7. arithmetic & number theoretic recreations a. Fibonacci numbers
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- A magic hexagon within a circle, pp. 110-112. This is really a pattern of magic triangles -- cf. 7.N.2.
I. Arrange 1 12 on the points and crossings of a Star of David so that the sum on each line = sum of vertices of each large triangle = sum on inner hexagon = sum of vertices on each parallelogram = 26 -- i.e. the problem of Ordish/Ouseley. Gives one solution. Quotes Escott: "There are only six solutions." "The first appeared in Knowledge, in 1895, and the second is due to Mr. S. Lloyd." [Error for Loyd??]II. Arrange 1 19 in a hexagon, consisting of six equilateral triangles "so that the sum on every side in the same". This gives just 12 sums of three points, which is Dudeney's problem. He gives solutions with the sum = 22 and 23 and notes that subtracting from 20 gives sums = 38 and 37.Ahrens. A&N. 1918. Chap. XII: "Die wunderbare 26", pp. 133 140. Consider a hexagram (or six pointed star) formed from two triangles. This has 12 vertices. He finds 6 ways to place 1, 2, ..., 12 on these vertices so that each set of four along a triangle edge adds up to the same value (which must then be 26), and the six corners of the inner hexagon also add to 26. I.e. this is Ordish/Ouseley's problem, but with a slightly different statement of conditions. On p. 134, a note says the puzzle "Wunderbare 26" is made and sold by Züllchower Anstalten, Züllchow bei Stettin, and they have registered designs 42,768 and 45,600 for it, though the 'Rustic 26' has already been on sale for many years. S&B, p. 39, shows an English example with no identification. Collins. Fun with Figures. 1928.
A nest of magic hexagons, pp. 109-110. Central point surrounded by hexagons of 12, 24, 36 points. The numbers 1, ..., 73 are placed on the points so the sides of each hexagon add up to 111, 185, 259 respectively and the diagonals and the midlines add up to 259. In fact, opposite points in each hexagon add to 74 and the central value is 37. Note that 111 = 3 x 37, 185 = 5 x 37, 259 = 7 x 37.A magic hexagon within a circle, pp. 110-112. This is really a pattern of magic triangles -- cf. 7.N.2.Perelman. 1934. See in 7.N.2 for the Star of David pattern with just one solution. Birtwistle. Math. Puzzles & Perplexities. 1971. Digital sum 1, pp. 29, 163 & 188. Take 12 points on the vertices and midpoints of a hexagon, so each edge contains three points, together with a point in the middle of the hexagon. Place the numbers 1, ..., 13 so each straight line of three points has the same sum. Standard arguments show the central point must be 7 and the magic sum is 21. Factoring out symmetry, there are four solutions. He gives one.Digital sum 2, pp. 29, 163 & 188. Magic star of David, as in the 26 Puzzle, but with no condition on the sum of the inner hexagon. This gives many more solutions -- see Singmaster, 1998. He gives one solution, which has the sum of the inner hexagon being 26.Mathe mit Energie -- Energie mit Mathe. Verlag Leipziger Volkszeitung, 1981. A Magische Rosette, p. 34 & solutions p. 17. Place numbers 1, ..., 13 on F L G B the centred Star of David pattern at the right so that each rhombus from K M H the centre to an outer vertex has the same sum 35. They give one E J I C solution. In 2000, I found three, after factoring out the symmetries of the D figure. A little work shows that the magic sum satisfies 20 S 36 and a simple program shows that 21 S 35 and for S = 21, 22, ..., there are 3, 5, 9, 20, 23, 18, 8, 26, 8, 18, 23, 20, 9, 5, 3 solutions, giving a total of 198 solutions. Bell & Cornelius. Board Games Round the World. Op. cit. in 4.B.1. 1988. Marvellous '26', p. 79. Says this was sold for 6d by T. Ordish "probably in about 1920". [Above we see that it was known in 1895.] They quote the instructions from the box: "each of the six sides as well as the six spaces around the centre total up to 26 with perhaps the finding of several additional 'twenty sixes'." This is Ordish/Ouseley's problem. Bauch. Op. cit. in 7.O, 1990. In this, he gives the Star of David problem, which is a subproblem of the Magic Hexagon. He asserts it has 96 'classical' solutions, but he gives no discussion or reference and it is not clear if this is for all possible magic sums -- he shows one solution with magic sum 26 and the central hexagon having total 26. David Singmaster. Magic Stars of David and the 26 Puzzle. Draft written in May 1998. There are 960 solutions for a magic Star of David. There is clearly an equivalence given by the symmetries of the regular hexagon, so these solutions fall into 80 equivalence classes. Six of these are solutions of the 26 Puzzle (i.e. the sum of the central hexagon is also 26) and six are solutions of the complemented problem (i.e. the sum of the outer points is 26). One can use complementation to reduce the number of classes to 40. However, there are more symmetries of the magic Star of David when one ignores the additional constraint of the 26 Puzzle -- indeed the pattern is isomorphic to labelling the edges of a cube so the sum of the edges around each face is 26. Hence one can use the symmetries of the cube to produce 20 equivalence classes of solutions, but these symmetries do not interact simply with the additional sums used in the 26 Puzzle. 7.P. DIOPHANTINE RECREATIONS See also 7.E, 7.R.1, 7.R.2, 7.U. 7.P.1. HUNDRED FOWLS AND OTHER LINEAR PROBLEMS See Tropfke 565, 569, 572 & 613. NOTATION: n for p at a, b, c means n items of three types, costing a, b, c were bought for a total of p. I.e. we want x + y + z = n; ax + by + cz = p, with the conditions that x, y, z are positive (or non-negative) integers. (a, b) solutions means a non negative solutions, including b positive solutions -- so a b. I have checked these with a computer program. I also have a separate numerical index to these problems which enables me to tell whether problems are the same. When there are just two types of fowl, one gets two equations in two unknowns, but I have generally omitted such problems except when they seem to be part of an author's development or they represent a different context. See: MS Ambros. P114; Tartaglia 17 25 & 26; Hutton; Ozanam-Hutton 9; Williams; Collins. Della Francesca gives two problems where the value of two different combinations of two fruits or three animals are given. These are not Hundred Fowls Problems, but in the second, elimination of one unknown leads to one equation in two unknowns, just as the Hundred Fowls Problem does. Pacioli gives a similar problem with no integral solutions. I don't recall seeing other examples of this type of problem. The medieval problems of alligation are related, but the solution need not be integral. See: Fibonacci; Lucca 1754; Bartoli; della Francesca; Borghi; Apianus; Tropfke 569 for discussion and examples. See: Devi for a modern version with integral solutions. See Williams for a history of this aspect. Some monetary problems naturally occur here. Paying a sum with particular values and a specified number of pieces is just our ordinary problem. Paying a sum with particular values, without specifying the number of pieces, leads to one equation in several unknowns. This is the same as asking for the number of ways to change the total value. Perhaps more interesting are problems where one person has to pay a debt to another and they only have certain values, which leads to problems like ax - by = c. Ordinary problem in monetary terms -- see: Riese (1524); Dodson (1747?); Ozanam Montucla (1778); Ozanam-Hutton (1803); Hall (1846); Colenso (1849); Perelman (1934); M. Adams (1939); Depew (1939); Hedges (1950?); Little Puzzle Book (1955); Ripley's (1966); Scott (1973); Holt (1977); Paying a sum or making change: ax + by + ... = n -- see: Simpson (1745 & 1790); Dodson (1747? & 1753); Euler (1770); Moss (1773); Ozanam-Montucla (1778); Bonnycastle (1782 & 1815); Hutton, 1798?; Ozanam-Hutton (1803); De Morgan (1831?); Bourdon (1834); Unger (1838); Hall (1846); Clark (1916); Paying a debt with limited values: usually ax - by = c -- see: Euler (1770); Ozanam Montucla (1778); Bonnycastle (1782); Stewart (1802); Ozanam-Hutton (1803); De Morgan (1831?); Unger (1838); Todhunter (1870); McKay (1940); McKay (1940); Little Puzzle Book (1955) use the context of buying postage stamps. G.F. (1993) uses the context of wheels of vehicles. Simpson (1745); Dodson (1747?); Euler (1770); Ozanam-Montucla (1778); Bonnycastle (1782 & 1815); Ozanam-Hutton (1803); Bourdon (1834); Todhunter (1870); Clark (1904); McKay (At Home Tonight, 1940) are the only examples here which consider problems with one equation in two unknowns. Problems involving heads and feet of a mixture of birds and beasts: Clark; Williams; Collins; Ripley's. Impossible problems -- sometimes a problem is impossible only if positive solutions are required. Abu Kamil; Fibonacci 1202 & 1225; Tartaglia; Buteo; Simpson; Euler; Ozanam Montucla; Bonnycastle; Perelman; Depew; Little Puzzle Book; Scott; Holt. I have recently realised that the relatively modern problem of asking how to hit target values to make a particular value is a problem of this general nature, especially if the number of shots is given. E.g. a target has areas of value 16, 17, 23, 24, 39; how does one achieve a total of 100? These occur in Loyd, Dudeney, etc., but I haven't recorded them. I may add some of them. Zhang Qiujian. Zhang Qiujian Suan Jing. Op. cit. in 7.E. 468. Chap. 3, last problem, pp. 37a ff (or 54 f ??). ??NYS. Hundred Fowls: 100 for 100 at 5, 3, ⅓. (Cocks, hens, chicks.) This has (4, 3) solutions -- he gives (3, 3) of them, but only states the relation between the solutions -- no indication of how he found a solution. (Translation in Needham 121-122 (problem only), Libbrecht 277, Mikami 43, Li & Du 99.) Ho Peng Yoke. The lost problems of the Chang Ch'iu chien Suan Ching, a fifth century Chinese mathematical manual. Oriens Extremus 12 (1965) 37 53. On pp. 46 48, he identifies prob. 31 of Yang Hui (below at 1275) as being from Chang (= Zhang). The solution involves choosing one value arbitrarily. Zhen Luan (= Chen Luan). Op. cit. in 7.N, also called Commentary on Hsü Yo's Shu Shu Chi I. c570. ??NYS. [See Li & Du, p. 100.] 100 for 100 at 5, 4, ¼. (Cocks, hens, chicks.) This has (2, 1) solutions -- he gives (1, 1). Mikami says Chen's method would give one solution to Chang's problem. Libbrecht, pp. 278 279, gives the method, which is indeed nonsense, and states that other scholars noted that Chen's method is fortuitous. He says Chen also gives 100 for 100 at 4, 3, ⅓, which has (3, 2) solutions, of which Chen gives (1, 1). Liu Hsiao sun. Chang ch'iu chien suan ching hsi ts'ao (Detailed solutions of [the problems] in the Chang ch'iu chien suan ching). c600. ??NYS. Described in Libbrecht pp. 279 280 as nonsense!! Bakhshali MS. c7C. Sūtra C7 (VII 11-12). Hayashi 648-650 studies this. General discussion and first example are too mutilated to restore.
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