7. arithmetic & number theoretic recreations a. Fibonacci numbers

No. 107, pp. 206-207. Two clock hands start together. One circles in 1 day, the other in 3 days. When do they meet? Also does the general problem with periods b, c, b < c, getting bc/(c-b).

No. 108, p. 207. Applies above to sun and moon to get synodic month.

Vyse. Tutor's Guide. 1771? Prob. 20, 1793: p. 79; 1799: p. 85 & Key p. 110-111. Island 73 in circumference; three persons set out in the same direction at speeds 5, 8, 10. When do they all meet again?

Bonnycastle. Algebra. 1782. P. 86, no. 23. Identical to Vyse.

Pike. Arithmetic. 1788. P. 353, no. 31. Island 50 in circumference. Three walkers start in the same direction at speeds 7, 8, 9. When and where do they meet again? = D. Adams; Scholar's Arithmetic; 1801, p. 208, no. 66.

Hutton. A Course of Mathematics. 1798? Prob. 37, 1833: 223; 1857: 227. Identical to Vyse.

Kaida Anmuyo. c1800. Problem given on pp. 139 140 of Shen Kangsheng, loc. cit. under Zhang Qiujian above. Assume 365¼ degrees in a circle. Five stars are in a line and travel at speeds of 28 13/16, 19 1/4, 13 5/12, 11 1/7, 2 7/9 degrees per day. When do they meet at the starting point again?

D. Adams. New Arithmetic. 1835. P. 244.

No. 84. Island of circumference 20 and three travellers set out from the same point in the same direction at rates 2, 4, 6. When do they meet?

No. 85. Same with just the travellers of rates 2, 6. = O-(2, 6; 20)

Anonymous. A Treatise on Arithmetic in Theory and Practice: for the use of The Irish National Schools. 3rd ed., 1850. Op. cit. in 7.H. P. 360, no. 48. Three walkers start to circle an island of circumference 73, at rates 6, 10, 16. When do they meet again?

James B. Thomson. Higher Arithmetic; or the Science and Application of Numbers; .... Designed for Advanced Classes in Schools and Academies. 120th ed., Ivison, Phinney & Co, New York, (and nine copublishers), 1862. Prob. 94, p. 308 & 422. Same as Anon: Treatise.

Daniel W. Fish, ed. The Progressive Higher Arithmetic, for Schools, Academies, and Mercantile Colleges. Forming a Complete Treatise of Arithmetical Science, and its Commercial and Business Applications. Ivison, Blakeman, Taylor & Co., NY, nd [but prefaces give: 1860; Improved Edition, 1875]. P. 418, no. 64. Island 120 in circumference. Seven men start walking around it from the same point at speeds 5, 25/4, 22/3, 33/4, 19/2, 41/4, 45/5 per day. When are they all together again? Answer: after 1440 days.

Mittenzwey. 1880. Prob. 117, pp. 23-24 & 76; 1895?: 135, pp. 27-28 & 79; 1917: 135, pp. 25-26 & 76. Seven colleagues return to a forest inn every 1, 2, 3, 4, 5, 6, 7 days. when will they all return again?

Lemon. 1890. The Maltese cross, no. 483, pp. 63 & 115. Walkers complete 6, 9, 12, 15 circuits per hour -- when are they all again at start?

Hoffmann. 1893. Chap. IV.

No. 38: When will they get it?, pp. 152 & 202 = Hoffmann-Hordern, p. 127. Guests come to restaurant with periods 1, 2, ..., 7 days. When do they all meet again?

No. 47: The walking match, pp. 154 & 207 = Hoffmann-Hordern, p. 130. Four men walk around a track of length 1 with speeds 5, 4, 3, 2 per hour. When do all meet at start again?

Clark. Mental Nuts. 1897, no. 51. When will we three meet again. Three bicycle riders can ride around a one mile track in 2 1/2, 2 3/5, 3 1/4 minutes. If they all start together, when will they all meet again at the starting point?

Dudeney. "The Captain" puzzle corner. The Captain 3:2 (May 1900) 97 & 179 & 3:4 (Jul 1900) 303. No. 2: The seven money boxes. Boy puts a penny in i-th box on i-th day, where day 1 is 1 Jan 1900. When he has to put in seven pennies, he will then open them all up. When is this and how much will he have? Answer: 420 days = 24 Feb 1901 and £4 10s 9d.

Depew. Cokesbury Game Book. 1939. Bicycle racers, p. 221. One can travel around the track in 6 minutes, the other in 9 minutes. When are they together again?
7.P.7. ROBBING AND RESTORING
Men keep money together and divide it into amounts x₁, x₂, ... -- usually by robbing the common fund. They put fractions a_ix_i into a pool and divide the pool in proportion b₁ : b₂ : .... They then have money in the proportion c₁ : c₂ : ..., or actual amounts d₁, d₂, ....

I use A for (a₁, a₂, a₃), etc.

Kurt Vogel. Ein unbestimmtes Problem al-Karağī in Rechenbüchern des Abendlands. Sudhoffs Archiv 61 (1977) 66-74. Gives the history of this problem, particularly the transmission to Fibonacci via John of Palermo and the different methods of solving it -- Fibonacci gives three methods. He mentions all the entries below except Gherardi and Calandri, which had not been published when he wrote, and Pacioli, which is sad because Pacioli is not very clear!

Jacques Sesiano, op. cit. under 7.R, 1985, discusses this problem along with the problems in 7.R. He calls it "The disloyal partners". He cites Abū Kāmil's Algebra, ff. 100r-101r, ??NYS, and says it is solved two ways there. Martin Levey's 1966 edition of the Algebra does not give the problem and states that the unique Arabic MS ends on f. 67r, and uses a 'better' Hebrew text. The Arabic MS actually continues and the third part of the book, ff. 79r-111r, was treated by Schub & Levey in 1968 & 1970. Sesiano; Les méthodes d'analyse indéterminée chez Abū Kāmil; Centaurus 21 (1977) 89-105 is scathing about the work of Levey and of Schub & Levey, saying the Hebrew MS is third-rate, and the translators have made serious mathematical and philological errors. Sesiano studies some of Abu Kamil's problems in this article, but unfortunately the problem of this section is not among them.

al Karkhi. c1010. Sect I, no. 45 & 47; sect III, no. 6; pp. 80 81 & 90.

I 45: Two men have d₁ = 40 and d₂ = 60. From the common sum, they take x₁ and x₂ = 100   x₁. The first gives a₁ = 1/4 of what he took to the second and the second gives a₂ = 1/5 of what he took to the first. Then they have correct amounts. Answer: (400, 700)/11.