7. arithmetic & number theoretic recreations a. Fibonacci numbers

P. 6, Quest. XV. III-(30, 36, 40).

P. 19, Quest. L. Find x/y such that (x+1)/y = 1/3, x/(y+1) = 1/4, i.e.

V-(1, 1/3; 1, 1/4). [(x+1)/y = 1/n; x/(y+1) = 1/(n+1) has solution x = n+1, y = (n+1)² - 1. In general, (x+1)/y  =  a/b, x/(y+1) = c/d, i.e. V-(1, a/b; 1, c/d) gives x = c(a+b)/(ad-bc), y = b(c+d)/(ad-bc). One would normally assume a/b > c/d. Cf Wolff in 7.R.3 for a different phrasing of the same problem.]

P. 38, Quest LXXXIX. I-(1, 2, 3) with purse worth 55. The context is three horses and a saddle. The first horse, with the saddle, is worth as much as the other two horses, etc. Answer: 5, 25, 35.

P. 64, Quest. CXX. Multiplicative form of type III. xy = 12, xz = 18, yz = 24. He doesn't multiply the equations to get (xyz)² = 72². Answer: 3, 4, 6.

P. 75, Quest. CXL. Like the previous, with four variables. wxy = 252, wxz = 432, wyz = 756, xyz = 336. Here he does multiply them together to find wxyz = 3024. Answer: 9, 4, 7, 12.

Hutton. A Course of Mathematics. 1798? Prob. 30, 1833: 222; 1857: 226. Two horses with saddle. I (2, 3) with saddle worth 50. Answer: 30, 40; 50. (= 10 times Mahavira v. 244.)

Unger. Arithmetische Unterhaltungen. 1838. Pp. 115-117 & 256, nos. 431-442. All of these have different purses except no. 435, which is I-(½, ⅓) with a purse of -10, i.e. 10 is taken away rather than gained. No. 432 uses the context of ages.

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Arithmetical puzzles, no. 2, pp. 172-173 (1868: 184). I-(1, 2, 3) with purse of 11. Answer: 1, 5, 7; 11. See AR 113 & 227.

Todhunter. Algebra, 5th ed. 1870. Section XIII, art. 185; Example XIII, nos. 1, 3, 5, pp. 99-100 & 578.

Art. 185. V-(6, 3/4; -2, 1/2). Answer: 9/20.

No. 1. V-(3, 1; 2, 1/2). Answer: 5/8.

No. 3. V-(1, 1/3; 1, 1/4. Answer: 4/15. = Dodson, p. 19.

No. 5. V-(36, 3; 5, 2). Answer: 42/26.

Mittenzwey. 1880.

Prob. 295, pp. 54 & 105; 1895?: 325, pp. 57 & 106; 1917: 325, pp. 52 & 100. III (89, 82, 97), in context of measuring pairs of edges of a triangle. Answer:  45, 52, 37.

1895?: prob. 93, pp. 21 & 69-70; 1917: 93, pp. 19 & 66-67. III-(23. 19, 10). Answer:  3, 7, 16. Algebraic solution by elimination given.

Clark. Mental Nuts. 1897, no. 75. The horses and saddle. I-(1, 2, 3) with the saddle acting as the purse and the total value of the horses and saddle being 220. Answer:  (1, 5, 7; 11) * 55/6. See AR 113 & 227; Parlour Pastime.

Haldeman-Julius. 1937. No. 23: The coin problem, pp. 5 & 21. Like two men finding a purse of 25. First says: If I had the purse, I'd have 3 times what you have. Second says: And if I had the purse, I'd have five less than half of what you have. Answer: 230, 85.

Ken Russell & Philip Carter. Intelligent Puzzles. Foulsham, Slough, 1992. Prob. 57, pp. 45 & Answer 75, p. 164: Money. Jack has 75p and ¾ of what Jill has; Jill has 50p and ½ of what Jack has. This is a straightforward problem, but it is also a concealed version of I (4/3, 2) with a purse of 100.
7.R.2. "IF I HAD 1/3 OF YOUR MONEY, I COULD BUY THE HORSE"
See Tropfke 608.

For related problems, see 7.H.4.

NOTATION. Again there are two forms.

I (a₁, a₂, ..., a_n) -- i-th says "If I had a_i of what the rest of you have, then I could buy the

horse".

II (a₁, a₂, ..., a_n) -- i-th says "If I had a_i of what the i+1-st has, then I could buy the horse".

The problems are the same when there are only two people and I will label two person cases as form I.

The problem often replaces horse by house, ship, etc. In some cases, the value of the horse, house, etc. is given. In some cases the value is given with no reference to anything bought and I say "with h = ..." to indicate the value. Some problems have different values of horses. If values are simply given, I say "with h₁, h₂ = ..." or "with h_i = ...". See: Fibonacci, Gherardi, Lucca 1754, AR, Benedetto da Firenze, della Francesca, Pacioli, Riese: Coss, Tartaglia, Buteo, Schott, Simpson for examples where there are different values of horses.

Solution notation as in 7.R.1 with the horse last.

Let h be the value of the horse and let T be the total of the amounts. Then I (a₁,a₂,...) with n people has n equations x_i + a_i(T   x_i) = h, so x_i = (h - a_iT)/(1-a_i). Adding these for all i gives one equation in the two unknowns T and h. However, letting C = T - h leads to the simplest equation: (n 1)T  =  [Σ 1/(1 a_i)] C.

For II (a₁,a₂,...), systematic elimination in the n equations x_i + a_ix_i+1 = h leads to x₁ [1 + (-1)ⁿ⁺¹a₁a₂...a_n] = h [1 - a₁ + a₁a₂ - ... + (-1)ⁿa₁a₂...a_n], and any other value can be found by shifting the starting point of the cycle.

In either case, the solution can be adapted to variable purses -- see 7.R. Taking negative values of h and all a_i converts this into 7.R.1 -- men find a purse, which is slightly easier to deal with.

INDEX OF SOME COMMON TYPES.

I have recently compiled this and was surprised to see how much repetition is present.

I (⅔, ¾). Riese: Rechnung; A Lover of the Mathematics; Euler IV.618; Vyse;

I (½, ⅔). Chiu Chang Suan Ching; Sun Zi; AR 176;

I-(½, ⅓). della Francesca 16r; Peurbach; Lacroix;

I-(⅓, ¼). al-Karkhi I-42; Fibonacci 228; BR 7; AR 171-175, 177, 221;

della Francesca 36v; Pacioli 190v; Riese: Rechnung; Schott 562-563;

I-(1/3, 1/5). Calandri; Calandri: Raccolta;

I (1/6, 1/7). Benedetto da Firenze;
I-(½, ⅔, ¾). della Francesca 21v;

I (½, ⅓, ¼). Gherardi 46-47; Lucca 1754 58r; Munich 14684; Folkerts;

Provençale Arithmétique; AR 178, 224, 340; Benedetto da Firenze;

della Francesca 17v-18r, 39r; Chuquet; Borghi; Pacioli 105v-106r, 192r, 192v;

Tonstall; Riese: Coss 122, 123; Buteo 81; Les Amusemens; Euler III.19, IV.622;

I (1/3, 1/4, 1/5). Diophantos 24; al-Karkhi III-26; Fibonacci 245; AR 341; Buteo 192-193;

Schott 563;
I-(2/3, 3/4, 4/5, 5/6). Chuquet;

I (1/2, 1/3, 1/4, 1/5). Fibonacci 245-248. de Nemore II-27; Provençale Arithmétique;

Riese: Coss 124-126; Pearson;

I (1/3, 1/4, 1/5, 1/6). Diophantos 25; al-Karkhi III-27;

I-(1/2, 2/3, 3/4, 4/5, 5/6). Chuquet;

I (1/2, 1/3, 1/4, 1/5, 1/6). Provençale Arithmétique;

II (3/4, 4/5, 5/6). Riese: Coss 121;

II (½, ⅔, ¾). AR 180; Chuquet; Tartaglia 22;

II (½, ⅓, ¼). Lucca 1754 58r; AR 179, 223, 339; della Francesca 36v-37r; Chuquet;

Riese: Rechnung; Riese: Coss 31, 47, 120; Peurbach; Tartaglia 40, 41; Buteo 81;

Euler IV.619-629;

II (1/3, 1/4, 1/5). Fibonacci 229; BR 114; AR 156, 157, 338; della Francesca 17v; Chuquet;

II (1/3, 1/5, 1/4). al-Karkhi III-33;
II (1/2, 1/3, 1/4, 1/5). de Nemore II-25; Gherardi 45-46

II (1/3, 1/4, 1/5, 1/6). Fibonacci 231-232.

II (1/2, 1/3, 1/4, 1/5, 1/6). Fibonacci 327-329.

II (1/3, 1/4, 1/5, 1/6, 1/7). Fibonacci 234.

II (2/3, 3/4, 4/5, 5/6, 6/7, 7/8, 8/9). Riese: Coss 140;
Buying a horse with a friend or a found purse. Lucca 1754 61r-61v; della Francesca 40v;

Pacioli 193v-194r; Cardan;

Jens Høyrup. Sub-scientific mathematics: Undercurrents and missing links in the mathematical technology of the Hellenistic and Roman world. Preprint from Roskilde University, Institute of Communication Research, Educational Research and Theory of Science, 1990, Nr. 3. (Written for: Aufsteig und Niedergang der römischen Welt, vol. II 37,3 [??].) P. 18 quotes Plato's Republic, 333b-c, "to buy in common or sell a horse" and feels this may be a reference to this type of problem. (This seems a bit far-fetched as there are many simpler types of commercial problem which could be described by this phrase.)

Chiu Chang Suan Ching. c 150. Chap. VIII.