7. arithmetic & number theoretic recreations a. Fibonacci numbers

P. 9. Monetary version getting £12 18s 11d. Cites Educational Times Reprints, giving the same examples, but specifically says the amount must be less than £12 and the number of pounds cannot be the same as the number of pence. He says to take the difference of the number and its reversal. At the end, he states: "The rule can be generalized to cover any system of monetary units", but he doesn't give the general solution. This version is dropped in the 12th ed., 1974.

P. 13. Notes that a number minus its reversal is divisible by 9.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. X: A subtraction sum, p. 10. A 3-digit number minus its reversal has the form a9c, where a + c = 9.

Somerville Gibney. An arithmetical flourish for drawing-room shows. The Boy's Own Paper 15 (No. 734) (4 Feb 1893) 299. Monetary version, getting £12 18s 11d. "An arithmetical curiosity which is quite new ...."

Somerville Gibney. Re The arithmetical flourish for drawing-room shows. The Boy's Own Paper 15 (No. 750) (27 May 1893) 556. Explains why it works. Correspondents from various places do it: in francs and centimes, getting 99 99; in thalers and grosschen (30  grosschen = a thaler), getting 29 29; and in avoirdupois, getting 28cwt 2qr 27lb.

Ball FitzPatrick. 1st ed., 1898, pp. 14 15. Shows the property of 1089. A remark of the translator shows the answer in base b is (b 1) * (b+1)², which is

b b² + (b-2) b + (b-1).

Carroll-Collingwood. 1899. P. 269 (Collins 194). = John Fisher; The Magic of Lewis Carroll; op. cit. in 1; pp. 244-245. = Carroll-Wakeling II, prob. 25: Pounds, shillings and pence, pp. 41 & 71, Monetary version, with number of pounds 'not more than twelve', giving £12 18s 11d. Carroll assumes the reversal is less than the original number and says it works "whatever numbers may have been selected." Mentioned in Carroll-Gardner, p. 76. Neither Collingwood nor Wakeling nor Gardner note the difficulty when numbers of pounds and pence are equal nor the question of whether the reversal is actually smaller than the original, but Fisher does, and also mentions 1089. Collingwood says he believes this was invented by Carroll and this seems reasonable, but see Orchard et al, 1890, above.

Clark. Mental Nuts. 1904, no. 81; 1916, no. 54. Mental telegraphy. As in Berkeley & Rowland.

Ball. MRE, 4th ed., 1905, p. 9. Adds a section on 1089 just before the monetary version.

Laisant. Op. cit. in 6.P.1. 1906. Chap. 18: Opérations curieuses: No. 1, p. 43. 1089. Notes that the two end digits must be different.

E. N. Barisien. ?? Suppl. al Periodico di Mat. 13 (1909) 20-21. ??NYS -- cited by Dickson I 462, item 53, but the interesting material is cited in the following.

E. Nannei. ?? Suppl. al Periodico di Mat. 14 (1910/11) 17-20. ??NYS -- cited by Dickson I 462, item 55, as treating a problem of Barisen. Takes a 6-digit number, subtracts its reversal and then adds the reversal of that to itself. Result is one of 13 values: 0,  9900,  ..., 1099989.

M. Adams. Indoor Games. 1912. Magic money, 354-355. Monetary version getting £12 18s 11d.

Clark. Mental Nuts. 1916, no. 53. Foretelling the answer. Gives the rule and an example, saying the answer will always be 1089, though he doesn't make any restriction on the original number.

Dudeney. AM. 1917. Prob. 33: A puzzle in reversals, pp. 5 & 151. Says most people know that the money process leads to £12 18s 11d -- provided you start with less than 12 in the £ place and the number of pounds exceeds the number of pence. If the number of pounds can be 12 or more, then what is the smallest sum for which the process fails and the largest sum for which it works?

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.] A mystery in money, pp. 178-180. Uses US money to get $10.89 and also gives the English result £12 18s 11d.

Peano. Giochi. 1924.

Prob. 48, p. 12. 1089.

Prob. 49, pp. 12-13. As in Berkeley & Rowland.

Hummerston. Fun, Mirth & Mystery. 1924.

An overheard conversation, p. 40. After getting 1089, he asks for it to be reversed and to multiply these, then do some other operations to get an 8-digit number when is translated to a message by 1 = A, 2 = B, ....

Two parlour tricks (second part), p. 146. Tell someone to take a number of three distinct digits, reverse it and subtract the smaller from the larger. Divide the result by nine and have him tell you one of the digits. You divine the other, which is in fact the same! This is based on the easily verified result that when C < A, ABC   CBA  =  99(A-C)

A. B. Nordmann. One Hundred More Parlour Tricks and Problems. Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 60: Mental arithmetic trick, pp. 51-52. Monetary version.

Kraitchik. Math. des Jeux. Op. cit. in 4.A.2. 1930. Chap. IV, no. 7, p. 55. Notes that the first result is a multiple of 99, but doesn't find 1089. Not in his Math. Recreations.

Perelman. 1934?? Guessing a number without asking anything. FFF (1934). Not in 1957 ed. 1979: prob. 14, pp. 33-35. = MCBF (1937), prob. 14, pp. 32-33. Basic property of 1089.

Rohrbough. Brain Resters and Testers. c1935. Brain Teaser, p. 15. Assumes digits are in descending order.

Freda Holmdahl, ed. The Rainy Day Book. Nelson, Edinburgh, 1936. Figure tricks, pp. 196 197. Monetary version.

Depew. Cokesbury Game Book. 1939. Descending numbers, p. 205. 1089.

"Willane". Willane's Wizardry. Academy of Recorded Crafts, Arts and Sciences, Croydon, 1947. Curious figures, p. 45. Version with English money, with £ less than 12.

Anonymous. Problems drive. Eureka 18 (Oct 1955) 15-17 & 21. No. 9. Usual 1089 problem.

P. M. Seeviour & M. Keates. Note 3077: Fibonacci again. MG 48 (No. 363) (Feb 1964) 78 79. Cites Kraitchik, Math. des Jeux. Determines all the possible results of the process for n digit numbers in base b, as being (b²   1) times certain n-digit numbers formed of 0s and 1s, corresponding to the borrows in the subtraction. They find the number t(n) of such values is given by t(2n)  =  t(2n+1)  =  F_2n+1 where F_m are the Fibonacci numbers. They do this via the case when the first and last digits are distinct for which it is stated that a certain recurrence holds, but this is not immediate -- I spent about half a day before I could see it and it would still require some work to make a careful proof. For convenience, I list the first solutions for the case of different end digits.

n = 2. 99.

n = 3. 1089.

n = 4. 10989, 9999, 10890.

n = 5. 109989, 99099, 109890.

n = 6. 1098900, 1099989, 1090089, 1099890, 1089990, 999999, 990099, 991089.

n = 7. 10998900, 10999989, 10891089, 10999890, 10890990, 10008999, 9900099, 9901089.

C. Dudley Langford. Note 3102: A party puzzle extended (Cf Note 3077). MG 48 (No. 366) (Dec 1964) 432. If a + b = 10, then 1089*a and 1089*b are reversals. 1089 can be replaced by 99 * 11...1 and he asserts that these, up to trailing 0s, are the numbers obtained in Note 3077. However, his numbers are: 99, 1089, 10989, 109989, 1099989, ..., 109...989, ..., and these do not immediately yield the above solutions.

Jerome S. Meyer. Arithmetricks. Scholastic Book Services, NY, 1965. The 2178 trick, pp. 3-4. Gets 1089, but he doubles the result at the end to produce 2178.

Michael Holt. What is the New Maths? Anthony Blond, London, 1967. Pp. 88-89. Gives a four digit version, saying to choose the least significant two digits as < 5 and the most significant as  5, then the result is always 10890.

Walter Gibson. Big Book of Magic for All Ages. Kaye & Ward, Kingswood, Surrey, 1982. Magic Math & Super Math, pp. 108-109. After doing 1089, he considers doing the same process with five digit numbers. For convenience, he assumes the digits are all distinct. Then only two possible answers can arise: 99099 or 109890. No proof given.

Paul Swinford. The Wondrous World of Numberplay & Wordplay. A Lecture by Paul Swinford. Published by the author, 1999. Pp. 1-4. Covers 1089 with a variety of ways to use the result. On pp. 3-4, he considers four digit versions and is unhappy to often get 10890. He modifies the process so that one only exchanges the end digits and finds that this always leads to 10989. (For a k-digit number, the process will lead to 1099..989, with k-3 9s in the middle.)

7.AS. CIGARETTE BUTTS
New section. There must be older examples. The usual form uses b butts to make a cigarette and the person has found B = b² butts which give him b+1 smokes. The Scotts are the only ones to realise that B can more generally be such that b-1 divides B-1, giving (B 1)/(b-1) smokes. I have just found the general result that the number of smokes is (B 1)/(b-1). Taking b = 2 gives the result that a knockout tennis tournament with B players has B-1 matches.

On 15 Nov 2001, Willy Moser mentioned that if the tramp borrows a butt, he gets B/(b-1) smokes with a butt left over to return to his friend! He thought this was well known, but I'd never heard or seen it before

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 192, pp. 74 & 158: Scrooge the miser. Starts with 125 cigarettes, but saves all his butts, using 5 to make a cigarette, so he gets 31 further cigarettes.

Rudin. 1936. No. 49, pp. 16 & 85. 36 butts, using 6 to make a cigarette.

J. R. Evans. The Junior Week End Book. Op. cit. in 6.AF. 1939. Prob. 47, pp. 266 & 272. Tramp has 49 butts and uses seven to make a cigarette. How many cigarettes can he make?

Harold Hart. The World's Best Puzzles. Home Recreation Library, NY, 1943. Smoke gets in your eyes, pp. 22 & 54. 125 butts and five make a cigarette.

Leopold. At Ease! 1943. Iffs and Butts, pp. 10-11 & 196. 36 butts, six butts to a cigarette.

Sullivan. Unusual. 1943. Prob. 9: A bum cigarette. As in Rudin.

J. S. R. Cameron. A tennis problem. Eureka 10 (Mar 1948) 18 & 11 (Jan 1949) 31. Complicated system of byes in a tennis tournament with 100 players. Solution notes that the system of byes is irrelevant! This must have been known much earlier.

Jonathan Always. Puzzles to Puzzle You. Op. cit. in 5.K.2. 1965. No. 131: Cigarettes this way, pp. 41 & 89. As in Evans.

[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers. 1973. Op. cit. in 5.E. Hard times made easier, pp. 35-36. 41 butts, six butts to a cigarette.

David Singmaster. Cigarette butts. MS 31:2 (1998/9) 40. The number of smokes is (B 1)/(b-1). Taking b = 2 gives the result that a knockout tennis tournament with B players has B-1 matches.

7.AT. BOOKWORM'S DISTANCE
New section. This must appear in the 19C??
Loyd. Cyclopedia. 1914. Pp. 327 & 383. = SLAHP, Timing a bookworm, pp. 69 & 112. 2  volumes.

Dudeney. AM. 1917. Prob. 420: The industrious bookworm, pp. 143-144 & 248-249. 3  volumes.

Peano. Giochi. 1924. Prob. 10, p. 4. 3 volumes.

King. Best 100. 1927. No. 3, pp. 8 & 39. = Foulsham's, no. 1, pp. 5 & 10. 3 volumes.

William P. Keasby. The Big Trick and Puzzle Book. Whitman Publishing, Racine, Wisconsin, 1929. Pp. 37 & 63.

Streeter & Hoehn. Op. cit. in 7.AE. Vol. 1, 1932, p. 300, no. 10: "Brain twister".

Haldeman-Julius. 1937. No. 33: The bookworm problem, pp. 6 & 22. Three volumes.

Evelyn August. The Black-Out Book. Op. cit. in 5.X.1. 1939. Bookworm's pilgrimage, pp. 163 & 215. 4 volumes.

McKay. At Home Tonight. 1940. Prob. 18: The bookworm, pp. 66 & 79-80. 2 volumes.

The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. P. 148, prob. 4. 2  volumes.

Leopold. At Ease! 1943. Way of a worm, pp. 9 & 195.

Owen Grant. Popular Party Games. Universal, London, nd [1940s?]. Prob. 10, pp. 37 & 50. 3 volumes.

Birtwistle. Math. Puzzles & Perplexities. 1971. Second, pp. 145 & 196. 4 volumes.
7.AU. NUMBER OF CUTS TO MAKE N PIECES
New section. A typical problem is that it takes twice as long to make three pieces as it does to make two pieces. This will include other situations where one has to change from n to n-1.

If one can overlay the material, then one can reduce the number of cuts. If one can fold the material, then one can reduce to one cut as in 6.BG.

I've just added the problem of cutting a cube into 27 pieces here -- this must be much older??

See also 7.AV.

Alcuin. 9C. Prob. 15: De homine. "How many furrows has a man made ..., when he has made three turnings at each end of the field?" The Alcuin text gives six, but Bede, Folkerts and Folkerts & Gericke give seven.

Child. Girl's Own Book. Arithmetical puzzles. 1832: No. 10, pp. 171 & 179; 1833: No. 10, pp. 185 & 193 (answer numbered 8); 1839: No. 10, pp. 156 & 173; 1842: No. 10, pp. 284 & 291; 1876: No. 8, pp. 232 & 244. "If you cut thirty yards of cloth into one-yard pieces, and cut one yard every day, how long will it take you?" The 1839, 1842 & 1876 texts omit the hyphen in one-yard. I didn't copy the exact text from the 1832.

= Fireside Amusements, 1850, Prob. 8, pp. 132 & 184.

The Sociable. 1858. Prob. 46: A dozen quibbles: part 9, pp. 300 & 318. "If you cut thirty yards of cloth into one yard pieces, and cut one yard every day, how long will it take?" = Book of 500 Puzzles, 1859, prob. 46: part 9, pp. 18 & 36. c= Depew; Cokesbury Game Book; 1939; Cutting up, p. 216. c= Magician's Own Book (UK version), 1871, Paradoxes [no. 5], p. 38. = Wehman, New Book of 200 Puzzles, 1908, p. 50.

Hoffmann. 1893. Chap. IX, no. 24: The draper's puzzle, pp. 318 & 326 = Hoffmann-Hordern, p. 211.

Lucas. L'Arithmétique Amusante. 1895. Prob. XII: La coupe du tailleur, p. 26. How long to cut a piece of length 16 into lengths of length 2? [This makes the problem more deceptive.]

H. D. Northrop. Popular Pastimes. 1901. No. 10: A dozen quibbles, no. 9, pp. 68 & 73. = The Sociable.

Benson. 1904. The tailor's puzzle, p. 227.

Hummerston. Fun, Mirth & Mystery. 1924. Some queer puzzles, Puzzle no. 76, part 4, pp. 164 & 183. Cut twenty yards into yard lengths.

King. Best 100. 1927. No. 24, pp. 15 & 44. Cut 90 yards of cloth into 90 lengths.

Meyer. Big Fun Book. 1940.