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directions indicated by the dotted lines there are three queens in a straight line. There is only one of the twelve fundamental ways that will solve the puzzle. Can you find it? (5/5)

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ninety-six, as would happen if all twelve were non-symmetrical. It is well to have a clear understanding on the matter of reversals and reflections when dealing with puzzles on the chessboard.

Can the reader place the eight queens on the board so that no queen shall attack another and so that no three queens shall be in a straight line in any oblique direction? Another glance at the diagram will show that this arrangement will not answer the conditions, for in the two (4/5)

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other sides at the bottom, you get another way that is not identical. Then if you reflect these two ways in a mirror you get two more ways. Now, all the other eleven solutions are non-symmetrical, and therefore each of them may be presented in eight ways by these reversals and reflections. It will thus be seen why the twelve fundamentally different solutions produce only ninety-two arrangements, as I have said, and not (3/5)

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literature of its own) to discover in just how many different ways this may be done. I show one way in the diagram, and there are in all twelve of these fundamentally different ways. These twelve produce ninety-two ways if we regard reversals and reflections as different. The diagram is in a way a symmetrical arrangement. If you turn the page upside down, it will reproduce itself exactly; but if you look at it with one of the (2/5)

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300. THE EIGHT QUEENS. The queen is by far the strongest piece on the chessboard. If you place her on one of the four squares in the centre of the board, she attacks no fewer than twenty-seven other squares; and if you try to hide her in a corner, she still attacks twenty-one squares. Eight queens may be placed on the board so that no queen attacks another, and it is an old puzzle (first proposed by Nauck in 1850, and it has quite a little (1/5)

SOLUTION TO 365. NEW MEASURING PUZZLE. (2/2) 

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9 .. 3 .. 4 .. 4
9 .. 3 .. 5 .. 3
9 .. 8 .. 0 .. 3
4 .. 8 .. 5 .. 3
4 .. 10 .. 3 .. 3

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SOLUTION TO 365. NEW MEASURING PUZZLE. (1/2) 

The following solution in eleven manipulations shows the contents of every vessel at the start and after every manipulation:--

10-quart. 10-quart. 5-quart. 4-quart.

10 .. 10 .. 0 .. 0
5 .. 10 .. 5 .. 0
5 .. 10 .. 1 .. 4
9 .. 10 .. 1 .. 0
9 .. 6 .. 1 .. 4
9 .. 7 .. 0 .. 4
9 .. 7 .. 4 .. 0

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365. NEW MEASURING PUZZLE. Here is a new poser in measuring liquids that will be found interesting. A man has two ten-quart vessels full of wine, and a five-quart and a four-quart measure. He wants to put exactly three quarts into each of the two measures. How is he to do it? And how many manipulations (pourings from one vessel to another) do you require? Of course, waste of wine, tilting, and other tricks are not allowed.

SOLUTION TO 238. ARRANGING THE JAM POTS. (2/2) 

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18-9). When you have made the interchanges within any pair of brackets,
all numbers within those brackets are in their places. There are five pairs of brackets, and 5 from 22 gives the number of changes required--17.

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SOLUTION TO 238. ARRANGING THE JAM POTS. (1/2) 

Two of the pots, 13 and 19, were in their proper places. As every interchange may result in a pot being put in its place, it is clear that twenty-two interchanges will get them all in order. But this number of moves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17),
(24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14,

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SOLUTION TO 129. THE BATTLE OF HASTINGS. (5/5) 

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way: Allowing one square foot of standing-room per man, each small square would have required all the space allowed by a globe three times the diameter of the earth.

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SOLUTION TO 129. THE BATTLE OF HASTINGS. (4/5) 

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The reason why I assumed that there must be something wrong with the figures in the chronicle is that we can confidently say that Harold's army did not contain over three trillion men! If this army (not to mention the Normans) had had the whole surface of the earth (sea included) on which to encamp, each man would have had slightly more than a quarter of a square inch of space in which to move about! Put another

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SOLUTION TO 129. THE BATTLE OF HASTINGS. (3/5) 

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side. The general problem, of which this is a particular case, is known as the "Pellian Equation"--apparently because Pell neither first propounded the question nor first solved it! It was issued as a challenge by Fermat to the English mathematicians of his day. It is readily solved by the use of continued fractions.

Next to 61, the most difficult number under 100 is 97, where 97 ×
6,377,352² + 1 = a square.

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SOLUTION TO 129. THE BATTLE OF HASTINGS. (2/5) 

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Now, of all the numbers from 2 to 99 inclusive, 61 happens to be the most awkward one to work, and the lowest possible answer to our puzzle is that Harold's army consisted of 3,119,882,982,860,264,400 men. That is, there would be 51,145,622,669,840,400 men (the square of 226,153,980) in each of the sixty-one squares. Add one man (Harold), and they could then form one large square with 1,766,319,049 men on every

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SOLUTION TO 129. THE BATTLE OF HASTINGS. (1/5) 

Any number (not itself a square number) may be multiplied by a square that will give a product 1 less than another square. The given number must not itself be a square, because a square multiplied by a square produces a square, and no square plus 1 can be a square. My remarks throughout must be understood to apply to whole numbers, because fractional soldiers are not of much use in war.

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form a square with 31 men on every side. Similarly in the case of the figures I have given for 62. Now, find the lowest answer for 61. (6/6)

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square. What is the smallest possible number of men there could have been?

In order to make clear to the reader the simplicity of the question, I will give the lowest solutions in the case of 60 and 62, the numbers immediately preceding and following 61. They are 60 × 4² + 1 = 31²,
and 62 × 8² + 1 = 63². That is, 60 squares of 16 men each would be 960 men, and when Harold joined them they would be 961 in number, and so (5/6)

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castle, impenetrable to the Normans;" while Robert Wace, a century after, tells us the same thing. So in this respect my newly-discovered chronicle may not be greatly in error. But I have reason to believe that there is something wrong with the actual figures. Let the reader see what he can make of them.

The number of men would be sixty-one times a square number; but when Harold himself joined in the fray they were then able to form one large (4/6)

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fray the Saxons were one mighty square of men, shouting the battle-cries, 'Ut!' 'Olicrosse!' 'Godemitè!'"

Now, I find that all the contemporary authorities agree that the Saxons did actually fight in this solid order. For example, in the "Carmen de Bello Hastingensi," a poem attributed to Guy, Bishop of Amiens, living at the time of the battle, we are told that "the Saxons stood fixed in a dense mass," and Henry of Huntingdon records that "they were like unto a (3/6)

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can hardly fail to interest those of my readers who have arithmetical predilections. Here is the passage in question.

"The men of Harold stood well together, as their wont was, and formed sixty and one squares, with a like number of men in every square thereof, and woe to the hardy Norman who ventured to enter their redoubts; for a single blow of a Saxon war-hatchet would break his lance and cut through his coat of mail.... When Harold threw himself into the (2/6)

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