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Remember that when you have made your five pieces, you must be able, as desired, to put them together to form either the single original triangle or to form two triangles or to form three triangles--all equilateral. (2/2)

(1/2)

SOLUTION TO 292. THE ABBOT'S WINDOW.

THE man who was "learned in strange mysteries" pointed out to Father John that the orders of the Lord Abbot of St. Edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch:--

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of lights. See thou that this be done without delay, lest the cellars be locked up for a month and other grievous troubles befall thee."

Father John was at his wits' end, but after consultation with one who was learned in strange mysteries, a way was found to satisfy the whim of the Lord Abbot. Which lights were blocked up, so that those which remained added up an even number in every line horizontally, vertically,

and diagonally, while the least possible obstruction of light was (4/5)

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"Dost thou not see that the sixty-four lights add up an even number vertically and horizontally, but that all the _diagonal_ lines, except fourteen are of a number that is odd? Why is this?"

"Of a truth, my Lord Abbot, it is of the very nature of things, and cannot be changed."

"Nay, but it _shall_ be changed. I command thee that certain of the lights be closed this day, so that every line shall have an even number (3/5)

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for Father John, and that venerable monk was soon at the bedside.

"Father John," said the Abbot, "dost thou know that I came into this wicked world on a Christmas Even?"

The monk nodded assent.

"And have I not often told thee that, having been born on Christmas Even, I have no love for the things that are odd? Look there!"

The Abbot pointed to the large dormitory window, of which I give a sketch. The monk looked, and was perplexed.

(2/5)

"devotions too strong for his head," fell sick and was unable to leave his bed. As he lay awake, tossing his head restlessly from side to side,

the attentive monks noticed that something was disturbing his mind; but nobody dared ask what it might be, for the abbot was of a stern disposition, and never would brook inquisitiveness. Suddenly he called (1/5)

SOLUTION TO 198. THE EIGHT STICKS. (3/3)

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be impossible, because an ordinary match is about twenty-one times as long as it is broad, and the enclosed rectangles would not be squares.

SOLUTION TO 198. THE EIGHT STICKS. (2/3)

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of them touch the table: to say that every one lies on the table would not be correct. To obtain a solution it is only necessary to have our sticks of proper dimensions. Say the long sticks are each 2 ft. in length and the short ones 1 ft. Then the sticks must be 3 in. thick,

when the three equal squares may be enclosed, as shown in the second diagram. If I had said "matches" instead of "sticks," the puzzle would

SOLUTION TO 198. THE EIGHT STICKS. (1/3)

The first diagram is the answer that nearly every one will give to this puzzle, and at first sight it seems quite satisfactory. But consider the conditions. We have to lay "every one of the sticks on the table." Now,

if a ladder be placed against a wall with only one end on the ground, it can hardly be said that it is "laid on the ground." And if we place the sticks in the above manner, it is only possible to make one end of two

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SOLUTION TO 76. THE BARREL OF BEER. (2/2)

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20 gallons of beer for his own use and sold one man 33 gallons (the 18-gallon and 15-gallon barrels) and sold the other man 66 gallons (the 16, 19, and 31 gallon barrels).

SOLUTION TO 76. THE BARREL OF BEER. (1/2)

Here the digital roots of the six numbers are 6, 4, 1, 2, 7, 9, which together sum to 29, whose digital root is 2. As the contents of the barrels sold must be a number divisible by 3, if one buyer purchased twice as much as the other, we must find a barrel with root 2, 5, or 8 to set on one side. There is only one barrel, that containing 20 gallons, that fulfils these conditions. So the man must have kept these

SOLUTION TO 98. ACADEMIC COURTESIES.

There must have been ten boys and twenty girls. The number of bows girl to girl was therefore 380, of boy to boy 90, of girl with boy 400, and of boys and girls to teacher 30, making together 900, as stated. It will be remembered that it was not said that the teacher himself returned the bows of any child.

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can you say exactly how many boys there were in the school? If you are not very careful, you are likely to get a good deal out in your calculation. (2/2)

SOLUTION TO 51. HOW OLD WAS MARY? (2/2)

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times as old as Ann (5½)." Now, check this backwards. When Mary was three times as old as Ann, Mary was 16½ and Ann 5½ (11 years younger). Then we get 49½ for the age Ann will be when she is three times as old as Mary was then. When Mary was half this she was 24¾. And at that time Ann must have been 13¾ (11 years younger). Therefore Mary is now twice as old--27½, and Ann 11 years younger--16½.

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Puzzles from Henry Ernest Dudeney's "Amusements in Mathematics"

Source: https://www.gutenberg.org/ebooks/16713

Joined Sep 2017