7. arithmetic & number theoretic recreations a. Fibonacci numbers

11 sources for x * 2 + 1 * 5 * 10 = 100x + 50.

Two sources for x * 2 + 5 * 5 * 10 = 100x + 250. Also cites Pseudo-Bede, Fibonacci and AR (see AR, pp. 122-124, 138 & 227-228 for examples and further references). Cf Pacioli, Effect VIII.

21 sources for PB1 or PB2, with two simple variants.

6 sources where the above ideas are extended to more values. Cf section 7.AO for one form of this.

Chuquet. 1484. Prob. 155. English in FHM 230-231. PB1. Text only indicates what happens when x is even. Marre notes that this appears in de la Roche, 1520: ff. 218v   219r; 1538: ff. 150v - 151r. FHM say Chuquet gives an example, but it is not in Marre.

Pacioli. De Viribus. c1500.

Effects I - VI, ff. 3v - 16v. = Peirani 25-39. Algebraic divinations of 2, 3, 3, 3, 4, 5 values which are parts of a given number. E.g. Effect I divines x, y such that x + y = a from a(a+1)   2x - ay. Divide this by a-1 to get x + y/(a-1).

Effects VII, IX, X. Agostini's descriptions are very inadequate.

Ff. 16v - 19v. Septimo effecto trovare un Nů pensato int^o (Seventh effect to find a number thought of). = Peirani 40-43. PB2, using 1 + ½. He does examples for all four cases of a, b.

Ff. 20v - 21v. Nono effecto a trovare un Nů senza rotto (Ninth effect to find a number without a fraction). = Peirani 45. Take 3/2 twice, but rounding down. This takes 4(c+1) - a - 2b to 9(c+1) - 3a - 5b, with a = 1 iff there is a rounding at the first stage and b = 1 iff there is a rounding at the second stage. Pacioli gives a simpler rule: given c, form 4c and add: 1 if there are roundings at both stages; 2 if there is rounding at only the second stage; 3 if there is rounding only at the first stage. He does examples of all four cases of a, b.

Ff. 21v - 23r. Decimo effecto de trovare un Numero senza rotto (Tenth effect to find a number without a fraction). = Peirani 46-47. PB1, clearly describing that the result is 6a + 9b.

Ff. 19v - 20v. Octavo effecto quando el Nů fosse con R(otto) (Eighth effect when the number can be a fraction). = Peirani 43-44. Divines x from

x * 2 + 5 * 5 + 10 * 10 = 100x + 350. Cf Folkerts.

Effects XI - XXI, XXVII - XXXI, ff. 23v - 34v, 47r - 63v. = Peirani 48-62, 77-97. More complex algebraic divinations. E.g. Effect XI divines a number a by asking he person to split it into two parts, x, y (so that x + y = a) and compute x² + y² + 2xy.

Ghaligai. Practica D'Arithmetica. 1521. Prob. 34-35, f. 67r-67v. Take 3/2 twice, but he rounds down each time and he is not at all clear how the roundings relate to finding the number, and it is not nearly as elegant as when one rounds up.

Apianus. Kauffmanss Rechnung. 1527. Ff. M.vii.r - M.viii.r. PB1.

Tartaglia. General Trattato. 1556. Book 16, art. 197-198, f. 263v-264r. Divination by 1 + 1/2 twice and by 3/2 twice, i.e. PB2.

Recorde-Mellis. Third Part. 1582. Ff. Yy.ii.r - Yy.ii.v (1668: 473-474). PB2, done with one example of 7.

Prévost. Clever and Pleasant Inventions. (1584), 1998.

Pp. 166-168. PB2.

Pp. 185-189. Divination from 16 marked counters by a two column version of the 21 card trick. The method of rearranging the 16 counters is not entirely clear, but the principle is clearly explained.

Io. Baptiste Benedicti (= Giambattista Benedetti). Diversarum Speculationum Mathematicarum, & Physicarum Liber. Turin, (1580), Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599). [Rara 364. Graves 141.f.16.] Theorema CXVI, pp. 78-79. PB2.

Bachet. Problemes. 1612.

Prob. I. 1612: 14-17; 1624: 53-55; 1884: 15-16. PB1. = Pacioli X, but with more explanation.

Prob. II. 1612: 17-27; 1624: 56-65; 1885: 17-22. PB2. = Pacioli IX, but with more explanation.

Prob. III. 1624: 66-74; 1885: 23-26. Take 3/2 twice and then tell all but one digit. The Advertissement in the 1624 ed. says this was invented by R. P. Jean Chastelier, S.J. Not in the 1612,

Prob. XVI, 1612: 87-92. Prob. 18, 1624: 143-151; 1884: 72-83. 15-card trick. His Advertissement mentions that other versions are possible and describes a two column version. Labosne adds several diagrams which make the process much clearer and discusses the general idea, illustrating it with 27 cards and then with 45 cards in 5 columns.

Hunt. 1631 (1651).

Pp. 217-218 (209-210): A sixth way to find out a number thought. PB2.

Pp. 221 (misprinted 212) - 222 (213-214): An eighth way to find out a number thought.

PB1.

Schott. 1674.

Art. I, p. 57. PB1.

Art. II, pp. 56-57. PB2.

Ozanam. 1694.

Prob. 14 [part 9], 1696: 58; 1708: 52. Prob. 17, part 10, 1725: 146. Prob. 1, part 1,  1778: 139-140; 1803: 137; 1814: ??NYS; 1840: 62. PB1. In 1840, the algebraic proof is given. In 1725: 170 173, he adds, as a remark to Prob. 17, PB2, with a detailed explanation.

Prob. 16, 1696: 63-64; 1708: 56-57. Prob. 19, 1725: 154-156. Prob. 1, part 3,  1778: 140-142; 1803: 138-139; 1814: ??NYS; 1840: omitted. Take 1 + 1/2 twice but then subtract 2x to get r, then divide r repeatedly by 2 until one gets down to 1. Observing when he has to discard gives you the binary expansion of r. Subtracting 2x yielded r = a + b + c, so x = 4r   (2b + 3c).

Prob. 31. 1696: 85; 1708: 76-77. Prob. 35, 1725: 220-221. 21-card trick done with 36 cards. He says the desired card will be in the middle of its row, i.e. in the 6th place!

Prob. 15. 1778: 164-166; 1803: 165-166. Prob. 14, 1840: 74-75. Replaces the above with the 16 counter binary version as in Prevost and notes that one can use other powers of two.

Henry Dean. The Whole Art of Legerdemain, or Hocus Pocus in Perfection. 11th ed., 1790? ??NX -- seen at UCL Graves 124.b.36. Pp. 89-90 (89 is misprinted as 87). 21 card trick. "This trick may be done by an odd number of cards that may may be divided by three."

Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 81-83, no. 129: To tell the Number a Person has fixed upon, without asking him any question. A variant of the 3/2 method. Pick a number from 1 to 15, add 1 to it. (Triple the number; if it is odd, add 1; then halve it) thrice. Finally, if the result is odd, add one, and again halve. From observing when the person has to add one, you determine the number. If n + 1  =  8a + 4b + 2c + d, where the coefficients are zero or one, the process is straightforward up to the third tripling which yields 54a + 27b + 15c + 9d which is odd if and only if b + c + d is odd. The next steps are no longer simply expressible in terms of the coefficients. One has to add one at the third stage if and only if n + 1  1, 2, 4, 7 (mod 8) and one has to add one at the fourth stage if and only if n + 1 = 1, 6, 8, 10, 11, 12, 13, 15. He gives tables to determine n from the observations.

Manuel des Sorciers. 1825. ??NX