SOLUTION TO 408. MAGIC SQUARES OF TWO DEGREES. (1/4)
The following is the square that I constructed. As it stands the constant is 260. If for every number you substitute, in its allotted place, its square, then the constant will be 11,180. Readers can write out for themselves the second degree square.
The main key to the solution is the pretty law that if eight numbers sum to 260 and their squares to 11,180, then the same will happen in the
260. THE HONEYCOMB PUZZLE. Here is a little puzzle with the simplest possible conditions. Place the point of your pencil on a letter in one of the cells of the honeycomb,
and trace out a very familiar proverb by passing always from a cell to one that is contiguous to it. If you take the right route you will have visited every cell once, and only once. The puzzle is much easier than it looks.
SOLUTION TO 158. THE GREAT MONAD. (1/5)
The areas of circles are to each other as the squares of their diameters. If you have a circle 2 in. in diameter and another 4 in. in diameter, then one circle will be four times as great in area as the other, because the square of 4 is four times as great as the square of 2. Now, if we refer to Diagram 1, we see how two equal squares may be cut into four pieces that will form one larger square; from which it is
158. THE GREAT MONAD. Here is a symbol of tremendous antiquity which is worthy of notice. It is borne on the Korean ensign and merchant flag, and has been adopted as a trade sign by the Northern Pacific Railroad Company, though probably few are aware that it is the Great Monad, as shown in the sketch below. This sign is to the Chinaman what the cross is to the Christian. It is the sign of Deity and eternity, while the two parts into which the (1/4)
SOLUTION TO 404. THE SIBERIAN DUNGEONS. (1/4)
| | | | |
| 8 | 5 | 10 | 11 |
| | | | |
| 16 | 13 | 2 | 3 |
| | | | |
| 1 | 12 | 7 | 14 |
| | | | |
| 9 | 4 | 15 | 6 |
| | | | |
404. THE SIBERIAN DUNGEONS. The above is a trustworthy plan of a certain Russian prison in Siberia. All the cells are numbered, and the prisoners are numbered the same as the cells they occupy. The prison diet is so fattening that these political prisoners are in perpetual fear lest, should their pardon arrive, they might not be able to squeeze themselves through the narrow doorways and get out. And of course it would be an unreasonable thing to (1/3)
131. THE SPANISH MISER. There once lived in a small town in New Castile a noted miser named Don Manuel Rodriguez. His love of money was only equalled by a strong passion for arithmetical problems. These puzzles usually dealt in some way or other with his accumulated treasure, and were propounded by him solely in order that he might have the pleasure of solving them himself. Unfortunately very few of them have survived, and when travelling (1/3)
SOLUTION TO 419. THE FIVE PENNIES.
First lay three of the pennies in the way shown in Fig. 1. Now hold the remaining two pennies in the position shown in Fig. 2, so that they touch one another at the top, and at the base are in contact with the three horizontally placed coins. Then the five pennies will be equidistant, for every penny will touch every other penny.
117. A FENCE PROBLEM. The practical usefulness of puzzles is a point that we are liable to overlook. Yet, as a matter of fact, I have from time to time received quite a large number of letters from individuals who have found that the mastering of some little principle upon which a puzzle was built has proved of considerable value to them in a most unexpected way. Indeed,
it may be accepted as a good maxim that a puzzle is of little real value (1/3)
SOLUTION TO 400. THE MAGIC STRIPS. (1/3)
There are of course six different places between the seven figures in which a cut may be made, and the secret lies in keeping one strip intact and cutting each of the other six in a different place. After the cuts have been made there are a large number of ways in which the thirteen pieces may be placed together so as to form a magic square. Here is one of them:--
The arrangement has some rather interesting features. It will be seen
400. THE MAGIC STRIPS. I happened to have lying on my table a number of strips of cardboard,
with numbers printed on them from 1 upwards in numerical order. The idea suddenly came to me, as ideas have a way of unexpectedly coming, to make a little puzzle of this. I wonder whether many readers will arrive at the same solution that I did.
Take seven strips of cardboard and lay them together as above. Then write on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so that (1/3)
SOLUTION TO 413. A CHESSBOARD FALLACY. (1/2)
The explanation of this little fallacy is as follows. The error lies in assuming that the little triangular piece, marked C, is exactly the same height as one of the little squares of the board. As a matter of fact,
its height (if we make the sixty-four squares each a square inch) will be 1+1/7 in. Consequently the rectangle is really 9+1/7 in. by 7 in., so that the area is sixty-four square inches in either case. Now, although
413. A CHESSBOARD FALLACY.
"Here is a diagram of a chessboard," he said. "You see there are sixty-four squares--eight by eight. Now I draw a straight line from the top left-hand corner, where the first and second squares meet, to the bottom right-hand corner. I cut along this line with the scissors, slide up the piece that I have marked B, and then clip off the little corner C by a cut along the first upright line. This little piece will exactly (1/4)
SOLUTION TO 152. ANOTHER JOINER'S PROBLEM. (1/4)
THE point was to find a general rule for forming a perfect square out of another square combined with a "right-angled isosceles triangle." The triangle to which geometricians give this high-sounding name is, of course, nothing more or less than half a square that has been divided from corner to corner.
The precise relative proportions of the square and triangle are of no consequence whatever. It is only necessary to cut the wood or material
152. ANOTHER JOINER'S PROBLEM. A joiner had two pieces of wood of the shapes and relative proportions shown in the diagram. He wished to cut them into as few pieces as possible so that they could be fitted together, without waste, to form a perfectly square table-top. How should he have done it? There is no necessity to give measurements, for if the smaller piece (which is half a square) be made a little too large or a little too small it will not affect the method of solution.
SOLUTION TO 315. THE HAT-PEG PUZZLE. (1/2)
The moves will be made quite clear by a reference to the diagrams, which show the position on the board after each of the four moves. The darts indicate the successive removals that have been made. It will be seen that at every stage all the squares are either attacked or occupied, and that after the fourth move no queen attacks any other. In the case of the last move the queen in the top row might also have been moved one
315. THE HAT-PEG PUZZLE. Here is a five-queen puzzle that I gave in a fanciful dress in 1897. As the queens were there represented as hats on sixty-four pegs, I will keep to the title, "The Hat-Peg Puzzle." It will be seen that every square is occupied or attacked. The puzzle is to remove one queen to a different square so that still every square is occupied or attacked,
then move a second queen under a similar condition, then a third queen, (1/2)
285. THE FOUR POSTAGE STAMPS.
| 1 | 2 | 3 | 4 |
| 5 | 6 | 7 | 8 |
| 9 | 10 | 11 | 12 |
"It is as easy as counting," is an expression one sometimes hears. But mere counting may be puzzling at times. Take the following simple example. Suppose you have just bought twelve postage stamps, in this form--three by four--and a friend asks you to oblige him with four (1/2)
Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"
A Mastodon instance for maths people. The kind of people who make \(\pi z^2 \times a\) jokes.
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