7. arithmetic & number theoretic recreations a. Fibonacci numbers
Part II, f. 55v, prob. 38. Dig a well, (10, 10; 6). Divides as T6 : T10 = 21 : 55
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- Section 8, ff. CC.i.r - CC.i.v (p. 138). Men dig a well, (34, 60; 20). Divides in ratio T 20 : T 34 = 6 : 17.
- Prob. 36, pp. 240-241. Dig a well, (100, 50; x) worth 28 22/101.
- If the cost per unit depth is an arithmetic progression: A, A+D, ..., A+(a 1)D and δ = D/A and d is the value of the partial well, then
- Problem 4: "Let 12 be set down in four Figures and let each Figure be the same."
- No. 2, pp. 1 51. "It is required to express 100 by four 9s."
- No. 40, pp. 10 62. Use three 3s to make 1/3, 1 and 3.
- Prob. 46: A dozen quibbles: part 12, pp. 18 36. As in Family Friend.
- Prob. 1, pp. 427 431. "Put down four nines, so that they shall make one hundred."
- Prob. 564-15, pp. 253 395. Write 100 with 4 nines. Write 1000 with no zeroes -- answer: 999 9/9.
- Vagaries, no. 217(b), pp. 33 105. Three 6s to make 7.
Part II, f. 55v, prob. 38. Dig a well, (10, 10; 6). Divides as T6 : T10 = 21 : 55.Part II, f. 55v, prob. 39. Dig a well, (10, 10; x) worth 4. He notes 4 : 10 = 22 : 55, so we want n(n+1)/2 = 22, which would give n = (-1 + 177)/2 = 6.15207.... He interpolates as 6 days plus 1/7 of the sixth day, i.e. n = 6.14286....Calandri. Raccolta. c1495. Prob. 38, pp. 33 34. If a well 24 deep is worth 24, how deep a well is worth 40? I.e. (24, 24; x) worth 40. He takes values proportional to Td. Tagliente. Libro de Abaco. (1515). 1541. Prob. 110, ff. 55r-55v. Dig a well, (9, 24; 5). Divides in ratio T5 : T9 = 15 : 45. Apianus. Kauffmanss Rechnung. 1527. F. D.vi.v. Mason to build a tower 100 high for 100. He falls ill after 84. Divides in ratio T84 : T100. Cardan. Practica Arithmetice. 1539. Chap. 66. Section 8, ff. CC.i.r - CC.i.v (p. 138). Men dig a well, (34, 60; 20). Divides in ratio T20 : T34 = 6 : 17.Section 10, ff. CC.i.v CC.ii.v (p. 138). Man building a wall 5 high at prices 10, 20, 40, 80, 160 per unit of height. He stops at height 2½. Takes half a unit of height as having 2 times the value of the previous half-unit, so the interval from 2 to 3 high worth 40 divides into two halves worth x and x2, giving x = 40/(1+2).Buteo. Logistica. 1559. Prob. 35, pp. 238-240. Dig a well, (100, 50; 50). Divides as T50 : T100 = 1275 : 5050. Also does (200, 50; 100). His depths are initially in cubits, but are converted to 'semipedes' -- 120 cubits is 50 semipedes.Prob. 36, pp. 240-241. Dig a well, (100, 50; x) worth 28 22/101.Ozanam. 1694, 1725. Prob. 7, question 1, 1696: 30; 1708: 27. Prob. 10, question 1, 1725: 60-61. Prob. 2, 1778: 65; 1803: 67-68; 1814: 60; 1840: 32-33. Man digging a well 20 feet deep, to receive 3, 5, 7, ..., for each successive foot. [This is not really in this section, but is included because later ed. use it as the basis of the next problem.]Prob. 51, question 1, 1725: 256 257. Prob. 3, 1778: 66-67; 1803: 68-69; 1814: 61-62; 1840: 33. Man digging a well, (20, b; 12) (1778 et seq. change 12 to 8). 1725 divides as T12 : T20 = 78 : 210. 1778 notes that the difficulty of the work increases in arithmetic progression, but that there are many such progressions. He then posits that the first unit is worth 1/4 when the agreed payment is 20 and this gives a difference of 30/11 for the arithmetic progression.If the cost per unit depth is an arithmetic progression: A, A+D, ..., A+(a 1)D and δ = D/A and d is the value of the partial well, thend/b = ((2A-D)c + Dc2)/((2A-D)a + Da2).Les Amusemens. 1749. Prob. 44, p. 176. Mason to dig a well, (10, b; 4). Divides as T4 : T10 = 10 : 55. Vyse. Tutor's Guide. 1771? Prob. 17, 1793: p. 136; 1799: pp. 144-145 & Key p. 188. Same as Ozanam, prob. 7, but with depth 30. Jackson. Rational Amusement. 1821. Curious Arithmetical Questions, no. 25, pp. 20 & 79. Dig 20 yards for £20. Man falls sick after 8 yards. "How much was then due to him, on a supposition that the labour increases in arithmetical proportion as the depth?" I.e. (20, 20; 8). Solution notes that the data does not determine what the arithmetic progression is and chooses 5s as cost of the first yard -- see Ozanam. Unger. Arithmetische Unterhaltungen. 1838. Pp. 182 & 263, no. 693. Man digging a well 49 feet deep. First foot costs 15, but each successive foot costs 6 more than the previous. Find cost of last foot and total cost. So this is really an arithmetic progression problem, but I haven't seen others of those using this context. Vinot. 1860. Art. CVII: Problème du puits et du Maçon, pp. 126-127, (20, 400; 10). He assumes the cost of digging up a unit depth is 5 and that lifting the i-th unit raises it from its centre of gravity, so is given by A, 3A, 5A, ..., 39A, where A has to be determined from the total cost. He finds A = 3/4 and d = 125. Pierre Berloquin. The Garden of the Sphinx, op. cit. in 5.N. 1981. Prob. 117: A bailout fee, pp. 66 & 166. Man contracts to bail out a 20 yard deep well for $400 and gives up at 10 yards. Answer says value is proportional to the depth d times the average distance lifted, i.e. d/2, hence value is proportional to d2/2 and this is the result that integration produces. 7.I. FOUR FOURS, ETC. Express an integer using four 4s, etc. Cupidus Scientiae, 1881, seems to be the first to ask for solutions to a lot of the integers, rather than a few specific examples. The next examples of the general form are Cunningham & Wiggins (1905), Pearson (1907), Ball (1911), Ball (1912). Dawson (1916) is the first to ask for four R's, where R is indeterminate, e.g. 3 = (R+R+R)/R. I have included examples where a set of numbers and operations is given and one has to obtain a given value. This overlaps a bit with 7.I.1, where the object is to find the maximum possible value, and with 7.AC.3-6, where one uses all nine or ten of the digits and I have included problems of inserting signs into 12...9 to make 100 in 7.AC.3. Dilworth. Schoolmaster's Assistant. 1743. Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168. Problem 4: "Let 12 be set down in four Figures and let each Figure be the same."Problem 9. "Says Jack to his brother Harry, I can place four threes in such manner that they shall just make 34; can you do so too?"Les Amusemens. 1749. Pp. 52-54. Several problems leading to: 7 7/7, 33 3/3, 55 5/5, 99 9/9, 77 77/77, 2222 2222/2222, 11 1/1, etc. See the entry in 7.AN. Vyse. Tutor's Guide. 1771? Prob. 1, 1793: p. 155; 1799: p. 165 & Key p. 206. "Four Figures of nine may be so placed and disposed of as to denote and read for 100, neither more nor less. Pray how is that to be done?" Pike. Arithmetic. 1788. P. 350, no. 16. "Said Harry to Edmund, I can place four 1's so that, when added, they shall make precisely 12; Can you do so too?" Jackson. Rational Amusement. 1821. Arithmetical Puzzles. No. 2, pp. 1 & 51. "It is required to express 100 by four 9's."No. 4, pp. 2 & 51. Use three 2s to make ½, 1 and 2.No. 5, pp. 2 & 51. Express 12 by four equal figures.No. 17, pp. 5 & 56. Express 6½ by four 5s. Answer is: 5.5 + 5/5.No. 33, pp. 8 & 59. Use four 2s to make 1/8, 1/2, 2 and 8.No. 40, pp. 10 & 62. Use three 3s to make 1/3, 1 and 3.No. 42, pp. 10 & 62. Use four 3s to make 1/243, 1/27, 1/3, 3, 27 and 243.No. 43, pp. 10 & 62. Use five 3s to make the same numbers as in no. 42.No. 44, pp. 10 & 63. Express 78 by six equal digits.Endless Amusement II. 1826? Prob. 23, p. 201. "Put down four nines, so that they will make one hundred." Child. Girl's Own Book. Arithmetical puzzles, no. 5. 1832: 170 & 179; 1833: 184 & 193; 1839: 164 & 173; 1842: 282 & 291; 1876: 231 & 244. "Place four nines together, so as to make exactly one hundred. In the same way, four may be made from three threes, three may be made from three twos, &c." The 1833 solution is printed rather oddly as 199 9 9, while the 1839 and 1842 solution is 99 9-9 and the 1876 solution is 99 99. Nuts to Crack III (1834), no. 211. "Write down four nines so as to make a hundred." Family Friend 1 (1849) 150 & 178. Problems, arithmetical puzzles, &c. -- 4. "Put four fives in such a manner, that they shall make 6½. -- D. F." Answer is 5 5/5 + .5. = The Sociable, 1858, Prob. 46: A dozen quibbles, part 12: pp. 300 & 318. = Book of 500 Puzzles, 1859, prob. 46: part 12, pp. 18 & 36. Boy's Own Book. To place four figures of 9 in such a manner as to represent 100. 1855: 601. Magician's Own Book. 1857. Quaint questions, p. 253. [No. 4] -- "Place three sixes together, so as to make seven." [No. 6] -- "Place four fives so as to make six and a half." [Boy's Own Conjuring Book, 1860, pp. 224-225, has the Quaint Questions, but omits these two questions!] Book of 500 Puzzles. 1859.
Quaint questions, p. 67. [Nos. 4 & 6] -- Identical to Magician's Own Book.Charades, Enigmas, and Riddles. 1860: prob. 30, pp. 60 & 64; 1862: prob. 31, pp. 136 & 142; 1865: prob. 575, pp. 108 & 155. "Write a Hundred with 4 nines." (1862 & 1865 have slightly different typography.) Illustrated Boy's Own Treasury. 1860. Prob. 1, pp. 427 & 431. "Put down four nines, so that they shall make one hundred."Prob. 26, pp. 429 & 433. "Put four fives in such a manner, that they shall make 6½."Prob. 38, pp. 430 & 434. "It is required to place four 2's in such a manner as to form four numbers in geometrical progression?" Uses four 2s to make each of 1/8, 1/2, 2, 8.Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 564-15, pp. 253 & 395. Write 100 with 4 nines. Write 1000 with no zeroes -- answer: 999 9/9.Prob. 564-18, pp. 253 & 395. Write 100 with no zeroes, using 4, 6 or 8 figures. Answers: 99 3/3; 99 44/44; 99 999/999.Magician's Own Book (UK version). 1871. Paradoxes [no. 3], p. 37. "With four fives make 6½? -- 5 5/5 .5." where the 5/5 is written as a 5 over a 5 with no fraction bar. Cf Jackson and Magician's Own Book. Hugh Rowley. More Puniana; or, Thoughts Wise and Other-Why's. Chatto & Windus, London, 1875. P. 300. "Write down one hundred with four nines." Mittenzwey. 1880. Prob. 1, pp. 1 & 58; 1895?: 1, pp. 7 & 62; 1917: 1, pp. 7 & 56. Write 100 with six equal digits. "Cupidus Scientiae" (possibly the editor, Richard A. Proctor). Four fours, singular numerical relation. Knowledge 1 (30 Dec 1881) 184, item 151. A bit vague as to what operations are permitted, but wants four 4s to make various values. Says he has not been able to make 19. H. Snell. Singular property of number 4. Knowledge 1 (6 Jan 1882) 209, item 178. 19 = 4! 4 - 4/4. Editor says 4! is not reasonable for the problem as posed. Solutions from various contributors. Four fours. Knowledge 1 (13 Jan 1882) 229, item 184. Numerous solutions for 1 through 20, except 19. Solutions for 19 are: 4/.4 + 4/.4; 4! - 4 - 4/4; 4/.4 4/4 ("manifestly erroneous"); (4 + 4 - .4)/.4; (x + x .x)/.x in general. Four 3s give same results as three 5s, except for 17. Albert Ellery Berg, ed. Op. cit. in 4.B.1. 1883. P. 373. "Place three sixes together so as to make seven." Lemon. 1890. Vagaries, no. 217(b), pp. 33 & 105. Three 6s to make 7.Arithmetical, no. 752, pp. 92 & 124. = Sphinx, no. 600, pp. 81 & 118. "Place four nines so as to make one hundred."Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. II: A curious addition sum, p. 2. Mentions "writing down 100 with four nines" as 99 9/9. (Sam Loyd.) One hundred pounds for correct answer to a puzzle. Tit-Bits (14 Oct 1893) 25. "Find How to Arrange the Figures · 4 · 5 · 6 · 7 · 8 · 9 · 0 · in an Arithmetical Sum which Adds up the Nearest to 82." "Mr. Loyd is confident that no one will find it out." Indeed, Loyd will be paid £100 only if no correct answer is received. (Sam Loyd.) Solution of Mr. Sam Loyd's one hundred pound puzzle. Tit-Bits (18 Nov 1893) 111. 80·5 + ·97 + ·46 = 82. (There are points over the 5, 9, 7, 4, 6, but my printer may not print these clearly.) Here the mid-line dot (·) is used for a decimal point. Because of the number of correct solutions, ten extra names were drawn from them for additional £5 prizes. "It seems that not a single person in the whole of America has sent the correct answer when a prize was offered there, but here we have received a very large number actually correct." [See MRE for another solution.] Report on the 82 puzzle appeared in 25 Nov and letter from Loyd appeared about two weeks later - photocopies on order. Hoffmann. 1893. Chap. IV, no. 18: Another way to make a hundred, pp. 148 & 193 = Hoffmann-Hordern, p. 120. Use six 9s to make 100. Ball. MRE, 3rd ed., 1896. P. 13. "... a question which attracted some attention in London in October, 1893, ...." [See Loyd above.] He says that the problem is to make 82 with the seven digits 9, 8, 7, 6, 5, 4, 0 and gives one solution as 80·69 + ·74 + ·5 (with points over the 9, 4, 5 -- there should also be points over the 6 and 7). H. D. Northrop. Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 12, pp. 68 & 73. Express 6½ by four 5s. Answer is: 5 5/5 . 5, which seems pretty poor to me. c= Jackson, no. 17. Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 10:5 (Sep 1903) 426-427. "How to arrange four 9's to make 100." Ball. MRE, 4th ed., 1905. P. 14. Repeats the material in the 3rd ed of 1896, again omitting two points, and adds further questions. Use the 10 digits to total 1 -- a solution is 35/70 + 148/296 -- or to total 100 -- a solution is 50 + 49 + 1/2 + 38/76. Use the 9 digits to make four numbers which total 100 -- a solution is 78 + 15 + 29 + 364. A. Cunningham & T. Wiggins. ?? Math. Quest. Educ. Times 7 (1905) 43 46. ??NYS -- cited in Dickson I 460, item 45d. Expressions using four 9s and four 4s. Pearson. 1907.
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