7. arithmetic & number theoretic recreations a. Fibonacci numbers

Pp. 152 154 (S: 240-242): De consolamine trium monetarum inter se [On the alloying of three monies]. A = (3, 4, 6), b = 5. Answers: 1, 1, 3 and 2, 5, 9.

Pp. 156 158 (S: 245-247): De consolamine septum monetarum [On an alloy of seven monies]. A = (1, 2, 3, 5, 6, 7, 8), b = 4. Answer has fractions.

Pp. 158 159 (S: 247-248) deals with 240 metals!

P. 160 (S: 249-250): De homine qui emit libras 7 trium canium per denarios 7 [On a man who buys 7 pounds of three meats for 7 denari]. 7 for 7 at 3, 2, ½. (Pork, beef, hyrax [an animal somewhat like a rabbit].) This has (0, 0) solutions! He gives 1, ⅓, 5⅓!

P. 165 (S: 256): De homine qui emit aves triginta trium generum pro denariis 30 [On a man who buys thirty birds of three kinds for 30 denari]. 30 for 30 at 3, 2, ½. (Partridges, doves, sparrows.) = Alcuin 33. He gives the (1, 1) answer.

P. 165 (S: 256-257): De eodem [On the same]. 12 for 12 at 2, ½, ¼. (Same birds.) = Alcuin 47. He gives the (1, 1) answer.

P. 165 166 (S: 257): De eodem cum genera avium sint quattuor [On the same when there are four kinds of birds]. 30 for 30 at 3, 2, ½, ¼. (Partridges, doves, turtle doves, sparrows.) (27, 19) solutions -- he gives (2, 2).

Pp. 322 323 (S: 452-453) give some examples with two and four metals done by false position.

Fibonacci. Epistola. c1225. In Picutti, pp. 332-336, numbers XI - XIII. One of the problems is briefly mentioned in: M. Cantor; Mathematische Beiträge zum Kulturleben des Völker; Halle, 1863; reprinted by Olms, Hildesheim, 1964; p. 345. Surprisingly, none of these problems have appeared elsewhere!

P. 247: De avibus emendis secundum proportionem datam. 30 for 30 at ⅓, ½, 2. (Sparrows, turtledoves, pigeons.) This has (3, 1) solutions -- he gives (1, 1).

Pp. 247-248: De eodem. 29 for 29 at ⅓, ½, 2. (Sparrows, turtledoves, pigeons.) This has (2, 2) solutions -- he gives both.

P. 248: Item de avibus. 15 for 15 at ⅓, ½, 2. (Sparrows, turtledoves, pigeons.) This has (0, 2) solutions -- he says "hoc esse non posse sine fractione avium demonstrabo." By eliminating the first variable, he gets y + 10z = 60 and he notes that 0, 10, 5 solves the problem. "Sed si volemus frangere aves", one can have 9/2, 5, 11/2. However he fails to find 9, 0, 6.

P. 248: above continued. 15 for 16 at ⅓, ½, 2. (Sparrows, turtledoves, pigeons.) This has (1, 1) solutions which he gives.

P. 248: above continued. 30 for 30 at ⅓, 2, 3. (Sparrows, pigeons, partridges.) This has (2, 1) solutions -- he gives (1, 1).

P. 249: above continued. 24 for 24 at 1/5, ⅓, 2, 3. (Sparrows, turtledoves, pigeons, partridges.) This has (6, 2) solutions -- he gives (2, 2).

Abbot Albert. c1240.

Prob. 2, pp. 332 333: 6 for 50 at 2, 9, 10. (1, 1) solutions. Actually he is using this for divination: if x + y + z = 6, then the value of 2x + 9y + 10z determines x, y, z. He gives a table of all the partitions of 6 into 3 non negative summands and computes 2z + 9y + 10z for each. This is more properly a problem for Section 7.AO.

Prob. 7, p. 334: 30 for 30 at 4, 2, ½. (Geese??, ducks??, fig-peckers.) (2, 1) solutions, he gives (1, 1).

Yang Hui. Xu Gu Zhai Qi Suan Fa (= Hsü Ku Chai Ch'i Suan Fa). 1275. Loc. cit. in 7.N, pp. 165 166, prob. 29 31.

29. 100 for 100 at 5,3, ⅓. (Same as Chang's problem, but only gives (1, 1) of the (4, 3) solutions.)

30. 100 for 100 at 7, 3, ⅓. (Three types of tangerine.) (6, 4) solutions, (1, 1) given.

31. same as 30 in terms of wines and with different measures.

Gherardi. Libro di ragioni. 1328.

Pp. 85 86. Chompera. 24 for 24 at ¼, 2, 3. (Sparrows, doves, geese.) This has (1, 1) solutions, which he gives.

P. 86. Chompera ucelli. 24 for 24 at 1/5, 1, 2, 3. (Sparrows, thrushes, doves, geese.) (1, 1) solution given of the (13, 6) solutions.

Lucca 1754. c1330.

Ff. 9v 10r, pp. 34 35. 40 for 40 at 3, 2, ¼. (Thrushes, larks, sparrows.) (2, 2) solutions, both given.

Ff. 10r 10v, pp. 35 36. 100 for 100 at 3, 1, 1/20. (Oxen, pigs, sheep.) (2, 1) solutions, (1, 1) given.

F. 46v, pp. 96 97. 60 for 600 at 7, 9, 11, 17. (Mixing grain.) (314, 272) solutions, he gives one: 35, 5, 5, 15.

F. 46v, p. 97. 60 for 480 at 5, 9, 7. (Mixing of metals.) (16, 14) solutions, he gives one: 0, 30, 30.

F. 48v, p. 103. 775 for 162.75 at .16, .18, .20, .27, .31. (Mixing of metals.) This has (3027289, 2966486) solutions! He gives one solution: 250, 150, 150, 100, 125.

(There are several similar problems here with solutions obtained by guessing.)

F. 56r, p. 125. 100 coins worth 2150 at 50, 33, 17, 25, 15. (2160, 1536) solutions, he gives one solution: 10, 10, 10, 10, 60.

Also the same with total value 3900. (526, 388) solutions, he gives one solution: 60, 10, 10, 10, 10.

F. 59v, pp. 135 136. 24 for 24 at ¼, 2, 3. (Sparrows, doves, geese.) (1, 1) solutions, which he gives. = Gherardi, pp. 85 86.

Munich 14684. 14C. Prob. XI, p. 79.

12 for 12 at 2, ¼, ½. (2, 1) solutions, he gives (1, 1). = Alcuin 47.

20 for 20 at 2, ¼, ½. (2, 2) solutions, he gives (1, 1).

Prob. XII, p. 79. 12 for 12 at 2, 1, ½, ¼. (11, 4) solutions, he gives (1, 1).

Narayana Pandita (= Nārāyaņa Paņdita [NOTE: ņ denotes n with an overdot and the d should have an underdot.]). 1356. Op. cit. in 7.N, p. 1, lines 2 5, p. 93. (Same as Mahavira's 152.) ??NYS -- see Bag, op. cit. under Bakhshali MS, p. 92.

Folkerts. Aufgabensammlungen. 13-15C.

12 for 12 at 2, ½, ¼. (Soldiers, girls, footsoldiers.) 16 sources. (2, 1) solutions, (1, 1) given. = Alcuin 47.

12 for 12 at 2, 1, ½, ¼. (Dividing a sum of money.) 9 sources. (11, 4) solutions, (1, 1) given. = Munich 14684, XII.

20 for 20 at 2, ½, ¼. (Birds or horses.) 8 sources. (2, 2) solutions, (1, 1) given. = Munich 14684, 2nd problem.

20 for 20 at 2, 1, ½. (Birds or horses.) 1 source. (7, 6) solutions, (1, 1) given.