Amusements in Mathematics, Henry Ernest Dudeney

  • Question 121 - HEADS OR TAILS

    Crooks, an inveterate gambler, at Goodwood recently said to a friend, "I'll bet you half the money in my pocket on the toss of a coin—heads I win, tails I lose." The coin was tossed and the money handed over. He repeated the offer again and again, each time betting half the money then in his possession. We are not told how long the game went on, or how many times the coin was tossed, but this we know, that the number of times that Crooks lost was exactly equal to the number of times that he won. Now, did he gain or lose by this little venture?
    Topics:
    Algebra -> Word Problems

  • Question 122 - THE SEE-SAW PUZZLE

    Necessity is, indeed, the mother of invention. I was amused the other day in watching a boy who wanted to play see-saw and, in his failure to find another child to share the sport with him, had been driven back upon the ingenious resort of tying a number of bricks to one end of the plank to balance his weight at the other.

    As a matter of fact, he just balanced against sixteen bricks, when these were fixed to the short end of plank, but if he fixed them to the long end of plank he only needed eleven as balance.

    Now, what was that boy's weight, if a brick weighs equal to a three-quarter brick and three-quarters of a pound?

  • Question 123 - A LEGAL DIFFICULTY

    "A client of mine," said a lawyer, "was on the point of death when his wife was about to present him with a child. I drew up his will, in which he settled two-thirds of his estate upon his son (if it should happen to be a boy) and one-third on the mother. But if the child should be a girl, then two-thirds of the estate should go to the mother and one-third to the daughter. As a matter of fact, after his death twins were born—a boy and a girl. A very nice point then arose. How was the estate to be equitably divided among the three in the closest possible accordance with the spirit of the dead man's will?"

  • Question 124 - A QUESTION OF DEFINITION

    "My property is exactly a mile square," said one landowner to another.

    "Curiously enough, mine is a square mile," was the reply.

    "Then there is no difference?"

    Is this last statement correct?

  • Question 125 - THE MINERS' HOLIDAY

    Seven coal-miners took a holiday at the seaside during a big strike. Six of the party spent exactly half a sovereign each, but Bill Harris was more extravagant. Bill spent three shillings more than the average of the party. What was the actual amount of Bill's expenditure?

  • Question 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 card (which in this case is to be a `3`) From the beginning of the row to the end?

  • Question 127 - SIMPLE DIVISION

    Sometimes a very simple question in elementary arithmetic will cause a good deal of perplexity. For example, I want to divide the four numbers, `701, 1,059, 1,417`, and `2,312`, by the largest number possible that will leave the same remainder in every case. How am I to set to work Of course, by a laborious system of trial one can in time discover the answer, but there is quite a simple method of doing it if you can only find it.

  • Question 128 - A PROBLEM IN SQUARES

    We possess three square boards. The surface of the first contains five square feet more than the second, and the second contains five square feet more than the third. Can you give exact measurements for the sides of the boards? If you can solve this little puzzle, then try to find three squares in arithmetical progression, with a common difference of `7` and also of `13`.

  • Question 129 - THE BATTLE OF HASTINGS

    All historians know that there is a great deal of mystery and uncertainty concerning the details of the ever-memorable battle on that fatal day, October `14, 1066`. My puzzle deals with a curious passage in an ancient monkish chronicle that may never receive the attention that it deserves, and if I am unable to vouch for the authenticity of the document it will none the less serve to furnish us with a problem that 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 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 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 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 `60xx4^2+1 = 31^2`, and `62xx8^2+1=63^2`. 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 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`.

  • Question 130 - THE SCULPTOR'S PROBLEM

    An ancient sculptor was commissioned to supply two statues, each on a cubical pedestal. It is with these pedestals that we are concerned. They were of unequal sizes, as will be seen in the illustration, and when the time arrived for payment a dispute arose as to whether the agreement was based on lineal or cubical measurement. But as soon as they came to measure the two pedestals the matter was at once settled, because, curiously enough, the number of lineal feet was exactly the same as the number of cubical feet. The puzzle is to find the dimensions for two pedestals having this peculiarity, in the smallest possible figures. You see, if the two pedestals, for example, measure respectively `3` ft. and `1` ft. on every side, then the lineal measurement would be `4` ft. and the cubical contents `28` ft., which are not the same, so these measurements will not do.