7. arithmetic & number theoretic recreations a. Fibonacci numbers
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146. Problem A 7 (= Bhaskara I).147. Problem A 5. First appearance.148. Problem A 11 (= Fibonacci).149. n 1 (2, 3, ..., 9).150. Problem D-7 (= Fibonacci).199. Problem E.Buteo. Logistica. 1559. Prob. 70, pp. 279-280. Problem D-7 (= Fibonacci). Discusses problem A 7 (= Bhaskara I) and Cardan. Baker. Well Spring of Sciences. 1562? Prob. 4, 1580?: ff. 190v-191r; 1646: pp. 300-301; 1670: pp. 342-343. Probs. A 5 (= Tartaglia) & A 7 (= Bhaskara I).Prob. 5, 1580?: ff. 191v-192r; 1646: pp. 301-302; 1670: p. 342. Problem D 5. First appearance.Prob. 6, 1580?: f. 192r; 1646: p. 302; 1670: pp. 343-344. Problem D-7 (= Fibonacci).Bachet. Problemes. 1612. Prob. V: Faire encore le même d'une autre façon, 1612: 37-45. Prob. VI, 1624: 84-93; 1884: 34 37. General solution for 3, 4, 5 used for divination. Labosne adds case 2, 3, 5, 7 and a general approach. 1612 cites Forcadel, Gemma Frisius, Tartaglia, etc. van Etten. 1624. Prob. 51 52 (46 47), pp. 46 48 (69 71). Problems A 7 (= Bhaskara I) and D-7 (= Fibonacci). French ed. refers to Bachet for more detailed treatment. Henrion's 1630 Notte to prob. 52, p. 18, says that Bachet has treated this problem. Seki Kōwa. Shūi Shoyaku no Hō. 1683. ??NYS -- described in Smith & Mikami, pp. 123 124. Studies ax - by = 1. n 1 (5), 2 (7) (= Bhaskara I). Then extends to any number of congruences. Then does the system 35 n 35 (42), 44 n 28 (32) and 45 n 35 (50). W. Leybourn. Pleasure with Profit. 1694. Prob. 17, pp. 40-41. Prob. C-9. Constructs a table of 2·n! - 1, n = 2, ..., 9, and says 2·9! - 1 is the least solution. But he then gives 2519 + 2520k, k = 0, ..., 7 and says these are some of the infinitely many other solutions Ozanam. 1694. Prob. 23, 1696: 74-77; 1708: 65 67. Prob. 27, 1725: 188 198. Prob. 10, 1778: 195-198; 1803: 192-195; 1814: 167-169; 1840: 86-87. 1696 gives many examples, too numerous to detail, and some general discussion. The following is common to all editions: n 1 (2, 3, 5), 0 (7). 1725 has problem A-7 (= Bhaskara I). 1778 et seq. has problem D-7 (= Fibonacci) and then notes that Ozanam would solve this as 119 (mod 5040) rather than 119 (mod 420) -- but the 1696 or 1725 ed. only have relatively prime moduli. Dilworth. Schoolmaster's Assistant. 1743. Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168. Problem 2. Problem A-5 (= Tartaglia). Simpson. Algebra. 1745. Section XIII.
Quest. 1, p. 170 (1790: prob. I, p. 182). n 7 (17), 13 (26).Quest. 9, pp. 175-176. (1790: prob. VIII, p. 187). n 19 (28), 15 (19), 11 (15).1790: Prob. X, pp. 189 190. General case for 28, 19, 15. This is for computing the Julian year. Cf Dodson.Gentlemen's Diary, 1747 -- see Vyse, below. C-20. Ladies' Diary, 1748 -- see Vyse, below. C-6. Les Amusemens. 1749. Prob. 171, p. 318. n 1 (2, 3, 4, 6), 4 (5), 0 (7). Euler. Algebra. 1770. II.I.
Art. 13, pp. 305 306. n 2 (6), 3 (13).Art. 14: Question 8, pp. 306 307. n 16 (39), 27 (56).Art. 20: Question 11, p. 310. n 3 (11), 5 (19).Art. 21: Question 12, pp. 310 311. n 3 (11), 5 (19), 10 (29).II.III: Questions for practice. No. 10, p. 321. Problem B-9.Vyse. Tutor's Guide. 1773? The following are in a supplement in the Key only, but some are referred to earlier sources. Prob. 1, Key p. 361. C-20. Attributed to the Gentlemen's Diary, 1747.Prob. 3, Key p. 362. A-7 in verse.Prob. 4, Key pp. 362-363. C-6. Attributed to the Ladies' Diary, 1748.Dodson. Math. Repository. 1775. P. 142, Quest. CCXXVI. A-7 (= Bhaskara I).Pp. 142-143, Quest CCXXVII. D-7 (= Fibonacci).Pp. 148-149, Quest. CCXXXVI. x n (28), m (19). Then gives an Example: "The cycle of the sun 17; and the cycle of the moon 13; being given; to find the year of the Dionysian period?" This is the case n = 17, m = 13 of the general problem.Pp. 150-151, Quest. CCXXXVII. x n (28), m (19), p (15). Then applies to Julian period where the first is the cycle of the sun, the second is the cycle of the moon and the third is the Roman indiction. Cf Simpson.Pp. 151-153. He continues the discussion to find the general solution for any number of moduli which are prime to each other. Then does x 1 (2), 2 (3), 3 (5), 4 (7), 5 (11).Bonnycastle. Algebra. 1782. Pp. 137-140 (c= 1815: pp. 159-162) discuss the general method and give the following examples and problems. No. 1. n 7 (17), 13 (26). (1815: no. 1, = Simpson).No. 2. n 3 (11), 5 (19), 10 (29). (1815: no. 2, = Euler).No. 3. n 7 (19), 13 (28).No. 4. n 2 (3), 4 (5), 6 (7), 0 (2).No. 5. n 6 (16), 7 (17), 8 (18), 9 (19), 10 (20).No. 6. Problem B-9. (1815: no. 6, = Euler).No. 7. n 1 (2), 2 (3), 3 (5), 4 (7), 5 (11) (= Dodson).P. 205, no. 35 (in 1805; 34 in 1788). n -1 (6, 5, 4, 3, 2) (C-6).Carlile. Collection. 1793. Prob. LXVIII, pp. 39-40. n 1 (2, 3, 4, 5, ..., 9) (= Tartaglia, C-9). He gives the answer: 2519. D. Adams. Scholar's Arithmetic. 1801. P. 208, no. 65. A-7. Answer: 721. Bonnycastle. Algebra. 10th ed., 1815. Pp. 159-162 is similar to the 1782 ed., but has the following different problems.
No. 3. n 2 (6), 3 (13) (= Euler).No. 4. n 5 (7), 2 (9).No. 5. n 16 (39), 27 (56) (= Euler).No. 6. n 5 (7), 7 (8), 8 (9).No. 8. n 0 (2, 3, 4, 5, 6), 5 (7).P. 230, no. 35. n -1 (6, 5, 4, 3, 2).Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 49-50, no. 76: A quantity of eggs being broken, to find how many there were, without remembering the number. Problem A-7 (= Bhaskara I). Jackson. Rational Amusement. 1821. Curious Arithmetical Questions. No. 15, pp. 17-18 & 74. Problem A-5 (= Tartaglia).Download 1.54 Mb. Share with your friends:
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