341. THE FOUR FROGS. (3/7) Show more
moves are indicated by lines, to obviate the necessity of the reader's understanding the nature of the knight's move in chess. But it will at once be seen that the two problems are identical. The central square can, of course, be ignored, since no knight can ever enter it. Now,
regard the toadstools as buttons and the connecting lines as strings, as in Diagram B. Then by disentangling these strings we can clearly present
SOLUTION TO 341. THE FOUR FROGS. (2/7) Show more
This is the familiar old puzzle by Guarini, propounded in 1512, and I give it here in order to explain my "buttons and string" method of solving this class of moving-counter problem.
Diagram A shows the old way of presenting Guarini's puzzle, the point being to make the white knights change places with the black ones. In
"The Four Frogs" presentation of the idea the possible directions of the
SOLUTION TO 341. THE FOUR FROGS. (1/7) Show more
The fewest possible moves, counting every move separately, are sixteen. But the puzzle may be solved in seven plays, as follows, if any number of successive moves by one frog count as a single play. All the moves contained within a bracket are a single play; the numbers refer to the toadstools: (1--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4,
4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1).
341. THE FOUR FROGS. In the illustration we have eight toadstools, with white frogs on 1 and 3 and black frogs on 6 and 8. The puzzle is to move one frog at a time,
in any order, along one of the straight lines from toadstool to toadstool, until they have exchanged places, the white frogs being left on 6 and 8 and the black ones on 1 and 3. If you use four counters on a simple diagram, you will find this quite easy, but it is a little more (1/2)
SOLUTION TO 151. THE JOINER'S PROBLEM. (2/2) Show more
correct; thus the triangle BEF is just a quarter of the square BCDF. Draw lines from B to D and from C to F and this will be clear.
SOLUTION TO 151. THE JOINER'S PROBLEM. (1/2) Show more
Nothing could be easier than the solution of this puzzle--when you know how to do it. And yet it is apt to perplex the novice a good deal if he wants to do it in the fewest possible pieces--three. All you have to do is to find the point A, midway between B and C, and then cut from A to D and from A to E. The three pieces then form a square in the manner shown. Of course, the proportions of the original figure must be
151. THE JOINER'S PROBLEM. I have often had occasion to remark on the practical utility of puzzles,
arising out of an application to the ordinary affairs of life of the little tricks and "wrinkles" that we learn while solving recreation problems.
The joiner, in the illustration, wants to cut the piece of wood into as few pieces as possible to form a square table-top, without any waste of material. How should he go to work? How many pieces would you require?
SOLUTION TO 197. THE COMPASSES PUZZLE. (4/4) Show more
the fact that HKI is an isosceles triangle by the construction, it can be proved that HI is half of HB. We can similarly prove that C is the centre of the square of which AIB are three corners.
I am aware that this is not the simplest possible solution.
SOLUTION TO 197. THE COMPASSES PUZZLE. (3/4) Show more
that twice the square of the line AB equals the square of the distance BG, from which it follows that HABN are the four corners of a square. To prove that I is the centre of this square, draw a line from H to P through QIB and continue the arc HK to P. Then, conceiving the necessary lines to be drawn, the angle HKP, being in a semicircle, is a right angle. Let fall the perpendicular KQ, and by similar triangles, and from
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
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