outward journey in the short time of ten minutes, though it took him an hour to get back to the starting point at Slocomb, with the wind dead against him. Now, how long would the ten miles have taken him if there had been a perfect calm? Of course, the hydroplane's engine worked uniformly throughout. (3/3)
Tommy replied, "it is true that in Ireland there are men of Cork and in Scotland men of Ayr, which is better still, but in England there are lightermen." Unfortunately it had to be explained to Mrs. Dobson, and this took the edge off the thing. The hydroplane flight was from Slocomb to the neighbouring watering-place Poodleville--five miles distant. But there was a strong wind, which so helped the airman that he made the (2/3)
72. THE HYDROPLANE QUESTION. The inhabitants of Slocomb-on-Sea were greatly excited over the visit of a certain flying man. All the town turned out to see the flight of the wonderful hydroplane, and, of course, Dobson and his family were there. Master Tommy was in good form, and informed his father that Englishmen made better airmen than Scotsmen and Irishmen because they are not so heavy. "How do you make that out?" asked Mr. Dobson. "Well, you see," (1/3)
SOLUTION TO 88. DIGITAL DIVISION. (7/7)
result in our getting either a nought or a second 5 in the denominator. Similarly 1 cannot ever appear in the same position, nor 6 in the fraction one-sixth, nor an even figure in the fraction one-fifth, and so on. The preliminary consideration of such points as I have touched upon will not only prevent our wasting a lot of time in trying to produce impossible forms, but will lead us more or less directly to the desired solutions.
SOLUTION TO 88. DIGITAL DIVISION. (6/7)
same figures may often be differently arranged, as shown in the two pairs of values for one-fifth that I have given in the last paragraph,
but here it will be found there is a general readjustment of figures and not a simple changing of the positions of pairs. There are other little points that would occur to every solver--such as that the figure 5 cannot ever appear to the extreme right of the numerator, as this would
SOLUTION TO 88. DIGITAL DIVISION. (5/7)
product there will, of course, always be a carry-over after multiplying the last figure to the left, and in every case higher than 4 we must carry over at least three times. Consequently in cases from one-fifth to one-ninth we cannot produce different solutions by a mere change of position of pairs of figures, as, for example, we may with 5832/17496 and 5823/17469, where the 2/6 and 3/9 change places. It is true that the
SOLUTION TO 88. DIGITAL DIVISION. (4/7)
arrangements, the roots of the numerator and denominator are respectively 6--3, and in the last two 3--6. The most curious case of all is, perhaps, one-eighth, for here the digital roots may be of any one of the five forms given above.
The denominators of the fractions being regarded as the numerators multiplied by 2, 3, 4, 5, 6, 7, 8, and 9 respectively, we must pay attention to the "carryings over." In order to get five figures in the
SOLUTION TO 88. DIGITAL DIVISION. (3/7)
cases of one-third, one-fourth, one-sixth, one-seventh, and one-ninth must be of the form 9--9; that is to say, the digital roots of both numerator and denominator will be 9. In the cases of one-half and one-fifth, however, the digital roots are 6--3, but of course the higher root may occur either in the numerator or in the denominator; thus 2697/13485, 2769/13845, 2973/14865, 3729/18645, where, in the first two
SOLUTION TO 88. DIGITAL DIVISION. (2/7)
The sum of the numerator digits and the denominator digits will, of course, always be 45, and the "digital root" is 9. Now, if we separate the nine digits into any two groups, the sum of the two digital roots will always be 9. In fact, the two digital roots must be either 9--9,
8--1, 7--2, 6--3, or 5--4. In the first case the actual sum is 18, but then the digital root of this number is itself 9. The solutions in the
SOLUTION TO 88. DIGITAL DIVISION. (1/7)
It is convenient to consider the digits as arranged to form fractions of the respective values, one-half, one-third, one-fourth, one-fifth,
one-sixth, one-seventh, one-eighth, and one-ninth. I will first give the eight answers, as follows:--
6729/13458 = 1/2
5823/17469 = 1/3
3942/15768 = 1/4
2697/13485 = 1/5
2943/17658 = 1/6
2394/16758 = 1/7
3187/25496 = 1/8
6381/57429 = 1/9
88. DIGITAL DIVISION. It is another good puzzle so to arrange the nine digits (the nought excluded) into two groups so that one group when divided by the other produces a given number without remainder. For example, 1 3 4 5 8 divided by 6 7 2 9 gives 2. Can the reader find similar arrangements producing 3, 4, 5, 6, 7, 8, and 9 respectively? Also, can he find the pairs of smallest possible numbers in each case? Thus, 1 4 6 5 8 divided (1/2)
11. THE CYCLISTS' FEAST.
'Twas last Bank Holiday, so I've been told,
Some cyclists rode abroad in glorious weather.
Resting at noon within a tavern old,
They all agreed to have a feast together.
"Put it all in one bill, mine host," they said,
"For every man an equal share will pay."
The bill was promptly on the table laid,
And four pounds was the reckoning that day.
But, sad to state, when they prepared to square, (1/2)
SOLUTION TO 134. THE BANKER'S PUZZLE. (3/3)
curious fact that any number up to 39, if added to its square and the sum increased by 41, makes a prime number. This was first discovered by Euler, the great mathematician. It has been suggested that the banker might desire the customer to transfer sufficient to raise the contents of the box to a given number; but this would not only make the thing an absurdity, but breaks the rule that neither knows what the other puts in.
SOLUTION TO 134. THE BANKER'S PUZZLE. (2/3)
add, he will desire the latter to transfer from the counter the square of the number next below what the customer put in. Thus, banker puts 40,
customer, we will say, adds 6, then transfers from the counter 25 (the square of 5), which leaves 71 in all, a prime number. Try again. Banker puts 40, customer adds 12, then transfers 121 (the square of 11), as desired, which leaves 173, a prime number. The key to the puzzle is the
SOLUTION TO 134. THE BANKER'S PUZZLE. (1/3)
In order that a number of sixpences may not be divisible into a number of equal piles, it is necessary that the number should be a prime. If the banker can bring about a prime number, he will win; and I will show how he can always do this, whatever the customer may put in the box, and that therefore the banker will win to a certainty. The banker must first deposit forty sixpences, and then, no matter how many the customer may
case not more than a pound in value), neither knowing what the other put in. Lastly, the customer was to transfer from the banker's counter to the box as many sixpences as the banker desired him to put in. The puzzle is to find how many sixpences the banker should first put in and how many he should ask the customer to transfer, so that he may have the best chance of winning. (2/2)
134. THE BANKER'S PUZZLE. A banker had a sporting customer who was always anxious to wager on anything. Hoping to cure him of his bad habit, he proposed as a wager that the customer would not be able to divide up the contents of a box containing only sixpences into an exact number of equal piles of sixpences. The banker was first to put in one or more sixpences (as many as he liked); then the customer was to put in one or more (but in his (1/2)
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
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