63. THE STOP-WATCH. We have here a stop-watch with three hands. The second hand, which travels once round the face in a minute, is the one with the little ring at its end near the centre. Our dial indicates the exact time when its owner stopped the watch. You will notice that the three hands are nearly equidistant. The hour and minute hands point to spots that are exactly a third of the circumference apart, but the second hand is a little too (1/2)
SOLUTION TO 133. THE FIVE BRIGANDS. (7/8)
one case being 1 and in the other -4. Thus in the case where Alfonso holds 6 doubloons one progression is 33 + 34 + 35 + 36 + ... + 43 + 44,
and the other 42 + 38 + 34 + 30 + ... + 6 + 2. The sum of the first series is 462, and of the second 242--results which again agree with the figures already given. The problem may be said to consist in finding the first and last terms of these progressions. I should remark that where
SOLUTION TO 133. THE FIVE BRIGANDS. (6/8)
200 doubloons 200 doubloons.
These will be found to work correctly. All the rest of the 704 answers,
where Alfonso always holds six doubloons, may be obtained in this way from the two tables by substituting the different numbers for the letters m and n.
Put in another way, for every holding of Alfonso the number of answers is the sum of two arithmetical progressions, the common difference in
SOLUTION TO 133. THE FIVE BRIGANDS. (5/8)
value of n. The former, therefore, produces 462 and the latter 242 answers, which together make 704, as already stated.
Let us take Table I., and say n = 5 and m = 2; also in Table II. take n
= 13 and m = 0. Then we at once get these two answers:--
A = 6 A = 6
B = 5 B = 13
C = 40 C = 1
D = 140 D = 168
E = 9 E = 12
SOLUTION TO 133. THE FIVE BRIGANDS. (4/8)
In the first table we may substitute for n any whole number from 1 to 12 inclusive, and m may be nought or any whole number from 1 to (31 + n) inclusive. In the second table n may have the value of any whole number from 13 to 23 inclusive, and m may be nought or any whole number from 1 to (93 - 4n) inclusive. The first table thus gives (32 + n) answers for every value of n; and the second table gives (94 - 4n) answers for every
SOLUTION TO 133. THE FIVE BRIGANDS. (3/8)
same amount. Let us take the cases where Alfonso has 6 doubloons, and see how we may obtain all the 704 different ways indicated above. Here are two tables that will serve as keys to all these answers:--
Table I. Table II.
A = 6. A = 6.
B = n. B = n.
C = (63 - 5n) + m. C = 1 + m.
D = (128 + 4n) - 4m. D = (376 - 16n) - 4m.
E = 3 + 3m. E = (15n - 183) + 3m.
SOLUTION TO 133. THE FIVE BRIGANDS. (2/8)
ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso held 11 doubloons, the remainder could be distributed in 3 different ways. More than 11 doubloons he could not possibly have had. It will scarcely be expected that I shall give all these 6,627 ways at length. What I propose to do is to enable the reader, if he should feel so disposed, to write out all the answers where Alfonso has one and the
SOLUTION TO 133. THE FIVE BRIGANDS. (1/8)
The sum of 200 doubloons might have been held by the five brigands in any one of 6,627 different ways. Alfonso may have held any number from 1 to 11. If he held 1 doubloon, there are 1,005 different ways of distributing the remainder; if he held 2, there are 985 ways; if 3,
there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways;
if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388
This problem, worded somewhat differently, was propounded by Tartaglia (died 1559), and he flattered himself that he had found one solution;
but a French mathematician of note (M.A. Labosne), in a recent work,
says that his readers will be astonished when he assures them that there are 6,639 different correct answers to the question. Is this so? How many answers are there? (3/3)
There are a good many equally correct answers to this question. Here is one of them:
A 6 × 12 = 72
B 12 × 3 = 36
C 17 × 1 = 17
D 120 × ½ = 60
E 45 × 1/3 = 15
The puzzle is to discover exactly how many different answers there are,
it being understood that every man had something and that there is to be no fractional money--only doubloons in every case.
133. THE FIVE BRIGANDS. The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban,
were counting their spoils after a raid, when it was found that they had captured altogether exactly 200 doubloons. One of the band pointed out that if Alfonso had twelve times as much, Benito three times as much,
Carlos the same amount, Diego half as much, and Esteban one-third as much, they would still have altogether just 200 doubloons. How many doubloons had each?
SOLUTION TO 386. A TRICK WITH DICE.
All you have to do is to deduct 250 from the result given, and the three figures in the answer will be the three points thrown with the dice. Thus, in the throw we gave, the number given would be 386; and when we deduct 250 we get 136, from which we know that the throws were 1, 3, and 6.
The process merely consists in giving 100a + 10b + c + 250, where a, b,
and c represent the three throws. The result is obvious.
386. A TRICK WITH DICE. Here is a neat little trick with three dice. I ask you to throw the dice without my seeing them. Then I tell you to multiply the points of the first die by 2 and add 5; then multiply the result by 5 and add the points of the second die; then multiply the result by 10 and add the points of the third die. You then give me the total, and I can at once tell you the points thrown with the three dice. How do I do it? As an (1/2)
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 112. A PUZZLING LEGACY.
As the share of Charles falls in through his death, we have merely to divide the whole hundred acres between Alfred and Benjamin in the proportion of one-third to one-fourth--that is in the proportion of four-twelfths to three-twelfths, which is the same as four to three. Therefore Alfred takes four-sevenths of the hundred acres and Benjamin three-sevenths.
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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