SOLUTION TO 397. THE MONTENEGRIN DICE GAME. Show more
The players should select the pairs 5 and 9, and 13 and 15, if the chances of winning are to be quite equal. There are 216 different ways in which the three dice may fall. They may add up 5 in 6 different ways and 9 in 25 different ways, making 31 chances out of 216 for the player who selects these numbers. Also the dice may add up 13 in 21 different ways, and 15 in 10 different ways, thus giving the other player also 31 chances in 216.
example, one player may select 7 and 15 and the other 5 and 13. Then if the first player throws so that the three dice add up 7 or 15 he wins,
unless the second man gets either 5 or 13 on his throw.
The puzzle is to discover which two pairs of numbers should be selected in order to give both players an exactly even chance. (2/2)
397. THE MONTENEGRIN DICE GAME. It is said that the inhabitants of Montenegro have a little dice game that is both ingenious and well worth investigation. The two players first select two different pairs of odd numbers (always higher than 3) and then alternately toss three dice. Whichever first throws the dice so that they add up to one of his selected numbers wins. If they are both successful in two successive throws it is a draw and they try again. For (1/2)
SOLUTION TO 119. RACKBRANE'S LITTLE LOSS. Show more
The professor must have started the game with thirteen shillings, Mr. Potts with four shillings, and Mrs. Potts with seven shillings.
was lost by Mr. Potts, and had the effect of doubling the money then held by his wife and the professor. It was then found that each person had exactly the same money, but the professor had lost five shillings in the course of play. Now, the professor asks, what was the sum of money with which he sat down at the table? Can you tell him? (2/2)
119. RACKBRANE'S LITTLE LOSS. Professor Rackbrane was spending an evening with his old friends, Mr. and Mrs. Potts, and they engaged in some game (he does not say what game) of cards. The professor lost the first game, which resulted in doubling the money that both Mr. and Mrs. Potts had laid on the table. The second game was lost by Mrs. Potts, which doubled the money then held by her husband and the professor. Curiously enough, the third game (1/2)
SOLUTION TO 126. SIMPLE MULTIPLICATION. Show more
The number required is 3,529,411,764,705,882, which may be multiplied by 3 and divided by 2, by the simple expedient of removing the 3 from one end of the row to the other. If you want a longer number, you can increase this one to any extent by repeating the sixteen figures in the same order.
126. SIMPLE MULTIPLICATION. If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on the table in this order:--
1 4 2 8 5 7
We can demonstrate that in order to multiply by 3 all that is necessary is to remove the 1 to the other end of the row, and the thing is done. The answer is 428571. Can you find a number that, when multiplied by 3 and divided by 2, the answer will be the same as if we removed the first (1/2)
SOLUTION TO 278. A DORMITORY PUZZLE. Show more
Arrange the nuns from day to day as shown in the six diagrams. The smallest possible number of nuns would be thirty-two, and the arrangements on the last three days admit of variation.
found that five times as many slept on the south side as on each of the other sides. Again she complained. On Wednesday she found four times as many on the south side, on Thursday three times as many, and on Friday twice as many. Urging the nuns to further efforts, she was pleased to find on Saturday that an equal number slept on each of the four sides of the house. What is the smallest number of nuns there could have been, (2/3)
278. A DORMITORY PUZZLE. In a certain convent there were eight large dormitories on one floor,
approached by a spiral staircase in the centre, as shown in our plan. On an inspection one Monday by the abbess it was found that the south aspect was so much preferred that six times as many nuns slept on the south side as on each of the other three sides. She objected to this overcrowding, and ordered that it should be reduced. On Tuesday she (1/3)
SOLUTION TO 279. THE BARRELS OF BALSAM. (3/3) Show more
n + 1
where 2n equals the number of barrels. The symbol C, of course, implies that we have to find how many combinations, or selections, we can make of 2n things, taken n at a time.
SOLUTION TO 279. THE BARRELS OF BALSAM. (2/3) Show more
the row) we get 42, which is the correct answer to the puzzle, for there are 42 different ways of arranging the barrels. Try this method of solution in the case of six barrels, three in each row, and you will find the answer is 5 ways. If you check this by trial, you will discover the five arrangements with 123, 124, 125, 134, 135 respectively in the top row, and you will find no others.
The general solution to the problem is, in fact, this:
SOLUTION TO 279. THE BARRELS OF BALSAM. (1/3) Show more
This is quite easy to solve for any number of barrels--if you know how. This is the way to do it. There are five barrels in each row Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10 together. Divide one result by the other, and we get the number of different combinations or selections of ten things taken five at a time. This is here 252. Now, if we divide this by 6 (1 more than the number in
barrel either beneath or to the right of one of less value. The arrangement shown is, of course, the simplest way of complying with this condition. But there are many other ways--such, for example, as this:--
1 2 5 7 8
3 4 6 9 10
Here, again, no barrel has a smaller number than itself on its right or beneath it. The puzzle is to discover in how many different ways the merchant of Bagdad might have arranged his barrels in the two rows (2/3)
279. THE BARRELS OF BALSAM. A merchant of Bagdad had ten barrels of precious balsam for sale. They were numbered, and were arranged in two rows, one on top of the other,
as shown in the picture. The smaller the number on the barrel, the greater was its value. So that the best quality was numbered "1" and the worst numbered "10," and all the other numbers of graduating values. Now, the rule of Ahmed Assan, the merchant, was that he never put a (1/3)
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.
\) for inline LaTeX, and
\] for display mode.