Chapter 39
But, as I have said, these are not the answers in the smallest numbers: for if we add 523 to the numbers 1 to 16, we get sixteen consecutive composites; and if we add 1,327 to the numbers 1 to 25, we get twenty-five consecutive composites, in each case the smallest numbers possible. Yet if we required to form a magic square of a hundred such numbers, we should find it a big task by means of tables, though by the process I have shown it is quite a simple matter. Even to find thirty-six such numbers you will search the tables up to 10,000 without success, and the difficulty increases in an accelerating ratio with each square of a larger order.
412.--THE MAGIC KNIGHT'S TOUR.
+----+----+----+----+----+----+----+----+ | 46 | 55 | 44 | 19 | 58 | 9 | 22 | 7 | +----+----+----+----+----+----+----+----+ | 43 | 18 | 47 | 56 | 21 | 6 | 59 | 10 | +----+----+----+----+----+----+----+----+ | 54 | 45 | 20 | 41 | 12 | 57 | 8 | 23 | +----+----+----+----+----+----+----+----+ | 17 | 42 | 53 | 48 | 5 | 24 | 11 | 60 | +----+----+----+----+----+----+----+----+ | 52 | 3 | 32 | 13 | 40 | 61 | 34 | 25 | +----+----+----+----+----+----+----+----+ | 31 | 16 | 49 | 4 | 33 | 28 | 37 | 62 | +----+----+----+----+----+----+----+----+ | 2 | 51 | 14 | 29 | 64 | 39 | 26 | 35 | +----+----+----+----+----+----+----+----+ | 15 | 30 | 1 | 50 | 27 | 36 | 63 | 38 | +----+----+----+----+----+----+----+----+
Here each successive number (in numerical order) is a knight's move from the preceding number, and as 64 is a knight's move from 1, the tour is "re-entrant." All the columns and rows add up 260. Unfortunately, it is not a perfect magic square, because the diagonals are incorrect, one adding up 264 and the other 256--requiring only the transfer of 4 from one diagonal to the other. I think this is the best result that has ever been obtained (either re-entrant or not), and nobody can yet say whether a perfect solution is possible or impossible.
413.--A CHESSBOARD FALLACY.
The explanation of this little fallacy is as follows. The error lies in assuming that the little triangular piece, marked C, is exactly the same height as one of the little squares of the board. As a matter of fact, its height (if we make the sixty-four squares each a square inch) will be 1+1/7 in. Consequently the rectangle is really 9+1/7 in. by 7 in., so that the area is sixty-four square inches in either case. Now, although the pieces do fit together exactly to form the perfect rectangle, yet the directions of the horizontal lines in the pieces will not coincide. The new diagram above will make everything quite clear to the reader.
414.--WHO WAS FIRST?
Biggs, who saw the smoke, would be first; Carpenter, who saw the bullet strike the water, would be second; and Anderson, who heard the report, would be last of all.
415.--A WONDERFUL VILLAGE.
When the sun is in the horizon of any place (whether in Japan or elsewhere), he is the length of half the earth's diameter more distant from that place than in his meridian at noon. As the earth's semi-diameter is nearly 4,000 miles, the sun must be considerably more than 3,000 miles nearer at noon than at his rising, there being no valley even the hundredth part of 1,000 miles deep.
416.--A CALENDAR PUZZLE.
The first day of a century can never fall on a Sunday; nor on a Wednesday or a Friday.
417.--THE TIRING-IRONS.
I will give my complete working of the solution, so that readers may see how easy it is when you know how to proceed. And first of all, as there is an even number of rings, I will say that they may all be taken off in one-third of (2^(n + 1) - 2) moves; and since n in our case is 14, all the rings may be taken off in 10,922 moves. Then I say 10,922 - 9,999 = 923, and proceed to find the position when only 923 out of the 10,922 moves remain to be made. Here is the curious method of doing this. It is based on the binary scale method used by Monsieur L. Gros, for an account of which see W.W. Rouse Ball's _Mathematical Recreations_.
Divide 923 by 2, and we get 461 and the remainder 1; divide 461 by 2, and we get 230 and the remainder 1; divide 230 by 2, and we get 115 and the remainder nought. Keep on dividing by 2 in this way as long as possible, and all the remainders will be found to be 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, the last remainder being to the left and the first remainder to the right. As there are fourteen rings and only ten figures, we place the difference, in the form of four noughts, in brackets to the left, and bracket all those figures that repeat a figure on their left. Then we get the following arrangement: (0 0 0 0) 1 (1 1) 0 (0) 1 (1) 0 1 (1). This is the correct answer to the puzzle, for if we now place rings below the line to represent the figures in brackets and rings on the line for the other figures, we get the solution in the required form, as below:--
O O O OO ------------------------- OOOO OO O O O
This is the exact position of the rings after the 9,999th move has been made, and the reader will find that the method shown will solve any similar question, no matter how many rings are on the tiring-irons. But in working the inverse process, where you are required to ascertain the number of moves necessary in order to reach a given position of the rings, the rule will require a little modification, because it does not necessarily follow that the position is one that is actually reached in course of taking off all the rings on the irons, as the reader will presently see. I will here state that where the total number of rings is odd the number of moves required to take them all off is one-third of (2^(n + 1) - 1).
With n rings (where n is _odd_) there are 2^n positions counting all on and all off. In (1/3)(2^(n + 1) + 2) positions they are all removed. The number of positions not used is (1/3)(2^n - 2).
With n rings (where n is _even_) there are 2^n positions counting all on and all off. In (2^(n + 1) + 1) positions they are all removed. The number of positions not used is here (1/3)(2^n - 1).
It will be convenient to tabulate a few cases.
+--------+------------+-----------+-----------+ | No. of | Total | Positions | Positions | | Rings. | Positions. | used. | not used. | +--------+------------+-----------+-----------+ | 1 | 2 | 2 | 0 | | 3 | 8 | 6 | 2 | | 5 | 32 | 22 | 10 | | 7 | 128 | 86 | 42 | | 9 | 512 | 342 | 170 | | | | | | | 2 | 4 | 3 | 1 | | 4 | 16 | 11 | 5 | | 6 | 64 | 43 | 21 | | 8 | 256 | 171 | 85 | | 10 | 1024 | 683 | 341 | +--------+------------+-----------+-----------+
Note first that the number of _positions used_ is one more than the number of _moves_ required to take all the rings off, because we are including "all on" which is a position but not a move. Then note that the number of _positions not used_ is the same as the number of _moves used_ to take off a set that has one ring fewer. For example, it takes 85 moves to remove 7 rings, and the 42 positions not used are exactly the number of moves required to take off a set of 6 rings. The fact is that if there are 7 rings and you take off the first 6, and then wish to remove the 7th ring, there is no course open to you but to reverse all those 42 moves that never ought to have been made. In other words, you must replace all the 7 rings on the loop and start afresh! You ought first to have taken off 5 rings, to do which you should have taken off 3 rings, and previously to that 1 ring. To take off 6 you first remove 2 and then 4 rings.
418.--SUCH A GETTING UPSTAIRS.
Number the treads in regular order upwards, 1 to 8. Then proceed as follows: 1 (step back to floor), 1, 2, 3 (2), 3, 4, 5 (4), 5, 6, 7 (6), 7, 8, landing (8), landing. The steps in brackets are taken in a backward direction. It will thus be seen that by returning to the floor after the first step, and then always going three steps forward for one step backward, we perform the required feat in nineteen steps.
419.--THE FIVE PENNIES.
First lay three of the pennies in the way shown in Fig. 1. Now hold the remaining two pennies in the position shown in Fig. 2, so that they touch one another at the top, and at the base are in contact with the three horizontally placed coins. Then the five pennies will be equidistant, for every penny will touch every other penny.
420.--THE INDUSTRIOUS BOOKWORM.
The hasty reader will assume that the bookworm, in boring from the first to the last page of a book in three volumes, standing in their proper order on the shelves, has to go through all three volumes and four covers. This, in our case, would mean a distance of 9½ in., which is a long way from the correct answer. You will find, on examining any three consecutive volumes on your shelves, that the first page of Vol. I. and the last page of Vol. III. are actually the pages that are nearest to Vol. II., so that the worm would only have to penetrate four covers (together, ½ in.) and the leaves in the second volume (3 in.), or a distance of 3½ inches, in order to tunnel from the first page to the last.
421.--A CHAIN PUZZLE.
To open and rejoin a link costs threepence. Therefore to join the nine pieces into an endless chain would cost 2s. 3d., whereas a new chain would cost 2s. 2d. But if we break up the piece of eight links, these eight will join together the remaining eight pieces at a cost of 2s. But there is a subtle way of even improving on this. Break up the two pieces containing three and four links respectively, and these seven will join together the remaining seven pieces at a cost of only 1s. 9d.
422.--THE SABBATH PUZZLE.
The way the author of the old poser proposed to solve the difficulty was as follows: From the Jew's abode let the Christian and the Turk set out on a tour round the globe, the Christian going due east and the Turk due west. Readers of Edgar Allan Poe's story, _Three Sundays in a Week_, or of Jules Verne's _Round the World in Eighty Days_, will know that such a proceeding will result in the Christian's gaining a day and in the Turk's losing a day, so that when they meet again at the house of the Jew their reckoning will agree with his, and all three may keep their Sabbath on the same day. The correctness of this answer, of course, depends on the popular notion as to the definition of a day--the average duration between successive sun-rises. It is an old quibble, and quite sound enough for puzzle purposes. Strictly speaking, the two travellers ought to change their reckonings on passing the 180th meridian; otherwise we have to admit that at the North or South Pole there would only be one Sabbath in seven years.
423.--THE RUBY BROOCH.
In this case we were shown a sketch of the brooch exactly as it appeared after the four rubies had been stolen from it. The reader was asked to show the positions from which the stones "may have been taken;" for it is not possible to show precisely how the gems were originally placed, because there are many such ways. But an important point was the statement by Lady Littlewood's brother: "I know the brooch well. It originally contained forty-five stones, and there are now only forty-one. Somebody has stolen four rubies, and then reset as small a number as possible in such a way that there shall always be eight stones in any of the directions you have mentioned."
The diagram shows the arrangement before the robbery. It will be seen that it was only necessary to reset one ruby--the one in the centre. Any solution involving the resetting of more than one stone is not in accordance with the brother's statement, and must therefore be wrong. The original arrangement was, of course, a little unsymmetrical, and for this reason the brooch was described as "rather eccentric."
424.--THE DOVETAILED BLOCK.
The mystery is made clear by the illustration. It will be seen at once how the two pieces slide together in a diagonal direction.
425.--JACK AND THE BEANSTALK.
The serious blunder that the artist made in this drawing was in depicting the tendrils of
the bean climbing spirally as at A above, whereas the French bean, or scarlet runner, the variety clearly selected by the artist in the absence of any authoritative information on the point, always climbs as shown at B. Very few seem to be aware of this curious little fact. Though the bean always insists on a sinistrorsal growth, as B, the hop prefers to climb in a dextrorsal manner, as A. Why, is one of the mysteries that Nature has not yet unfolded.
426.--THE HYMN-BOARD POSER.
This puzzle is not nearly so easy as it looks at first sight. It was required to find the smallest possible number of plates that would be necessary to form a set for three hymn-boards, each of which would show the five hymns sung at any particular service, and then to discover the lowest possible cost for the same. The hymn-book contains 700 hymns, and therefore no higher number than 700 could possibly be needed.
Now, as we are required to use every legitimate and practical method of economy, it should at once occur to us that the plates must be painted on both sides; indeed, this is such a common practice in cases of this kind that it would readily occur to most solvers. We should also remember that some of the figures may possibly be reversed to form other figures; but as we were given a sketch of the actual shapes of these figures when painted on the plates, it would be seen that though the 6's may be turned upside down to make 9's, none of the other figures can be so treated.
It will be found that in the case of the figures 1, 2, 3, 4, and 5, thirty-three of each will be required in order to provide for every possible emergency; in the case of 7, 8, and 0, we can only need thirty of each; while in the case of the figure 6 (which may be reversed for the figure 9) it is necessary to provide exactly forty-two.
It is therefore clear that the total number of figures necessary is 297; but as the figures are painted on both sides of the plates, only 149 such plates are required. At first it would appear as if one of the plates need only have a number on one side, the other side being left blank. But here we come to a rather subtle point in the problem.
Readers may have remarked that in real life it is sometimes cheaper when making a purchase to buy more articles than we require, on the principle of a reduction on taking a quantity: we get more articles and we pay less. Thus, if we want to buy ten apples, and the price asked is a penny each if bought singly, or ninepence a dozen, we should both save a penny and get two apples more than we wanted by buying the full twelve. In the same way, since there is a regular scale of reduction for plates painted alike, we actually save by having two figures painted on that odd plate. Supposing, for example, that we have thirty plates painted alike with 5 on one side and 6 on the other. The rate would be 4¾d., and the cost 11s. 10½d. But if the odd plate with, say, only a 5 on one side of it have a 6 painted on the other side, we get thirty-one plates at the reduced rate of 4½d., thus saving a farthing on each of the previous thirty, and reducing the cost of the last one from 1s. to 4½d.
But even after these points are all seen there comes in a new difficulty: for although it will be found that all the 8's may be on the backs of the 7's, we cannot have all the 2's on the backs of the 1's, nor all the 4 on the backs of the 3's, etc. There is a great danger, in our attempts to get as many as possible painted alike, of our so adjusting the figures that some particular combination of hymns cannot be represented.
Here is the solution of the difficulty that was sent to the vicar of Chumpley St. Winifred. Where the sign X is placed between two figures, it implies that one of these figures is on one side of the plate and the other on the other side.
d. £ s. d. 31 plates painted 5 X 9 @ 4½ = 0 11 7½ 30 " 7 X 8 @ 4¾ = 0 11 10½ 21 " 1 X 2 @ 7 = 0 12 3 21 " 3 X 0 @ 7 = 0 12 3 12 " 1 X 3 @ 9¼ = 0 9 3 12 " 2 X 4 @ 9¼ = 0 9 3 12 " 9 X 4 @ 9¼ = 0 9 3 8 " 4 X 0 @ 10¼ = 0 6 10 1 " 5 X 4 @ 12 = 0 1 0 1 " 5 X 0 @ 12 = 0 1 0 149 plates @ 6d. each = 3 14 6 ---------- £7 19 1
Of course, if we could increase the number of plates, we might get the painting done for nothing, but such a contingency is prevented by the condition that the fewest possible plates must be provided.
This puzzle appeared in _Tit-Bits_, and the following remarks, made by me in the issue for 11th December 1897, may be of interest.
The "Hymn-Board Poser" seems to have created extraordinary interest. The immense number of attempts at its solution sent to me from all parts of the United Kingdom and from several Continental countries show a very kind disposition amongst our readers to help the worthy vicar of Chumpley St. Winifred over his parochial difficulty. Every conceivable estimate, from a few shillings up to as high a sum as £1,347, 10s., seems to have come to hand. But the astonishing part of it is that, after going carefully through the tremendous pile of correspondence, I find that only one competitor has succeeded in maintaining the reputation of the _Tit-Bits_ solvers for their capacity to solve anything, and his solution is substantially the same as the one given above, the cost being identical. Some of his figures are differently combined, but his grouping of the plates, as shown in the first column, is exactly the same. Though a large majority of competitors clearly hit upon all the essential points of the puzzle, they completely collapsed in the actual arrangement of the figures. According to their methods, some possible selection of hymns, such as 111, 112, 121, 122,211, cannot be set up. A few correspondents suggested that it might be possible so to paint the 7's that upside down they would appear as 2's or 4's; but this would, of course, be barred out by the fact that a representation of the actual figures to be used was given.
427.--PHEASANT-SHOOTING.
The arithmetic of this puzzle is very easy indeed. There were clearly 24 pheasants at the start. Of these 16 were shot dead, 1 was wounded in the wing, and 7 got away. The reader may have concluded that the answer is, therefore, that "seven remained." But as they flew away it is clearly absurd to say that they "remained." Had they done so they would certainly have been killed. Must we then conclude that the 17 that were shot remained, because the others flew away? No; because the question was not "how many remained?" but "how many still remained?" Now the poor bird that was wounded in the wing, though unable to fly, was very active in its painful struggles to run away. The answer is, therefore, that the 16 birds that were shot dead "still remained," or "remained still."
428.--THE GARDENER AND THE COOK.
Nobody succeeded in solving the puzzle, so I had to let the cat out of the bag--an operation that was dimly foreshadowed by the puss in the original illustration. But I first reminded the reader that this puzzle appeared on April 1, a day on which none of us ever resents being made an "April Fool;" though, as I practically "gave the thing away" by specially drawing attention to the fact that it was All Fools' Day, it was quite remarkable that my correspondents, without a single exception, fell into the trap.
One large body of correspondents held that what the cook loses in stride is exactly made up in greater speed; consequently both advance at the same rate, and the result must be a tie. But another considerable section saw that, though this might be so in a race 200 ft. straight away, it could not really be, because they each go a stated distance at "every bound," and as 100 is not an exact multiple of 3, the gardener at his thirty-fourth bound will go 2 ft. beyond the mark. The gardener will, therefore, run to a point 102 ft. straight away and return (204 ft. in all), and so lose by 4 ft. This point certainly comes into the puzzle. But the most important fact of all is this, that it so happens that the gardener was a pupil from the Horticultural College for Lady Gardeners at, if I remember aright, Swanley; while the cook was a very accomplished French chef of the hemale persuasion! Therefore "she (the gardener) made three bounds to his (the cook's) two." It will now be found that while the gardener is running her 204 ft. in 68 bounds of 3 ft., the somewhat infirm old cook can only make 45+1/3 of his 2 ft. bounds, which equals 90 ft. 8 in. The result is that the lady gardener wins the race by 109 ft. 4 in. at a moment when the cook is in the air, one-third through his 46th bound.
The moral of this puzzle is twofold: (1) Never take things for granted in attempting to solve puzzles; (2) always remember All Fools' Day when it comes round. I was not writing of _any_ gardener and cook, but of a _particular_ couple, in "a race that I witnessed." The statement of the eye-witness must therefore be accepted: as the reader was not there, he cannot contradict it. Of course the information supplied was insufficient, but the correct reply was: "Assuming the gardener to be the 'he,' the cook wins by 4 ft.; but if the gardener is the 'she,' then the gardener wins by 109 ft. 4 in." This would have won the prize. Curiously enough, one solitary competitor got on to the right track, but failed to follow it up. He said: "Is this a regular April 1 catch, meaning that they only ran 6 ft. each, and consequently the race was unfinished? If not, I think the following must be the solution, supposing the gardener to be the 'he' and the cook the 'she.'" Though his solution was wrong even in the case he supposed, yet he was the only person who suspected the question of sex.
429.--PLACING HALFPENNIES.
Thirteen coins may be placed as shown on page 252.
430.--FIND THE MAN'S WIFE.
There is no guessing required in this puzzle. It is all a question of elimination. If we can pair off any five of the ladies with their respective husbands, other than husband No. 10, then the remaining lady must be No. 10's wife.