Part 32

61.--The Silver Cubes.

There is no limit to the number of different dimensions that will give two cubes whose sum shall be exactly seventeen cubic inches. Here is the answer in the smallest possible numbers. One of the silver cubes must measure 2-23278/40831 inches along each edge, and the other must measure 11663/40831 inch. If the reader likes to undertake the task of cubing each number (that is, multiply each number twice by itself), he will find that when added together the contents exactly equal seventeen cubic inches. See also No. 20, "The Puzzle of the Doctor of Physic."

THE ADVENTURES OF THE PUZZLE CLUB

62.--The Ambiguous Photograph.

One by one the members of the Club succeeded in discovering the key to the mystery of the Ambiguous Photograph, except Churton, who was at length persuaded to "give it up." Herbert Baynes then pointed out to him that the coat that Lord Marksford was carrying over his arm was a lady's coat, because the buttons are on the left side, whereas a man's coat always has the buttons on the right-hand side. Lord Marksford would not be likely to walk about the streets of Paris with a lady's coat over his arm unless he was accompanying the owner. He was therefore walking with the lady.

As they were talking a waiter brought a telegram to Baynes.

"Here you are," he said, after reading the message. "A wire from Dovey: 'Don't bother about photo. Find lady was the gentleman's sister, passing through Paris.' That settles it. You might notice that the lady was lightly clad, and therefore the coat might well be hers. But it is clear that the rain was only a sudden shower, and no doubt they were close to their destination, and she did not think it worth while to put the coat on."

63.--The Cornish Cliff Mystery.

Melville's explanation of the Cornish Cliff Mystery was very simple when he gave it. Yet it was an ingenious trick that the two criminals adopted, and it would have completely succeeded had not our friends from the Puzzle Club accidentally appeared on the scene. This is what happened: When Lamson and Marsh reached the stile, Marsh alone walked to the top of the cliff, with Lamson's larger boots in his hands. Arrived at the edge of the cliff, he changed the boots and walked backwards to the stile, carrying his own boots.

This little manoeuvre accounts for the smaller footprints showing a deeper impression at the heel, and the larger prints a deeper impression at the toe; for a man will walk more heavily on his heels when going forward, but will make a deeper impression with the toes in walking backwards. It will also account for the fact that the large footprints were sometimes impressed over the smaller ones, but never the reverse; also for the circumstance that the larger footprints showed a shorter stride, for a man will necessarily take a smaller stride when walking backwards. The pocket-book was intentionally dropped, to lead the police to discover the footprints, and so be put on the wrong scent.

64.--The Runaway Motor-Car.

Russell found that there are just twelve five-figure numbers that have the peculiarity that the first two figures multiplied by the last three--all the figures being different, and there being no 0--will produce a number with exactly the same five figures, in a different order. But only one of these twelve begins with a 1--namely, 14926. Now, if we multiply 14 by 926, the result is 12964, which contains the same five figures. The number of the motor-car was therefore 14926.

Here are the other eleven numbers:--24651, 42678, 51246, 57834, 75231, 78624, 87435, 72936, 65281, 65983, and 86251.

Compare with the problems in "Digital Puzzles," section of A. in M., and with Nos. 93 and 101 in these pages.

65.--The Mystery of Ravensdene Park.

The diagrams show that there are two different ways in which the routes of the various persons involved in the Ravensdene Mystery may be traced, without any path ever crossing another. It depends whether the butler, E, went to the north or the south of the gamekeeper's cottage, and the gamekeeper, A, went to the south or the north of the hall. But it will be found that the only persons who could have approached Mr. Cyril Hastings without crossing a path were the butler, E, and the man, C. It was, however, a fact that the butler retired to bed five minutes before midnight, whereas Mr. Hastings did not leave his friend's house until midnight. Therefore the criminal must have been the man who entered the park at C.

66.--The Buried Treasure.

The field must have contained between 179 and 180 acres--to be more exact, 179.37254 acres. Had the measurements been 3, 2, and 4 furlongs respectively from successive corners, then the field would have been 209.70537 acres in area.

One method of solving this problem is as follows. Find the area of triangle APB in terms of x, the side of the square. Double the result=xy. Divide by x and then square, and we have the value of y^{2} in terms of x. Similarly find value of z^{2} in terms of x; then solve the equation y^{2}+z^{2}=3^{2}, which will come out in the form x^{4}-20x^{2}=-37. Therefore x^{2}=10+(sqrt{63})=17.937254 square furlongs, very nearly, and as there are ten acres in one square furlong, this equals 179.37254 acres. If we take the negative root of the equation, we get the area of the field as 20.62746 acres, in which case the treasure would have been buried outside the field, as in Diagram 2. But this solution is excluded by the condition that the treasure was buried in the field. The words were, "The document ... states clearly that the field is square, and that the treasure is buried in it."

THE PROFESSOR'S PUZZLES

67.--The Coinage Puzzle.

The point of this puzzle turns on the fact that if the magic square were to be composed of whole numbers adding up 15 in all ways, the two must be placed in one of the corners. Otherwise fractions must be used, and these are supplied in the puzzle by the employment of sixpences and half-crowns. I give the arrangement requiring the fewest possible current English coins--fifteen. It will be seen that the amount in each corner is a fractional one, the sum required in the total being a whole number of shillings.

68.--The Postage Stamps Puzzles.

The first of these puzzles is based on a similar principle, though it is really much easier, because the condition that nine of the stamps must be of different values makes their selection a simple matter, though how they are to be placed requires a little thought or trial until one knows the rule respecting putting the fractions in the corners. I give the solution.

I also show the solution to the second stamp puzzle. All the columns, rows, and diagonals add up 1s. 6d. There is no stamp on one square, and the conditions did not forbid this omission. The stamps at present in circulation are these:--1/2d., 1d., 1-1/2d., 2d., 2-1/2d., 3d., 4d., 5d., 6d., 9d., 10d., 1s., 2s. 6d., 5s., 10s., £1, and £5. In the first solution the numbers are in arithmetical progression--1, 1-1/2, 2, 2-1/2, 3, 3-1/2, 4, 4-1/2, 5. But any nine numbers will form a magic square if we can write them thus:--

1 2 3 7 8 9 13 14 15

where the horizontal differences are all alike and the vertical differences all alike, but not necessarily the same as the horizontal. This happens in the case of the second solution, the numbers of which may be written:--

0 1 2 5 6 7 10 11 12