7. arithmetic & number theoretic recreations a. Fibonacci numbers

Pp. 74-75: The card addition puzzle. If one views cards as 1 x 2 rectangles, then a 6 x 6 frame can be formed by having three vertical cards on each side and two horizontal cards filling in the top and bottom. Arrange the cards 1 - 10 so that the total along each side is the same. Note that this adds four cards along the top and bottom but three along the sides. Gives one solution and implies it is unique, but I have found ten solutions.

Pp. 82-84: Adding, subtracting, multiplying, dividing and fractional magic squares. For a subtraction square, if a, b, c are the elements in a row, then c   (b   a)  =  a   b + c is a constant. An easy solution is obtained by replacing the odd or the even terms of a magic square by ten minus themselves. He asserts this idea was invented by Dudeney. He says the multiplying magic square dates from the last quarter of the 18C but was first published by Dudeney. For a division square, we have c/(b/a) = ac/b is a constant. One can find an easy solution from a multiplication square. He says this was invented by Dudeney. Fractional squares are ordinary additive squares with constant of one.

Pp. 90-92: A magic circle. This is basically an 8 x 8 (semi?) magic square spread into a circular pattern. He uses the numbers 12 - 75 to get a sum of 348 and then sticks some values of 12 in so as to yield 360.

Pp. 93-95: A magic pentagon. Five 4 x 4 squares skewed to rhombi and fit together at a point to make a five-pointed star. Each rhombus has constant 162 and various sums add up to 324.

Meyer. Big Fun Book. 1940. Mathematical fun, pp. 98 & 731. Make the ABC

figure at the right magical. I've called this a Magic Hourglass -- see below. D  

standard arguments show that the magic constant must be 12 and D = 4. EFG

The solution is essentially unique, with one horizontal line containing

7, 3, 2.

Anonymous. The problems drive. Eureka 12 (Oct 1949) 7-8 & 15. No. 3. Place the numbers 1, ..., 20 at the vertices of a dodecahedron so that the sum of the numbers at the corners of each face is the same. Answer: it cannot be done! [Similar argument shows that the only regular polyhedron that can be so labelled is the cube. Then one sees that the sum on an edge is the same as on the edge symmetric with respect to the centre. Then one finds that an edge must have the numbers 1, 8 on it and hence all the edges parallel to it have a sum of 9, hence must have the pairs 2, 7; 3, 6; 4, 5. Putting these vertically and putting 1 in the top face, we find the top face must contain {1, 4, 6, 7} and there are three distinct ways to place these. The cubo-octahedron cannot be done, but I'm not sure about the rhombic dodecahedron.]

Ripley's Puzzles and Games. 1966. P. 48. Magic cross-cube. Consider a 2 x 2 x 2 cube. This has the numbers 1 - 12, 14 - 25 on the facelets so that each face totals 52 and the three facelets at each corner total 39. Such an arrangement requires the total of all numbers be a multiple of 24, but 1 + ... + 24 = 300 is an odd multiple of 12. 312 is the next multiple of 24 and leads to the numbers used.

Doubleday - 2. 1971. Ups and downs, pp. 121-122. Gives the figure at 3 3 3

right, with intermediate lines making three rows horizontally, 5 5 5  

one row vertically and four rows diagonally. Rearrange the 7 7 7

numbers present so all these rows total 15. I wondered if one

can put the numbers 1, ..., 9 on this figure to get the same

sum on all these lines. However, the magic sum must then be 15 and the middle number of the top and bottom rows must occur in four different sums 15 and only the digit 5 occurs in four such sums. Indeed, letting B be the middle digit in the top row and adding these four sums gives 4 x 15 = 45 + 3B, so B = 5. Similarly the middle digit of the bottom row must also be 5. So the magic figure is impossible. Further, this argument works for Doubleday's problem, forcing the vertical line to be all 5s. There are then just two solutions, depending on whether the top row is 3 5 7 or 7 5 3. Doubleday gives one solution, saying you may be able to find others.

Birtwistle. Math. Puzzles & Perplexities. 1971. A       

Pp. 11 & 13. Consider the pattern at the right. Place the digits B       

1, 2, ..., 8 so that the vertical and horizontal quadruples and the E F G H

inner and outer circles (i.e. A, H, D, E and B, G, C, F) all have the C       

same sum. By exchanging C, D with G, H, we have a cube with four D       

magic faces, but the pattern has more automorphisms than the cube. One

sees that A + D = F + G, B + C = E + H, so these pairs can be interchanged. Also, one can interchange B and C, etc. We can then assume A = 1, B < C, B < E < H, F < G and then there are 6 solutions. Each of these gives 16 solutions with A = 1 and hence 128 solutions allowing any value of A. Hence there are 768 solutions in total. He gives one.

Shakuntala Devi. Puzzles to Puzzle You. Orient Paperbacks (Vision Press), Delhi, India, 1976. Prob. 102: The circular numbers, pp. 65 & 125. The case n = 5 of the arithmetical star of The Secret Out, 1859. Asks for and gives just one solution. See Singmaster, 1992.

Gareth Harries. Going round in triangles. M500 128 (Jul 1992) 11-12. Consider the lattice triangle of side two. This has 4 triangles, 6 vertices and 9 edges. Place the numbers 1, .., 15 on the vertices and edges so that each edge number is the difference of the numbers at its ends. He says his computer found 19 solutions. For the side one problem, there are just two solutions.

David Singmaster. Braintwister: Correct sum, rounded up. The Weekend Telegraph (27 Jun 1992) xxx & (4 Jul 1992) xxviii. Based on the version in Devi, I asked for all the solutions for n = 5 and n = 4. Using the argument I gave under The Secret Out, there are no solutions for n = 4 or any even n. For n = 5, the common difference d = │a_i - a_n+i│ must be either 1 or 5 and in either case, the antipodal pairs are determined. Fixing a₁ = 1 gives 4!/2 distinct solutions for each value of d, where the divisor factors out mirror images. In a note to my solution, which was not published, I showed that for general n, d must be a divisor of n, and each such divisor gives (n-1)!/2 distinct solutions. Now that I have found the version in The Secret Out, I am somewhat surprised not to have found more examples of this problem.

David Singmaster. Braintwister: Give the hour-glass some time. The Weekend Telegraph (6 Feb 1993) xxxii & (13 Feb 1993) xxxvi. The Magic Hourglass problem, as in Meyer, though I don't recall where I saw the problem -- possibly in one of Meyer's other books. I recall that my source gave the value of D or S.

Mirko Dobnik (Rošnja 5, 62205 Starše, Slovenia). Letter and example of 31 Aug 1994. Take a solid cube and make each face into a 2 x 2 array of squares. He puts the numbers 1, 2, ..., 24 into these cells as shown below. Each face adds up to 50 and numerous bands of 8 around the cube (at least 18) add up to 100. As he notes, the roundness of these values is notable.

10 13

11 16

1 22 8 17 3 24 20 5

4 23 19 6 2 21 7 18

14 9

15 12

Mirko Dobnik (Rošnja 5, 62205 Starše, Slovenia). Letter and example of 24 Jan 1995. Amends two faces of the above array to the following

10 13

11 16

1 22 8 19 3 24 6 17

4 23 5 18 2 21 7 20

14 9

15 12
7.O. MAGIC HEXAGON

The unique solution is given at the right. It is often presented 18 17 3        

with a point up. I will describe versions by saying which point or 11 1 7 19    

edge value is up and whether it is a reflection or not. 9 6 5 2 16

E.g. This has 17 up. 14 8 4 12    

15 13 10        
[Ernst] von Haselburg (of Stralsund). MS of problem and solution in the City Archives of Stralsund, dated 5 May 1887, with note of 11 May 1887 saying it was sent to Illustrierten Zeitung, Leipzig. Xerox kindly provided by Heinrich Hemme.

[Ernst] von Haselburg (of Stralsund), proposer and solver. Prob. 795. Zeitschrift für mathematischen und naturwissenschaftlichen Unterricht 19 (1888) 429 & 20 (1889) 263 264. Poses the problem for side 3 hexagon. Solution deals with the three sums of six symmetric points to show the central number is at most 8 and then finds only 5 is feasible and it gives a unique solution. He has 18 up. Reported by Martin Gardner, 1988, without a diagram; see also Hemme, Bauch.

William Radcliffe. 38 Puzzle. UK & US patents, 1896. ??NYS -- cited by Gardner, 1984. I couldn't find the US patent in Marcel Gillen's compilation of US puzzle patents, but a version is reproduced in Tapson and in Hemme, which has 3 up.

The Pathfinder. (A weekly magazine from Washington, DC.) c1910. ??NYS Trigg says Clifford Adams saw the problem here, but with a row sum of 35!

Tom Vickers. Magic Hexagon. MG 42 (No. 342) (Dec 1958) 291. Simply gives the solution, with 13 up, reflected.

M. Gardner. SA (Aug 1963) = 6th Book, chap. 3. Describes Clifford Adams' discovery of it. Shows 15 up, reflected.

C. W. Trigg. A unique magic hexagon. RMM 14 (Jan Feb 1964) 40 43. Shows the uniqueness in much the same way as von Haselburg, but in a bit more detail. Shows 15 up, reflected.

Ross Honsberger. Mathematical Gems. MAA, 1973. Chap. 6, section 2, pp. 69-76. Describes Adams' work, as given in Gardner, and outlines Trigg's proof of the uniqueness. A postscript adds that a Martin Kühl, of Hannover, found a solution c1940, but it was never published and cites Vickers note.

M. Gardner. Puzzles from Other Worlds. Vintage (Random House), NY, 1984, p. 141. Describes Radcliffe's work -- see above. Gardner says he was a school teacher at the Andres School on the Isle of Man and discovered the hexagon in 1895 and patented it in the US and UK. He shows the pattern with 15 up, reflected.

Frank Tapson. Note 71.25: The magic hexagon: an historical note. MG 71 (No. 457) (Oct 1987) 217 220. Says he has a book of correspondence of Dudeney's containing: two letters from Radcliffe in 1902; Dudeney's copy of Frénicle's letter to Fermat (or Mersenne?? -- see 7.O.1 & 7.N.2) and a reproduction of Radcliffe's published solution, labelled 'Discovered 1895 Entered at Stationers Hall 1896'. He shows 13 up, reflected, and a reproduction of Radcliffe's reflected form, which is 3 up. Radcliffe's letters refer to the similar problem discussed by Dudeney in Harmsworth's Magazine (Jan 1902) and London Magazine (Feb 1902) -- see under Dudeney in 7.O.1. [These are the same magazine -- it changed its name.]

Martin Gardner. Letter: The history of the magic hexagon. MG 72 (No. 460) (Jun 1988) 133. Describes von Haselburg, with no diagram, as communicated to him by Hemme.

Heinrich Hemme. Das Kabinett: Das magische Sechseck. Bild der Wissenschaft (Oct 1988) 164-166. Shows the solution is unique. Describes Adams's discovery and Gardner's article. Says R. A. Cooper then discovered Vickers' note, then Tapson discovered Radcliffe's version in 1973. Hemme says Tapson said Radcliffe was a teacher at the Andreas-Schule on the Isle of Man and got a UK patent. Then in the mid 1980s, Ivan Paasche of Stockdorf saw the German translation of Gardner and recalled Haselburg's work which he was able to locate. Paasche found that there was a 'Stadtbaurat' with an interest in mathematics named von Haselberg in Stralsund at the time. Stadtbaurat has several meanings -- it could be a member of the local planning board or a city architect, but it also was an honorific for a distinguished architect.

Hans F. Bauch. Zum magischen Sechseck von Ernst v. Haselberg. Wissenschaft und Fortschritt 40:9 (1990) 240-242 & a cover side. This is the first source to give von Haselberg's given name and to give a picture of him. He was born in 1827 and died in 1905. The original MS of the problem and solution have been located in the City Archives of Stralsund (see above) and Bauch sketches the original method -- there is a fair amount of trial and error -- and reproduces some of the MS. There are a number of sub configurations which must add up to the constant 38 and Bauch shows these, including a Star of David configuration -- see 7.O.1. Von Haselberg submitted his MS to the Illustrierten Zeitung of Leipzig, but they didn't use it. He was 'Stadtbaumeister' (= City Architect) in Stralsund and restored the facade of the City Hall. He also published a five volume work on local architectural monuments.

Ed Barbeau. After Math. Wall & Emerson, Toronto, 1995. A magic hexagon, pp. 151-152. When he gave this problem in his puzzle column, a solution came in only two weeks later and a second solution arrived six weeks after that. Neither solver seems to have used a computer. He cites only Honsberger. The first solver obtained a number of identities which would simplify the solution.

7.O.1. OTHER MAGIC HEXAGONS
Frenicle de Bessy. Loc. cit. in 7.N.2. 1640. Discusses a magic hexagon of White's form II (below).

I. G. Ouseley. Letter: The puzzle of "26". Knowledge 18 (1895) 255-256. 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 = 26. Says this has come from Mr. T. Ordish, who has arranged the numbers to sum to 26 in 30 different ways. A number of these are consequences of the above conditions, e.g. the sum on the inner hexagon = the sum of the vertices of a rhombus which is 26, as is the sum along certain angles, but I can't understand what is meant by '6 obtuse angles' or '4 rhomboids'. "I believe there are at least six ways ...."

In the next issue, p. 278, are three letters with a comment by Ouseley. Un Vieil Étudiant sends four solutions with the sum of the vertices of the large triangles being 13, and hence the sum on the inner hexagon being 52. These are complementary to the solutions requested, though neither he nor Ouseley notes this. These complementary solutions are rather easier to find though and I have indeed found six solutions, as does Ahrens below. J. Willis sends one solution of the original form, possibly implying that it is unique. T. sends one different solution. Ouseley notes that in a solution, the value at a point is the sum of the values at the two opposite crossings. ?? -- possibly more letters in the next issue.

In the next volume, pp. 35 & 84, are two notes saying that the publishers (T. Ordish & Co., London) and the proprietors (Joseph Wood Horsfield & Co., Dewsbury) of the puzzle have complained that the above notes are an infringement of their copyright in the "26" Puzzle and Knowledge apologizes for this.

Dudeney. Puzzling times at Solvamhall Castle. London Magazine 7 (No. 42) (Jan 1902) 580 584 & 8 (No. 43) (Feb 1902) 53-56. The archery butt. = CP, prob. 35, pp. 60-61 & 187-188. Hexagon of 19 numbers so that the 6 radii from the centre to the corners and the 6 sides each add to 22. Problem is to rearrange them so each adds to 23. Solution says one can get any number from 22 to 38, except 30.

William F. White. Op. cit. in 5.E. 1908. Magic squares -- magic hexagons, pp. 187 188.