7. arithmetic & number theoretic recreations a. Fibonacci numbers
Pp. 153-154, no. 578. Exchange fewer than 120 coins worth 1½ for coins worth 5⅔ with a value of 14 left over. I.e. 1½ x = 5⅔ y + 14. Finds two answers
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- Pp. 165-166, no. 631. 30 for 105 at 5, 3, 2 (unspecified goods). (8, 7) answers -- he gives the positive ones.
- P. 132, ex. 4. 55 crowns (= 5s) and shillings make £7 3s.
- No. 16. 40 for 40 at 5, 1, ¼. (Calves, pigs, geese.) (3, 2) solutions, (2, 2) given.
- Examples XLVI, no. 13, pp. 393 596. Same as Hall, 1846, pp. 232-233. Says there are (5, 4) answers.
- Examples XLVI, no. 34, pp. 395 597. 100 for 100 at 5, 1, 1/20. (Oxen, sheep, ducks.) = Alcuin 39.
- Easy come, easy go, pp. 25 91. 100 for 1000 at 100, 30, 5. (2, 1) solutions -- he gives (1, 1).
Pp. 153-154, no. 578. Exchange fewer than 120 coins worth 1½ for coins worth 5⅔ with a value of 14 left over. I.e. 1½ x = 5⅔ y + 14. Finds two answers.P. 163 & 261, no. 616. Change 50 into coins worth 3/8 and 2/9. (9, 8) answers -- he gives the positive ones. No. 617 also deals with coins.Pp. 165-166, no. 631. 30 for 105 at 5, 3, 2 (unspecified goods). (8, 7) answers -- he gives the positive ones.PP. 167-168, no. 634. 50 for 395 at 5, 7, 12 (numbers). (6, 5) answers -- he gives the positive ones.Thomas Grainger Hall. The Elements of Algebra: Chiefly Intended for Schools, and the Junior Classes in Colleges. Second Edition: Altered and Enlarged. John W. Parker, London, 1846. P. 132, ex. 4. 55 crowns (= 5s) and shillings make £7 3s.P. 133, ex. 18. 25 coins worth 1/30 and 1/15 to make 1.Pp. 232-233, ex. 4. How many ways can one pay £100 with sovereigns (= £1) and half-guineas (= 21/40 £)? He gives all (5, 4) answers.P. 233, ex. 5. = Simpson 4.John William Colenso (1814-1883). The Elements of Algebra Designed for the Use of Schools. Part I. Longman, Brown, Green, Longmans & Roberts, London, (1849), 13th ed., 1858 [Advertisement dated 1849]. Exercises 63, p. 114 & Answers, p. 14. No. 11. Change a pound into 18 coins comprising half crowns (= 2½ s), shillings (= 1s) and six pences (= ½ s). I.e. 20 for 18 at 5/2, 1, ½. (4, 4) solutions, all given.No. 16. 40 for 40 at 5, 1, ¼. (Calves, pigs, geese.) (3, 2) solutions, (2, 2) given.Family Friend (Dec 1858) 357. Arithmetical puzzles -- 3. Same as Alcuin's 39 with (oxen, sheep, geese). I haven't got the answer. Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 564-44, pp. 256 & 397: Arithmetisches Rätsel. 100 for 100 at 10, 3, ½. (Geese, hares, partridges.) (2, 1) answers, only (1, 1) given. Todhunter. Algebra, 5th ed. 1870. Many straightforward examples, of which a I give a few.
Examples XLVI, no. 17, pp. 393 & 596. Pay 10s 6d (=10½s) with guineas (= 21s) to one who only has half-crowns (= 2½s). Gives least solution.Examples XLVI, no. 18, pp. 393 & 596. Pay 44s with sovereigns (= 20s) to one who only has francs (= 4/5 s). Gives least solution.Examples LV, no. 6, pp. 503 & 601. Pay £24 15s using shillings and francs, where 26 francs equals 21 shillings. (24, 23) solutions. His solution is: x = 26t; y = 495 21t.Examples XLVI, no. 28, pp. 395 & 597. Pay £4 16s with 16 coins, using guineas (= 21s), crowns (= 5s) and shillings. I. e. 16 for 96 at 21, 5, 1. (4, 2) answers which he describes.Examples XLVI, no. 34, pp. 395 & 597. 100 for 100 at 5, 1, 1/20. (Oxen, sheep, ducks.) = Alcuin 39.Mittenzwey. 1880. Prob. 111, pp. 22 & 74; 1895?: 129, pp. 27 & 77; 1917: 129, pp. 24-25 & 75. 14 X + 9 Y = 391. He gives one of the (3, 3) answers.Prob. 121, pp. 25 & 76; 1895?: 139, pp. 29 & 79; 1917: 139, pp. 26 & 77. 100 for 100 at 1/2, 3, 10, (Hares, does, stags.) = Leske. (2, 1) answers, only (1, 1) given.1895?: 147, pp. 30 & 79-80; 1917: 147, pp. 28 & 77. Reduces to 10 for 40 at 10, 5, 1. (Payment for grades.) (1, 1) solution, which he gives.Editorial answer to F. Chapman, a correspondent. Knowledge 2 (17 Nov 1882) 409. 100 for 100 at 10, 3, ½. (Bullocks, sheep, pigs.) (2, 1) answers, only (1, 1) given. c= Leske. Hoffmann. 1893. Chap. IV, no. 35: Well laid out, pp. 152 & 200 201 = Hoffmann-Hordern, pp. 125-126. 21 for 24 at 2, 3, ½, 4. (6, 2) solutions -- he gives (2, 2). Clark. Mental Nuts. 1904, no. 49; 1916, no. 63. Turkeys and sheep. "A drove of turkeys and sheep have 100 heads and feet. How many are there of each?" This leads to 3T + 5S = 100, which has (7, 6) answers. He says there are 6 answers and gives one example. His 1897, no. 12; 1904, no. 20; 1916, no. 93 and 1904, no. 67 are ordinary versions, with only one positive answer in each case. Pearson. 1907. Part II, no. 155, pp. 144 & 222. = Alcuin 39. Only the positive answer is given. Loyd. Cyclopedia. 1914. Sam Loyd's candy puzzle, pp. 121 & 354. = SLAHP: Assorted postcards, pp. 45 & 101. = Pacioli 18. Only the positive solution is given. Williams. Home Entertainments. 1914. The menagerie, p. 127. Menagerie of birds and beasts has 36 heads and 100 feet. I.e. 36 for 100 at 2, 4. Clark. Mental Nuts.
1916, no. 5. How old is dad. Dad, Ma, Bro & I. D + M + B + I = 83; 6D = 7M; M = 3I. This yields 15I + 2B = 166, This has (6, 5) solutions, but only one is possible.1916, no. 13. Change a quarter. How many ways can you change a quarter (= 25¢) using 1¢, 5¢, 10¢ coins? (12, 2) solutions; he says 12.Hummerston. Fun, Mirth & Mystery. 1924. The cinema puzzle, Puzzle no. 36, pp. 92 & 177. How many ways to change 6d using 6d, 3d, 1d, ½d, ¼d? (67,0) solutions -- he says 67. Collins. Fun with Figures. 1928. Four out of five have it, p. 183. Hunter has shot birds and rabbits and has 36 heads and 100 feet. I.e. 36 for 100 at 2, 4. Loyd Jr. SLAHP. 1928. He gives a number of examples, usually with some extra feature.
Poultry profits, pp. 24 & 91. 100 for 100 at .62, 1.02, 1.34. If he makes profits of .12, .22, .25 on each one, how does he maximize his profit? (11, 10) solutions -- he says 9, 86, 5 is maximal, but 5, 95, 0 is better.Easy come, easy go, pp. 25 & 91. 100 for 1000 at 100, 30, 5. (2, 1) solutions -- he gives (1, 1).Shooting mathematically, pp. 38 & 97. Find least number a + b + c such that 6a + 12b + 30c = 17 (a + b + c).A puzzle in pants, pp. 42 & 99. 147 for 147 at .49, .98, 2.45, but he wants to maximize the minimum of the three numbers. (37, 36) solutions in general.An observing waiter, pp. 66 & 111. This has men, women and couples, leading to a + b + 2c = 20, .2a + .3b + 3c = 20, which has (1, 1) solutions which he gives.Perelman. FFF. 1934. Hundred rubles for five. 1957: prob. 37, pp. 54 & 57-58; 1979: prob. 40, pp. 69 70 & 72 73. = MCBF: prob. 40, pp. 67 & 70-71. Magician asks for 20 for 500 or 300 or 200 at 50, 20, 5. These are all impossible! M. Adams. Puzzle Book. 1939. He has several straightforward problems, which I omit, and the following. Prob. C.10, pp. 125 & 173. Use 26d in florins (= 24d), shillings (= 12d), sixpence (= 6d), pennies (= 1d) and half pences (= ½ d) to measure 5⅝ inches. The widths of the coins, in 16ths of an inch, are: 18, 15, 12, 19, 16, respectively. This leads to: 24a + 12b + 6c + d + e/2 = 26, 18a + 15b + 12c + 19d + 16e = 90. (1, 0) solution, which is given. Depew. Cokesbury Game Book. 1939.
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