## Part I, no. 43: The nimble nines, pp. 125 & 187. Verse asking for three 9s to make 16 -- solution is 96/6 !! ## Part II: On all fours, p. 107. Four fours in general, with a few examples.
Wehman. New Book of 200 Puzzles. 1908. P. 26. = Magician's Own Book, no. 4.
Ball. MRE, 5th ed., 1911. Pp. 13-14. Briefly restates the material in the 4th ed. as "questions which have been propounded in recent years. ... To the making of such questions of this kind there is no limit, but their solution involves little or no mathematical skill."
"Another traditional and easy recreation .... I have never seen this recreation in print, but it seems to be an old and well-known question." Deals just with four 4s and says one can get up through 170. G. N. Watson has pointed out that one can get further by using factorials and subfactorials. (The subfactorial of n is n¡ = n![1/0! 1/1! + 1/2! - 1/3! + ... ± 1/n!].) The topic is not in earlier editions.
W. W. Rouse Ball. Four fours. Some arithmetical puzzles. MG 6 (No. 98) (May 1912) 289 290. "An arithmetical amusement, said to have been first propounded in 1881, ...." [This would seem to refer to Knowledge, above.] Studies various forms of the problem. Says it occurs in his MRE -- see above. MRE 6th ed., 1914, p. 14, cites this article.
Ball. MRE, 6th ed., 1914. Pp. 13-14. He now splits the material into three sections.
Empirical Problems. Restates the material in the 5th ed. as "... numerous empirical problems, ..." and omits Loyd's problem. "To the making of such questions there is no limit, but their solution involves little or no mathematical skill."
He then introduces the "Four Digits Problem". "I suggest the following problem as being more interesting." Using the digits 1, 2, ..., n, express the integers from 1 up using four different digits and the operations of sum, product, positive integral power and base-10 notation (or also allowing iterated square roots and factorials). With n = 4, he can get to 88 or to 264. With n = 5, he can get to 231 or 790. Using 0, 1, 2, 3, he can get to 36 (or 40).
Under Four Fours Problem, he discusses what operations are permitted and says he can get to 112, or to 877 if subfactorials are permitted (citing his MG article for this). Mentions four 9s and four 3s problems.
Williams. Home Entertainments. 1914. The six 9's, p. 119. "Express the number 100 by means of six 9's."
Thomas Rayner Dawson. 1916. ??NYS. Cited in: G&PJ 3 (Jan 1988) 45 & 4 (Mar 1988) 61. Asks for four R's, where R is indeterminate, e.g. 3 = (R+R+R)/R.
Ball. MRE, 7th ed., 1917. Pp. 13-14. The material of the first two sections is repeated, but under "Four Fours Problem", he discusses the operations in more detail. With +, -, x, , brackets and base-10 notation, he can get to 22. Allowing also finitely iterated square roots, he can get to 30. Allowing also factorials, he can get to 112. Allowing also integral indices expressible by 4s and infinitely iterated square roots, he can get to 156. Allowing also subfactorials, he can get to 877. (In the 11th ed., 1939, pp. 15-16, two footnotes are added giving expressions for 22 in the first case and 99 in the third case.) Gives some results for four 2s, four 3s, four 5s, four 9s. Mentions the general problem of n ds.
Smith. Number Stories. 1919. Pp. 112 113 & 140 141. Use four 9s to make 19, 2 and 20.
Ball. MRE, 9th ed., 1920. Pp. 13-14. In the "Four Digits Problem", he considers n = 4, i.e. using 1, 2, 3, 4, and discusses the operations in more detail. Using sum, product, positive integral power and base-10 notation, he can get to 88. Allowing also finitely iterated square roots and factorials, he can get to 264. Allowing also negative integral indices, he can get to 276. Allowing also fractional indices, he can get to 312. He then mentions using 0, 1, 2, 3 or four of the five digits 1, ..., 5.
Under "Four Fours Problem", he repeats the material of the 7th ed., but adds some extra results so he has results for four ds, d = 1, 2, 3, 5, 6, 7, 8, 9.
Ball. MRE, 10th ed., 1922. Pp. 13-14. In the "Four Digits Problem", he repeats the material of the 9th ed., but at the end he adds that using all of the five digits, 1, ..., 5, he has gotten to 3832 or 4282, depending on whether negative and fractional indices are excluded or allowed.
Hummerston. Fun, Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 2, pp. 164 & 183. "Express 100 by using the same figure six times."
Dudeney. MP. 1926. Prob. 58: The two fours, pp. 23 24 & 114. = 536, prob. 109, pp. 34 & 248 249, with extensive comments by Gardner.
King. Best 100. 1927.
## No. 44, pp. 20-21 & 48. = Foulsham's no. 14, pp. 8 & 11. "Can you put down four fifteens so that they come to 16,665? ## No. 45, p. 21 & 49. "Arrange the figures 1 to 7 so that they will amount to 100, when added together." Arrange four 9s to make 100. Gives two answers for the first part.
Heinrich Voggenreiter. Deutsches Spielbuch Sechster Teil: Heimspiele. Ludwig Voggenreiter, Potsdam, 1930. P. 112. Write 1000 with seven or five equal digits.
Perelman. FFF. 1934. 1957: probs. 98, 100 & 102, pp. 137 & 143-144; 1979: probs. 101, 103 & 105, pp. 166-167 & 174-175. = MCBF, probs. 101, 103 & 105, pp. 167 & 176 178.
## 101: Two digits: "What is the smallest integer that can be written with two digits?" 1/1 = 2/2 = .... [Though I think 0/1 might be counted.] ## 103: Five 9's: "Write 10 with five 9's. Do it in at least two ways." ## 105: Four ways: "Show four different ways of writing 100 with five identical digits."
Perelman. MCBF. 1937. Any number via three twos. Prob. 202, pp. 398-399. "A witty algebraic brain-teaser that amused the participants of a congress of physicists in Odessa." n = log_{2} log_{2} ^{n}2, where ^{n} means n fold iterated square root.
Haldeman-Julius. 1937. No. 16: Adding fives, pp. 5 & 21. Use four 5s to make 6½. Answer is: 5 + 5/5 + .5.
M. Adams. Puzzle Book. 1939. Prob. B.83: Figure juggling, part 3, pp. 78 & 107. Use a digit 8 times to make 1000. Answer uses 8s.
Evelyn August. The Black-Out Book. Op. cit. in 5.X.1. 1939. Number, please!, pp. 20 & 210.
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