7. arithmetic & number theoretic recreations a. Fibonacci numbers

Nos. 54-57: Ten is the number, pp. 10 & 35. Express 10 using five 9s, in four different ways.

Nos. 58-59: 3 + 3 + 3 = 30, pp. 10 & 35. Express 30 using three 3s, in two different ways.

No. 97: Eight to one thousand, pp. 13 & 37. Use a digit eight times to express 1000.

P. Grammer, I. McFiggans, N. Blacknell, T. Joyce, J. Anstey & A. Devonald. Counting in fours. MiS 9:4 (Sep 1980) 21-22. Uses four 4s to express 1, ..., 50. Says 51 - 100 will appear in next issue, but they didn't.

J. Bellhouse. Four fours. MiS 14:1 (Jan 1985) 15. Says the promised table for 51 - 100 (see Sep 1980 above) had not appeared, so his students found their own.

Anne Williamson. 1985. MiS 14:4 (Sep 1985) 7. Use the four digits 1, 9, 8, 5 to express integers 1 - 100. Unhappy with expressions for 24, 31, 65 which use !.

Ken Lister. Letter. MiS 15:2 (Mar 1986) 47. Responding to Bellhouse (Jan 1985). Corrects and improves some values, but says 71 and 73 have not been done. Expresses a/b, for single digits a, b, by use of four 4s.

Angie Aurora. Letter. MiS 15:3 (May 1986) 48. Improvements for Williamson's problem -- Sep 1985 above.

Joyce Harris. Letter: Four fours. MiS 15:3 (May 1986) 48. Responding to Lister (Mar 1986), gives expressions for 71 and 73.

Bob Wasyliw. Letter: Four 4's -- the ultimate solution. MiS 15:5 (1986) 39. Adapts Everett's 1971 method to include non positive integers.

Simon Gray & Colin Abell. Letters: Four fours again. MiS 16:2 (1987) 47. Gray notes that 4 = 4 * 4, so that 'four 4s' is the same as 'at most four 4s'. He gives π  =  4 * sin ¹(4/4) and more complex forms. Abell gives π =   ( 4/4) * log ( 4/4) [The first minus sign is ambiguous??] and π =   (4*4) * Tan^-1(4/4) [The minus sign is wrong.]

Tim Sole. The Ticket to Heaven and Other Superior Puzzles. Penguin, 1988.

Number play (ii)   (iv), pp. 15, 29 & 178 180.

(ii). Using +,  , *, /, ., ! and brackets, (, ), one can express 1   48 with four 3s. Allowing also SQR, one can express 1   64 with four 3s.

(iii). Using all 7 symbols in (ii) and brackets, one can express 1   112 with four 4s. Gives solutions with two and three 4s.

(iv). If we also allow Σ(n) [= T_n, the n-th triangle number], he finds solutions with one, two or three 4s. Using log, all integers can be expressed with three 4s.

Three of the best -- (iii), pp. 17 & 32. Some solutions with one 4, using !, , [n] = INT(n) and Σ(n).

Tony Forbes. Fours. M500 116 (Nov 1989) 4 5. Says someone (possibly Marion Stubbs?) gave a simple variation of King Hele's and Everett's formulae to use exactly four 4s to yield n. Forbes suggests using one 4 and the three operations: !,  and INT. He already gets stuck at 12.

Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. Take two, pp. 96 & 139. Use five 2s to express 1, 2, ..., 26, particularly 17 and 26.

P. H. R. Fawlty keys. Mathematical Pie, 129 (Summer 1993) 1023 & Notes, p. 1. Calculator whose only keys are 5, 7, +, -, x,  and =. Make numbers from 0 to 20. Solution notes you can make any number by adding enough terms of the form 5  5 and then gives short solutions for 1 through 20.

Clifford A. Pickover. Phi in four 4's. Theta (Crewe) 7:2 (Autumn 1993) 5-8. In Sep 1991 he asked for good approximations to φ using four 4s, either with as many symbols as you want or with each symbol used at most four times. Says he was inspired by Conway & Guy's paper of 1962. Brian Boutel produced φ = (4 + {4!-4})/4. Pickover then extended the question and various solvers got φ in five 5s, seven 6s, eight 8s, nine  9s and 2k-5 ks.

John Seldon. Fours. M500 136 (Jan 1994) 15 16. Answers Forbes' 1989 problem of expressing 1   100 with one 4 and any number of factorials, x!, and integer square roots, [x].

David Crawford and students. 1999 the end of an era. MiS 28:4 (Sep 1999) 25. Uses 1, 9, 9, 9 to make all integers from 1 to 100. Notes that 2000, 2001, ... are not going to be very useful for such puzzles!

Derek Ball. Four 4s. MTg 173 (Dec 2000) 18. Says his fifth year teacher discovered the following for n in terms of four 4s: n = log_4/4 log₄ ⁿ 4, where ⁿ denotes n-fold repeated square root. Cf Perelman, 1937; King-Hele, 1955; Everett, 1971 -- Everett is very close to this and the others are not quite that close.

7.I.1. LARGEST NUMBER USING FOUR ONES, ETC.
Mittenzwey. 1880. Prob. 142, pp. 30 & 79-80; 1895?: 162, pp. 34 & 82; 1917: 162, pp. 31 & 79-80. Find largest number using four digits. Gets 9^9^9^9 and tries to contemplate its size. 9^9 is given as 387,420,488 (last digit should be 9), so 9^9^9 has 369,693,100 digits.

James Joyce. Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934, apparently printed 1946. P. 684 (Gardner says the 1961 ed. has p. 699). Bloom estimates that 9^(9^9) would occupy "33 closely printed volumes of 1000 pages each", but he erroneously phrases the number as "the 9th power of the 9th power of 9", which is only 9⁸¹.

King. Best 100. 1927. No. 35, pp. 19 & 46. Largest number using two 4s. Gives 4⁴ = 256.

Perelman. FFF. 1934. Four 1's. 1957: prob. 103, pp. 137 & 144-145; 1979: prob. 106, pp. 167 & 175. = MCBF, prob. 106, pp. 167 & 178 = MCBF, prob. 131, p. 217. "What is the biggest number that can be written with four 1's?

Perelman. MCBF. 1937. Probs. 128-132, pp. 214-219. Largest numbers with: Three twos; Three threes; Three fours; Four ones; Four twos.

Sullivan. Unusual. 1947. Prob. 30: Not 999. Largest number that can be written with three integers [sic!]. Answer: 9^(9^9).

G. C. S[hephard, ed.] The problems drive. Eureka 11 (Jan 1949) 10-11 & 30. No. 5. Find the largest numbers expressible using four 2s or four 4s, no symbols allowed. Answers: 2^2^22; 4^4^4^4.

Leroy F. Meyers. An integer construction problem. AMM 66:7 (Aug/Sep 1959) 556-561. This deals with Ball's "Four Digits Problem" (see MRE, 6th ed., 1914 in 7.I) and generalizations. In particular, he shows that if one uses 1, 2, 3, 4, with operations +,  , x and brackets, then one can obtain precisely the following: 1, 2, ..., 28, 30, 32, 36. In general he obtains the largest integer expressible using a given multiset of integers (i.e. one is allowed a fixed number of repeats of a value) using the operations +, x and brackets. He also shows that allowing also -, for both negation and subtraction, does not increase the maximum obtainable value. He conjectures that allowing also , for both reciprocation and division, does not increase the maximum obtainable, but Meyers has written that a student once showed him a counterexample, but he cannot remember it. He applies his general results to show that no other values are obtainable when using 1, 2, 3, 4.

Problematical Recreations 4. Problem 1 and its answer, pp. 3 & 36. (This is one of a series of booklets issued by Litton Industries, Beverly Hills, California, nd [c1963], based on the series of the same name in Aviation Week and Electronic News during 1959-1971. Unfortunately, neither the date nor location nor author is given and the booklet is unpaginated. The answer simply states the maximum value with no argument.) Reproduced with a proper solution in: Angela Dunn; Mathematical Bafflers; (McGraw-Hill, 1964, ??NYS); revised and corrected 2nd ed., Dover, 1980, pp. 119 & 132 and with just the answer in: James F. Hurley; Litton's Problematical Recreations; Van Nostrand Reinhold, NY, 1971, chap. 7, prob. 8, pp. 238 & 329. "What is the largest number which can be obtained as the product of positive integers which add up to 100?" (This type of problem must be much older than this?? Meyers writes that he first encountered such problems as an undergraduate in 1947. If one looks at maximizing the LCM instead of the product of the terms, this is the problem of finding a permutation of 100 letters with maximum order.)

Sol Golomb. Section 13.8 The minimization of the cost of a digital device (the juke-box problem). IN: Ben Nobel; Applications of undergraduate mathematics in engineering. MAA & Macmillan, 1967, pp. 284-286. This considers the problem of Problematical Recreations in the inverse form. We have r k-state devices which allow k^r choices and the cost is proportional to rk. Minimize the ratio of cost to capacity or maximize the ratio of capacity to cost. [Another way to express this is to ask which base is best for a computer to use? I recall this formulation from when I was a student in the early 1960s. The answer is e, but here only integer values are used. One can generalise to: given a value, find the smallest sum of numbers whose product is the given value.] The connection with juke-boxes is that they typically have two rows of 12 buttons and one has to press two buttons to make a selection of one from 144 records. One can do much better with five rows of three buttons, but asking a customer to punch five buttons may be unreasonable, so perhaps three rows of five or six buttons might be best.

The Fortieth William Lowell Putnam Mathematical Competition, 1 Dec 1979. Problem A-1. Reproduced in: Gerald L. Alexanderson, Leonard F. Klosinski & Loren C. Larson; The William Lowell Putnam Mathematical Competition Problems and Solutions: 1965 1984; pp. 33 & 109. "Find positive integers n and a₁, a₂, ..., a_n such that a₁ + a₂ + ... + a_n = 1979 and the product a₁a₂...a_n is as large as possible."

Cliff Pickover & Ken Shirriff. The terrible twos problem. Theta (Crewe) 6:2 (Autumn 1992) 3-7. They study the problem of making numbers using just +, -, x, ^, and 1s (or 1s and 2s). For a given n, what is the least number of digits required? They later permit concatenation, e.g. 11 or 12 is permitted. They report results from various programs and mention some related problems.

Bryan Dye. 1, 2, 3, 4 -- four digits that dwarfed the universe. Micromath 10:3 (Aut 1994) 12 13. Says a version appeared in SA a few years ago and is discussed in: Clifford A. Pickover; Computers and the Imagination; Alan Sutton, 1991. Dye's version is to make the largest number using 1, 2, 3, 4 once and the signs +, -, x, , ( ), . (i.e. decimal point). Exponentiation was not considered a sign and was permitted. Pickover's version allowed only the signs -, ( ), . (i.e. decimal point). The largest value found actually fits Pickover's conditions: .3^-(.2^-{.1^-4}) has 10⁶⁹⁹⁰ digits. The largest number using just exponentiation was 2^(3^41)) with 10¹⁹ digits.
7.J. SALARY PUZZLE
It is better to get a rise of 5 every six months than a rise of 20 every year. The interpretation of the first phrase is somewhat ambiguous -- see Mills (1993). If the salary is S every six months, the usual interpretation of the first phrase is that the half-yearly payments are: S, S + 5, S + 10, S + 15, S + 20, S + 25, ..., while the second phrase gives payments of: S, S, S + 10, S + 10, S + 20, S + 20, ..., and the former gets 5 extra every year.
Ball. MRE, 3rd ed., 1896. pp. 26 27. £20 per year versus £5 every half year. He says this is a question "which I have often propounded in past years." It is not in the 1st ed.

Workman. Op. cit. in 7.H.1. 1902. Section IX (= Chap. XXXI in c1928 ed.), examples CXLV, prob. 16, pp. 425 & 544 (431 & 577 in c1928 ed.). Compares rise of £15 per year every 3 years with £5 every year. This represents a precursor of the puzzle version.

A. Sonnenschein & H. A. Nesbit. The New Science and Art of Arithmetic For the Use of Schools. A. & C. Black, London, 1903. Prob. 12 & 14, pp. 341 & 489. "A youth entered an office at the age of 15 at a salary of £40 a year, with an annual rise of £12. ..." "... What total sum would he have received in 30 years? and what would he have received if the increase had been at the rate of £1 per month?"

Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 10:4 (Aug 1903) 336-337. Not too obvious. Half-yearly rise of £5 versus £20 a year. No solution given.

Susan Cunnington. The Story of Arithmetic. Swan Sonnenschein,, London, 1904. Prob. 35, p. 217. £5 raise each six months versus £10 raise each year.

Dudeney?? Breakfast Table Problems No. 330: Smith's salary. Daily Mail (30 & 31 Jan 1905) both p. 7. Raise of 10 per year versus 2½ every six months.

Pearson. 1907. Part II, no. 87, pp. 132 & 208. As in Ball.

Loyd. Salary puzzle. Cyclopedia, 1914, pp. 312 & 381. = MPSL1, prob. 84, pp. 81 & 150 151. = SLAHP: The stenographer's raise, pp. 60 & 108. Raise of 100 per year versus 25 every half year. Interpreting the raise of 25 as worth only 12.50 in a half year, this option loses, contrary to all other approaches.

Clark. Mental Nuts. 1916, no. 6. The two clerks. $25 rise each six months versus $100 rise each year.

Dudeney. AM. 1917. Prob. 26: The junior clerk's puzzle, pp. 4 & 150. Two clerks getting £50 per year with one getting a raise of £10 per year versus the other getting a raise of £2 10s every six months with a complicated further process of savings at different rates for five years.

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 276, pp. 103 & 171: A problem in salaries. £20 rise every six months versus £80 rise each year.

T. O'Conor Sloane. Rapid Arithmetic. Quick and Special Methods in Arithmetical Calculation Together with a Collection of Puzzles and Curiosities of Numbers. Van Nostrand, 1922. [Combined into: T. O'Conor Sloane, J. E. Thompson & H. E. Licks; Speed and Fun With Figures; Van Nostrand, 1939.] The two clerks, p. 168. Raise of $50 every six months versus $200 each year.

F. & V. Meynell. The Week End Book, op. cit. in 7.E. 1924. Prob. one, p. 274 (2nd ed.), pp. 406 407 (5th? ed.). Raise of 20 per year versus 5 every half year.

Peano. Giochi. 1924. Prob. 16, p. 5. 1000 per year with rise of 20 each year versus 500 each half-year with rise of 5 each half-year.

Wood. Oddities. 1927. Prob. 31: A matter of incomes, p. 31. $1000 per year with $20 per year increase versus "$5 each half year increase".

Collins. Fun with Figures. 1928. Do figures really lie?, pp. 35-36. $50 every six months versus $200 per year.

R. Ripley. Believe It Or Not! Book 2. (Simon & Schuster, 1931); Pocket Books, NY, 1948. P. 123: Figure your raise in pay. A raise of one every day is better than a raise of 35 every week. (Assumes a six day week.)

Phillips. Week End. 1932. Time tests of intelligence, no. 20, pp. 16 & 189 190. Raise of 2000 per year versus 500 half yearly.

Phillips. Brush. 1936. Prob. G.1: The two clerks, pp. 20 & 87. Raise of 200 annually versus 50 half yearly.

McKay. At Home Tonight. 1940. Prob. 4: A choice of rises, pp. 63 & 76. £5 per half-year versus £15 per year. Solution is unclear and seems to be wrong. "£5 each six months is £5 in the first half-year and £10 in the second -- that is, £15 per year. But the man who gets £5 per six months gets £5 in the first year, and of course he keeps this advantage year by year." I get that the first case is ahead by £5n in the n-th year.

Sullivan. Unusual. 1943. Prob. 5: Raising the raise question. Raise of $20 per year versus $5 every half year.

Birtwistle. Calculator Puzzle Book. 1978. Prob. 56: The interview, pp. 39 & 99-100. £3050 yearly plus £100 each year versus £1500 half-yearly plus £50 each half-year.

D. J. Hancox, D. J. Number Puzzles For all The Family. Stanley Thornes, London, 1978. Puzzle 11, pp. 4 & 48.. £1 rise per month every month versus £144 rise per year every year. Says the first gives £66 extra in the first year and £210 extra in the second year, while the second gives no extra in the first years and £144 extra in the second year. He then says: "Hence he would never get the £66 he would have received in the first year." In fact, he loses £66 every year.

Stuart E. Mills. Dollars and sense. CMJ 24 (1993) 446-448. Raise of $1000 per year versus $300 every half year. Discusses various interpretations of the second phrase and gives some recent references.

Comments by myself and various others seem to have appeared (??NYS) as they are included in the collection of these columns: Edward J. Barbeau; Mathematical Fallacies, Flaws, and Flimflam; Spectrum Series, MAA, 2000, pp. 11-14. This gives references to various recent appearances of the problem in 1943, 1983, 1992.

John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all Ages. Keystone Agencies, Radnor, Ohio, 1997. P. 80: The best job offer. $20,000 per year plus a $500 raise every six months versus $20,000 per year plus a raise of $1000 each year. He says the six-monthly payments of the first are 10,000, 10,500, 11,000, 11,500, ... and the payments for the second are 10,000, 10,000, 11,000, 11,000, .... But the second is a raise of $2,000 per year!

7.K. CONGRUENCES
The Friday Night Book (A Jewish Miscellany). Soncino Press, London, 1933. Mathematical Problems in the Talmud: The Divisibility Test, p. 137. Hebrew law requires fields to lie fallow every seventh year, and this is to hold for all fields at once! Rabbi Huna gave the following rule. Write the year as y = 100a + b. Form 2a + b. Then the year is a Sabbatical year if 7 divides 2a + b. No explanation is given in the Talmud, but we clearly have y  2a + b (mod 7). [The Talmud was compiled in the period -300 to 500. This source says Rabbi Huna is one of the few mathematicians mentioned in the Talmud, but gives no dates and he is not mentioned in the EB. From the text of another problem attributed to him (cf in Section 6.AD), the problem would seem to be sometime in the 1-5 C.]
7.K.1. CASTING OUT NINES
See Smith, History II 151 154 for a detailed discussion. He says it appears in al Khowarizmi and al Karkhi and that it is generally assumed to come from India, but his earliest Indian source is Lilavati, 1150. G. R. Kaye; References to Indian mathematics in certain Mediæval works; J. Asiatic Society of Bengal (NS) 7:11 (Dec 1911) 801-816 notes the appearances in al-Khwârizmî, Avicenna and Maximus Planudes [Arithmetic after the Indian method; c1300; op. cit. in 7.E.1] but asserts it does not occur in early Indian sources -- but cf Aryabhata II, 950.

Dickson I, chap. XII, pp. 337 346, especially p. 337, gives a concise history. He says al Karkhi was the first to use a (mod 11) check.

See Tropfke, pp. 165-167.

I have recently realised that certain puzzle problems should be listed here, but so far I have only noted Boy's Own Book, Boy's Own Book (Paris), Carroll, Peano, Parlour Games for Everyone -- there must be many more 19C and even 18C examples. Basically these involve getting someone to produce a number divisible by nine and asking him to delete one digit and tell you the others -- you tell him the missing digit. There are many of these and I probably won't try to record all of them, but subtracting a number from its reversal may be the forerunner of the 1089 puzzle of 7.AR.

St. Hippolytus. Κατα Πασωv Αιρεσεωv Ελεγχoσ (??= Philosphumena) (= Refutatio Omnium Haeresium = Refutation of all Heresies). c200. Part iv, c. 14. ??NYS. Discusses adding up digits corresponding to letters (mod 9) and mentions considering it (mod 7). (HGM I 115 117) See also: Smith II 152, Dickson I 337 and Saidan (below), p. 472. Hippolytus doesn't use the method to check any arithmetic. (St. Hippolytus may be the only antipope to be counted a saint! A reference says he was Bishop of Portus and the MS was discovered at Mt. Athos in 1842.)

Iamblichus. On Nicomachus's Introduction to Arithmetic. c325. ??NYS. In: SIHGM I 108 109. Special case. (See also HGM I 114 117.)

Muhammad [the h should have an underdot] ibn Mûsâ al Khwârizmî. c820. Untitled Latin MS of 13C known as Algorismus or Arithmetic, Cambridge Univ. Lib. MS Ii.6.5. Facsimile ed., with transcription and commentary by Kurt Vogel as: Mohammed ibn Musa Alchwarizmi's Algorismus, Das früheste Lehrbuch zum Rechnen mit indischen Ziffern; Otto Zeller Verlagsbuchhandlungen, Aalen, 1963. English translation by John N. Crossley & Alan S. Henry as: Thus spake al Khwārizmī: A translation of the text of Cambridge University Library Ms.Ii.vi.5; HM 17 (1990) 103 131. [Crossley & Henry name this author Abū Ja‘far Muhammad [the h should have an underdot] ibn Mūsā al Khwārizmī, but I have seen no other authority giving Abū Ja‘far -- several give Abū ‘Abdallāh and Rosen's translation The Algebra of Mohammed ben Musa specifically says our author "must therefore be distinguished from Abu Jafar Mohammed ben Musa, likewise a mathematician and astronomer, who flourished under the Caliph Al Motaded" (c900). F. 108r = Vogel p. 25 = Crossley & Henry p. 117. Describes casting out 9s in doubling and in multiplication.

Aryabhata II. Mahâ siddhânta. 950. Edited by M. S. Dvivedi, Braj Bhushan Das & Co., Benares, 1910. English Introduction, pp. 21 23; Sanskrit text, p. 245. Casting out 9s for multiplication, division, squaring, cubing, and taking square and cube roots. (Datta & Singh I 181 give the text in English.)

Abû al-Hassan [the H should have an underdot] Ahmad[the h should have an underdot] Ibn Ibrâhîm al-Uqlîdisî. Kitâb al Fuşûl [NOTE: ş denotes an s with an underdot.] fî al Hisâb[the H should have an underdot] al Hindî. 952/953. MS 802, Yeni Cami, Istanbul. Translated and annotated by A. S. Saidan as: The Arithmetic of Al Uqlīdisī; Reidel, 1978. Book II, chap. 13, pp. 153 155 and Book III, chap. 7 8, pp. 195 201 deal with checking by casting out 9s, which is given only briefly, apparently being well known. He applies it to division and square roots. The method is also mentioned in Book II, chap. 2. On pp. 468 472, Saidan discusses the appearance of various rules in early texts. His earliest Indian example is Lilavati, 1150, but he gives no reference.

Kūshyār ibn Labbān = Abū ăl-Hasan [the H should have an underdot] Kūšyār ibn Labbān ibn Bāšahri al-Ğīlī. Kitāb fī usūl Hisāb al-Hind [Principles of Hindu Reckoning]. c1000. Facsimile with translation by Martin Levey & Marvin Petruck. Univ. of Wisconsin Press, Madison, 1965.