17)100(.05882385——150136——140136——4034——6051——9
Now, since all powers of 10 can only contain factors of the powers of 2 and 5, it clearly follows that your decimal never will come to an end if any other factor than these occurs in the denominator of your vulgar fraction. Thus, ½, 1/4, and 1/8 give us the exact decimals, .5, .25, and .125; 1/5 and 1/25 give us .2 and .04; 1/10 and 1/20 give us .1 and .05: because the denominators are all composed of 2 and 5 factors. But if you wish to convert 1/3, 1/6, or 1/7, your division sum will never end, but you will get these decimals, .33333, etc., .166666, etc., and .142857142857142857, etc., where, in the first case, the 3 keeps on repeating for ever and ever; in the second case the 6 is the repeater, and in the last case we get the recurring period of 142857. In the case of 1/17 (in "The Ribbon Problem") we find the circulating period to be .0588235294117647.
Now, in the division sum above, the successive remainders are1, 10, 15, 14, 4, 6, 9, etc., and these numbers I have inserted around the inner ring of the diagram. It will be seen that every number from 1 to 16 occurs once, and that if we multiply our ribbon number by any one of the numbers in the inner ring its position indicates exactly the point at which the product will begin. Thus, if we multiply by 4, the product will be 235, etc.; if we multiply by 6, 352, etc. We can therefore multiply by any number from 1 to 16 and get the desired result.
The kernel of the puzzle is this: Any prime number, with the exception of 2 and 5, which are the factors of 10, will exactly divide without remainder a number consisting of as many nines as the number itself, less one. Thus 999999 (six 9's) is divisible by 7, sixteen 9's are divisible by 17, eighteen 9's by 19, and so on. This is always the case, though frequently fewer 9's will suffice; for one 9 is divisible by 3, two by 11, six by 13, when our ribbon rule for consecutive multipliers breaks down and another law comes in. Therefore, since the 0 and 7 at the ends of the ribbon may notbe removed, we must seek a fraction with a prime denominator ending in 7 that gives a full period circulator. We try 37, and find that it gives a short period decimal, .027, because 37 exactly divides 999; it, therefore, will not do. We next examine 47, and find that it gives us the full period circulator, in 46 figures, at the beginning of this article.
If you cut any of these full period circulators in half and place one half under the other, you will find that they will add up all 9's; so you need only work out one half and then write down the complements. Thus, in the ribbon above, if you add 05882352 to 94117647 the result is 99999999, and so with our long solution number. Note also in the diagram above that not only are the opposite numbers on the outer ring complementary, always making 9 when added, but that opposite numbers in the inner ring, our remainders, are also complementary, adding to 17 in every case. I ought perhaps to point out that in limiting our multipliers to the first nine numbers it seems just possible that a short period circulator might give a solution in fewer figures, but there are reasons for thinking it improbable.
If the squares had not to be all the same size, the carpet could be cut in four pieces in any one of the three manners shown. In each case the two pieces marked A will fit together and form one of the three squares, the other two squares being entire. But in order to have the squares exactly equal in size, we shall require six pieces, as shown in the larger diagram. No. 1 is a complete square, pieces 4 and 5 will form a second square, and pieces 2, 3, and 6 will form the third—all of exactly the same size.
If with the three equal squares we form the rectangle IDBA, then the mean proportional of the two sides of the rectangle will be the side of a square of equal area. Produce AB to C, making BC equal to BD. Then place the point of the compasses at E (midway between A and C) and describe the arc AC. I am showing the quite general method for converting rectangles to squares, but in this particular case we may, of course, at once place our compasses at E, which requires no finding. Produce the line BD, cutting the arc in F, and BF will be the required side of the square. Now mark off AG and DH, each equal to BF, and make the cut IG, and also the cut HK from H, perpendicular to ID. The six pieces produced are numbered as in the diagram on last page.
It will be seen that I have here given the reverse method first: to cut the three small squares into six pieces to form a large square. In the case of our puzzle we can proceed as follows:—
Make LM equal to half the diagonal ON. Draw the line NM and drop from L a perpendicular on NM. Then LP will be the side of all the three squares of combined area equal to the large square QNLO. The reader can now cut out without difficulty the six pieces, as shown in the numbered square on the last page.
It might have struck the reader that the story of the bear impaled on the North Pole had no connection with the problem that followed. As a matter of fact it is essential to a solution. Eleven bears cannot possibly be arranged to form of themselves seven rows of bears with four bears in every row. But it is a different matter when Captain Longbow informs us that "they had so placed themselves thatthere were" seven rows of four bears. For if they were grouped as shown in the diagram, so that three of the bears, as indicated, were in line with the North Pole, that impaled animal would complete the seventh row of four, which cannot be obtained in any other way. It obviously does not affect the problem whether this seventh row is a hundred miles long or a hundred feet, so long as they were really in a straight line—a point that might perhaps be settled by the captain's pocket compass.
It was required to show how a resident at the town marked A might visit every one of the towns once, and only once, and finishup his tour at Z. This puzzle conceals a little trick. After the solver has demonstrated to his satisfaction that it cannot be done in accordance with the conditions as he at first understood them, he should carefully examine the wording in order to find some flaw. It was said, "This would be easy enough if he were able to cut across country by road, as well as by rail, but he is not."
Now, although he is prohibited from cutting across country by road, nothing is said about his going by sea! If, therefore, we carefully look again at the map, we shall find that two towns, and two only, lie on the sea coast. When he reaches one of these towns he takes his departure on board a coasting vessel and sails to the other port. The annexed illustration shows, by a dark line, the complete route.
This problem should be compared with No. 250, "The Grand Tour," inA. in M.It can be simplified in practically anidentical manner, but as there is here no choice on the first stage from A, the solutions are necessarily quite different. See also solution to No.94.
The solution is as follows. You may accept the invitation to "try to do it in twenty moves," but you will never succeed in performing the feat. The fewest possible moves are twenty-six. Play the cars so as to reach the following positions:—
E5678————= 10 moves.1234E56————= 2 moves.123 87 456————= 5 moves.E312 87 4E————= 9 moves.87654321
Twenty-six moves in all.
The smallest possible number of eggs that Mrs. Covey could have taken to market is 719. After selling half the number and giving half an egg over she would have 359 left; after the second transaction she would have 239 left; after the third deal, 179; and after the fourth, 143. This last number she could divide equally among her thirteen friends, giving each 11, and she would not have broken an egg.
The two words that solve this puzzle are BLUEBELL and PEARTREE. Place the letters as follows: B 3-1, L 6-8, U 5-3, E 4-6, B 7-5, E 2-4, L 9-7, L 9-2. This means that you take B,jump from 3 to 1, and write it down on 1; and so on. The second word can be inserted in the same order. The solution depends on finding those words in which the second and eighth letters are the same, and also the fourth and sixth the same, because these letters interchange without destroying the words. MARITIMA (or sea-pink) would also solve the puzzle if it were an English word.
Compare with No. 226 inA. in M.
Here is the way of arranging the seven men:—
ABCDEFGACDBGEFADBCFGEAGBFECDAFCEGDBAEDGFBCACEBGFDADGCFEBABFDEGCAEFDCGBAGEBDFCAFGCBEDAEBFCDGAGCEDBFAFDGBCE
Of course, at a circular table, A will be next to the man at the end of the line.
I first gave this problem for six persons on ten days, in theDaily Mailfor the 13th and 16th October 1905, and it has since been discussed in various periodicals by mathematicians. Of course, it is easily seen that the maximum number of sittings fornpersons is (n- 1)(n- 2)/2 ways. The comparatively easy methodfor solving all cases wherenis a prime+1 was first discovered by Ernest Bergholt. I then pointed out the form and construction of a solution that I had obtained for 10 persons, from which E. D. Bewley found a general method for all even numbers. The odd numbers, however, are extremely difficult, and for a long time no progress could be made with their solution, the only numbers that could be worked being 7 (given above) and 5, 9, 17, and 33, these last four being all powers of 2+1. At last, however (though not without much difficulty), I discovered a subtle method for solving all cases, and have written out schedules for every number up to 25 inclusive. The case of 11 has been solved also by W. Nash. Perhaps the reader will like to try his hand at 13. He will find it an extraordinarily hard nut.
The solutions for all cases up to 12 inclusive are given inA. in M., pp. 205, 206.
There are twelve ways of arranging the boxes without considering the pictures. If the thirty pictures were all different the answer would be 93,312. But the necessary deductions for cases where changes of boxes may be made without affecting the order of pictures amount to 1,728, and the boxes may therefore be arranged, in accordance with the conditions, in 91,584 different ways. I will leave my readers to discover for themselves how the figures are to be arrived at.
The number of ways in which the four pigs may be placed in the thirty-six sties in accordance with the conditions is seventeen, including the example that I gave, not counting the reversals and reflections of these arrangements as different. Jaenisch, in hisAnalyse Mathématique au jeu des Échecs(1862), quotes the statement that there are just twenty-one solutions to the little problem on which this puzzle is based. As I had myself only recorded seventeen, I examined the matter again, and found thathe was in error, and, doubtless, had mistaken reversals for different arrangements.
Here are the seventeen answers. The figures indicate the rows, and their positions show the columns. Thus, 104603 means that we place a pig in the first row of thefirstcolumn, in no row of thesecondcolumn, in the fourth row of thethirdcolumn, in the sixth row of thefourthcolumn, in no row of thefifthcolumn, and in the third row of thesixthcolumn. The arrangement E is that which I gave in diagram form:—
A.104603B.136002C.140502D.140520E.160025F.160304G.201405H.201605I.205104J.206104K.241005L.250014M.250630N.260015O.261005P.261040Q.306104
It will be found that forms N and Q are semi-symmetrical with regard to the centre, and therefore give only two arrangements each by reversal and reflection; that form H is quarter-symmetrical, and gives only four arrangements; while all the fourteen others yield by reversal and reflection eight arrangements each. Therefore the pigs may be placed in (2 × 2) + (4 × 1) + (8 × 14) = 120 different ways by reversing and reflecting all the seventeen forms.
Three pigs alone may be placed so that every sty is in line with a pig, provided that the pigs are not forbidden to be in line with one another; but there is only one way of doing it (if we do not count reversals as different), as follows: 105030.
Arrange the blocks so as to form the two multiplication sums 915 × 64 and 732 × 80, and the product in both cases will be the same: 58,560.
The smallest possible number of moves is twenty-two—that is, eleven for the foxes and eleven for the geese. Here is one way of solving the puzzle:
10—511—612—75—126—17—6————————————1—82—93—48—39—104—912—71—86—17—28—3——————————3—410—59—104—115—12
Of course, the reader will play the first move in the top line, then the first move in the second line, then the second move in the top line, and so on alternately.
InA. in M., p. 230, I have explained fully my "buttons and string" method of solving puzzles on chequered boards. In Diagram A is shown the puzzle in the form in which it may be presented on a portion of the chessboard with six knights. A comparison with the illustration on page 141 will show that I have there dispensed with the necessity of explaining the knight's move to the uninstructed reader by lines that indicate those moves. Thetwo puzzles are the same thing in different dress. Now compare page 141 with Diagram B, and it will be seen that by disentangling the strings I have obtained a simplified diagram without altering the essential relations between the buttons or discs. The reader will now satisfy himself without any difficulty that the puzzle requires eleven moves for the foxes and eleven for the geese. He will see that a goose on 1 or 3 must go to 8, to avoid being one move from a fox and to enable the fox on 11 to come on to the ring. If we play 1—8, then it is clearly best to play 10—5 and not 12—5 for the foxes. When they are all on the circle, then they simply promenade round it in a clockwise direction, taking care to reserve 8—3 and 5—12 for the final moves. It is thus rendered ridiculously easy by this method. See also notes on solutions to Nos.13and85.
The diagram shows how the piece of wood should be cut in two pieces to form the square table-top. A, B, C, D are the corners of the table. The way in which the piece E fits into the piece F will be obvious to the eye of the reader. The shaded part is the wood that is discarded.
The number must be the least common multiple of 1, 2, 3, etc., up to 15, that, when divided by 7, leaves the remainder 1, by 9 leaves 3, by 11 leaves 10, by 13 leaves 3, and by 14 leaves 8. Such a number is 120. The next number is 360,480, but as we have no record of a tree—especially a very young one—bearing anything like such a large number of apples, we may take 120 to be the only answer that is acceptable.
The rectangular closed cistern that shall hold a given quantity of water and yet have the smallest possible surface of metal must be a perfect cube—that is, a cistern every side of which is a square. For 1,000 cubic feet of water the internal dimensions will be 10 ft. × 10 ft. × 10 ft., and the zinc required will be 600 square feet. In the case of a cistern without a top the proportions will be exactly half a cube. These are the "exact proportions" asked for in the second case. The exact dimensions cannot be given, but 12.6 ft. × 12.6 ft. × 6.3 ft. is a close approximation. The cistern will hold a little too much water, at which the buyer will not complain, and it will involve the plumber in a trifling loss not worth considering.
If you take a sheet of paper and mark it with a diagonal line, as in Figure A, you will find that when you roll it into cylindrical form, with the line outside, it will appear as in Figure B.
It will be seen that the spiral (in one complete turn) is merely the hypotenuse of a right-angled triangle, of which the length and width of the paper are the other two sides. In the puzzle given, the lengths of the two sides of the triangle are 40 ft. (one-fifth of 200 ft.) and 16 ft. 8 in. Therefore thehypotenuse is 43 ft. 4 in. The length of the garland is therefore five times as long—216 ft. 8 in. A curious feature of the puzzle is the fact that with the dimensions given the result is exactly the sum of the height and the circumference.
All that is necessary is to add the two distances at which they meet to twice their difference. Thus 720 + 400 + 640 = 1760 yards, or one mile, which is the distance required. Or, put another way, three times the first distance less the second distance will always give the answer, only the first distance should be more than two-thirds of the second.
Just six different rings may be formed without breaking the conditions. Here is one way of effecting the arrangements.
ABCDEFGHIJKLMACEGIKMBDFHJLADGJMCFILBEHKAEIMDHLCGKBFJAFKCHMEJBGLDIAGMFLEKDJCIBH
Join the ends and you have the six rings.
Lucas devised a simple mechanical method for obtaining thenrings that may be formed under the conditions by 2n+1 children.
The only set of three numbers, of two, three, and five figures respectively, that will fulfil the required conditions is 27 × 594 = 16,038. These three numbers contain all the nine digits and 0, without repetition; the first two numbers multiplied together make the third, and the second is exactly twenty-two times the first. Ifthe numbers might contain one, four, and five figures respectively, there would be many correct answers, such as 3 × 5,694 = 17,082; but it is a curious fact that there is only one answer to the problem as propounded, though it is no easy matter to prove that this is the case.
It will be seen that in the arrangement given every number is different, and all the columns, all the rows, and each of the two diagonals, add up 179, whether you turn the page upside down or not. The reader will notice that I have not used the figures 3, 4, 5, 8, or 0.
There are 640 different routes. A general formula for puzzles of this kind is not practicable. We have obviously only to consider the variations of route between B and E. Here there are nine sections or "lines," but it is impossible for a train, under the conditions, to traverse more than seven of these lines in any route. In the following table by "directions" is meant the order of stationsirrespective of "routes." Thus, the "direction" BCDE gives nine "routes," because there are three ways of getting from B to C, and three ways of getting from D to E. But the "direction" BDCE admits of no variation; therefore yields only one route.
2two-linedirectionsof3routes—61three-line""1"—11"""9"—92four-line""6"—122"""18"—366five-line""6"—362"""18"—362six-line""36"—7212seven-line""36"—432——Total640
We thus see that there are just 640 different routes in all, which is the correct answer to the puzzle.
Each of the three pieces was clearly three cables long. But Simon persisted in assuming that the cuts were made transversely, or across, and that therefore the complete length was nine cables. The skipper, however, explained (and the point is quite as veracious as the rest of his yarn) that his cuts were made longitudinally—straight from the tip of the nose to the tip of the tail! The complete length was therefore only three cables, the same as each piece. Simon was not asked the exact length of the serpent, but how long itmusthave been. It must have been at least three cables long, though it might have been (the skipper's statement apart) anything from that up to nine cables, according to the direction of the cuts.
If there were twelve ladies in all, there would be 132 kisses among the ladies alone, leaving twelve more to be exchanged with the curate—six to be given by him and six to be received. Therefore, of the twelve ladies, six would be his sisters. Consequently, if twelve could do the work in four and a half months, six ladies would do it in twice the time—four and a half months longer—which is the correct answer.
At first sight there might appear to be some ambiguity about the words, "Everybody kissed everybody else, except, of course, the bashful young man himself." Might this not be held to imply that all the ladies immodestly kissed the curate, although they were not (except the sisters) kissed by him in return? No; because, in that case, it would be found that there must have been twelve girls, not one of whom was a sister, which is contrary to the conditions. If, again, it should be held that the sisters might not, according to the wording, have kissed their brother, although he kissed them, I reply that in that case there must have been twelve girls, all of whom must have been his sisters. And the reference to the ladies who might have worked exclusively of the sisters shuts out the possibility of this.
At the end of seventeen days the snail will have climbed 17 ft., and at the end of its eighteenth day-time task it will be at the top. It instantly begins slipping while sleeping, and will be 2 ft. down the other side at the end of the eighteenth day of twenty-four hours. How long will it take over the remaining 18 ft.? If it slips 2 ft. at night it clearly overcomes the tendency to slip 2 ft. during the daytime, in climbing up. In rowing up a river we have the stream against us, but in coming down it is with us and helps us. If the snail can climb 3 ft. and overcome the tendency to slip 2 ft. in twelve hours' ascent, it could with the same exertion crawl 5 ft. aday on the level. Therefore, in going down, the same exertion carries it 7 ft. in twelve hours—that is, 5 ft. by personal exertion and 2 ft. by slip. This, with the night slip, gives it a descending progress of 9 ft. in the twenty-four hours. It can, therefore, do the remaining 18 ft. in exactly two days, and the whole journey, up and down, will take it exactly twenty days.
When Montucla, in his edition of Ozanam'sRecreations in Mathematics, declared that "No more than three right-angled triangles, equal to each other, can be found in whole numbers, but we may find as many as we choose in fractions," he curiously overlooked the obvious fact that if you give all your sides a common denominator and then cancel that denominator you have the required answer in integers!
Every reader should know that if we take any two numbers,mandn, thenm2+n2,m2-n2, and2mnwill be the three sides of a rational right-angled triangle. Heremandnare called generating numbers. To form three such triangles of equal area, we use the following simple formula, wheremis the greater number:—
mn+m2+n2=am2-n2=b2mn+n2=c
Now, if we form three triangles from the following pairs of generators,aandb,aandc,aandb+c, they will all be of equal area. This is the little problem respecting which Lewis Carroll says in his diary (see hisLife and Lettersby Collingwood, p. 343), "Sat up last night till 4 a.m., over a tempting problem, sent me from New York, 'to find three equal rational-sided right-angled triangles.' I found two ... but could not find three!"
The following is a subtle formula by means of which we may always find a R.A.T. equal in area to any given R.A.T. Letz= hypotenuse,b= base,h= height,a= area of the given triangle; thenall we have to do is to form a R.A.T. from the generatorsz2and 4a, and give each side the denominator 2z(b2-h2), and we get the required answer in fractions. If we multiply all three sides of the original triangle by the denominator, we shall get at once a solution in whole numbers.
The answer to our puzzle in smallest possible numbers is as follows:—
First Prince51813201418Second Prince28024422458Third Prince23129602969Fourth Prince11161606161
The area in every case is 341,880 square furlongs. I must here refrain from showing fully how I get these figures. I will explain, however, that the first three triangles are obtained, in the manner shown, from the numbers 3 and 4, which give the generators 37, 7; 37, 33; 37, 40. These three pairs of numbers solve the indeterminate equation,a3b-b3a= 341,880. If we can find another pair of values, the thing is done. These values are 56, 55, which generators give the last triangle. The next best answer that I have found is derived from 5 and 6, which give the generators 91, 11; 91, 85; 91, 96. The fourth pair of values is 63, 42.
The reader will understand from what I have written above that there is no limit to the number of rational-sided R.A.T.'s of equal area that may be found in whole numbers.
The following is the simple solution of the three nines puzzle:—
9 + 9——.9
To divide 18 by .9 (or nine-tenths) we, of course, multiply by 10 and divide by 9. The result is 20, as required.
The solution is as follows: Between two players who thoroughly understand the play every game should be drawn. Neither player could ever win except through the blundering of his opponent. If Nought (the first player) takes the centre, Cross must take a corner, or Nought may beat him with certainty. If Nought takes a corner on his first play, Cross must take the centre at once, or again be beaten with certainty. If Nought leads with a side, both players must be very careful to prevent a loss, as there are numerous pitfalls. But Nought may safely lead anything and secure a draw, and he can only win through Cross's blunders.
The solution here is: The first player can always win, provided he plays to the centre on his first move. But a good variation of the game is to bar the centre for the first move of the first player. In that case the second player should take the centre at once. This should always end in a draw, but to ensure it the first player must play to two adjoining corners (such as 1 and 3) on his first and second moves. The game then requires great care on both sides.
Sir Isaac Newton has shown us, in hisUniversal Arithmetic, that we may divide the bullocks in each case in two parts—one part to eat the increase, and the other the accumulated grass. The first will vary directly as the size of the field, and will not depend on the time; the second part will also vary directly as the size of the field, and in addition inversely with the time. We find from the farmer's statements that 6 bullocks keep down the growth in a 10-acre field, and 6 bullocks eat the grass on 10 acres in 16 weeks. Therefore, if 6 bullocks keep down the growth on 10 acres, 24 will keep down the growth on 40 acres.
Again, we find that if 6 bullocks eat the accumulated grass on 10 acres in 16 weeks, then