7. arithmetic & number theoretic recreations a. Fibonacci numbers

Sissy's scissors, pp. 120 & 739. Three sheets of paper, 20 x 40, to be cut into 2 x 20, without folding. He does it in 19 cuts by stacking the three sheets and cutting the side of 40 into 20 pieces. One can easily reduce this to 10 by first cutting the 20 x 40 into two 20 x 20, stacking them and then cutting the stack into 10 parts. One can reduce further by cutting the 10 edge in the middle, getting 5, 5. Stack these and cut the 5 into 1, 2, 2 with two cuts. Then stack the 2s and cut once more. This takes a total of 5 cuts, which seems minimal to me. There is a problem in that the last 2 x 10 pack is 24 sheets thick!

No. 9, pp. 162 & 752. Draw four equidistant horizontal lines and then four equidistant verticals. How many squares are formed? This gives a 3 x 3 array of squares, but he counts all sizes of squares, as in 5.X.2, getting 9 + 4 + 1 = 14.

Birtwistle. Math. Puzzles & Perplexities. 1971. Fifth, pp. 58 & 190. How many cuts to make 27 cubes from a cube?

Putnam. Puzzle Fun. 1978. Nos. 93-96: Cube and 27 cubes, pp. 13 & 37. How many cuts to make 27 cubes from a cube? Other questions deal with colourings.

7.AV. HOW LONG TO STRIKE TWELVE?
New section. I have now started adding rate problems of this type.

See also 7.AU.

King. Best 100. 1927. No. 36, pp. 19 & 47. If a clock takes 8 sec to strike 8, how long does it take to strike 12?

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 186, pp. 72 & 157: Big Ben. "If it takes Big Ben ten seconds to strike five o'clock, how long will it take to strike twelve o'clock?"

Evelyn August. The Black-Out Book. Op. cit. in 5.X.1. 1939. Is your brain working? -- no. 2, pp. 148 & 215. If Big Ben takes 4 seconds to strike 4, how long does it take to strike midnight?

Depew. Cokesbury Game Book. 1939. The clock, p. 217. If it takes 7 seconds to strike 7, how long does it take to strike 11.

Sullivan. Unusual. 1943. Prob. 25: Caution. If it takes 8 minutes to pass 8 stop lights, how long does it take to pass 12?

W. A. Bagley. Puzzle Pie. Op. cit. in 5.D.5. 1944. No. 25: The patriot's paradox, pp. 30-31. 60 shots in an hour is not the same as one shot per minute, since the latter requires 61 shots to be fired in a 60 minute interval from the first shot.

Ripley's Puzzles and Games. 1966. P. 70. "If 10 sheep jump over a fence in 10 minutes how many will jump over in an hour?" Answer: 55.

Barbara Lee. Six short problems. M500 162 (?? 1998) ??NYS -- cited by solver. A, J. Welton. Solutions to 'Six short problems'. M500 163 (Aug 1998) 12-13. No. 3: "A clock strikes six in 5 seconds. How long does it take to strike twelve?" "Six hours and five seconds."

7.AW. 28/7 = 13
There are three (or more) ways of demonstrating this result -- e.g. 7 doesn't go into 2, but it goes into 8 once, leaving us with 21 and 7 goes into 21 3 times. The other ways are based on multiplying 13 by 7 and on adding seven 13s. This is the basis of a vaudeville routine which was used in the Abbott and Costello movie 'In the Navy', 1949. I can only recall seeing it in the following.
Rohrbough. Brain Resters and Testers. c1935. Tricks for the Toastmaster (no. 2), pp. 12-13.

W. A. Bagley. Paradox Pie. Op. cit. in 6.BN. 1944. No. 2: Financial skullduggery, pp. 7 8.

Bud Abbott and Lou Costello. In the Navy. Universal Pictures, 1941. Text and stills in Richard J. Anobile, ed.; Who's On First? Verbal and Visual Gems from the Films of Abbott & Costello; Darien House, NY, 1972; Studio Vista, London, 1973. Pp. 128 139.

William R. Ransom. Op. cit. in 6.M. 1955. Fake arithmetic, pp. 9 11.  Shows there are 22 examples of this phoney arithmetic.

Robert Harbin. Party Lines. Op. cit. in 5.B.1. 1963. The Christmas club, p. 62.
7.AX. SUM = PRODUCT, ETC.
The archetypal example is to determine A, B such that A + B  =  A * B, usually with some additional condition. I have not noticed early examples of such problems until recently reading Riese's Die Coss and later Muscarello, but I recall other later examples such as Briggs & Bryan.
Muscarello. 1478. Ff. 69r-69v, p. 180. A + B = A * B. He gives examples: 6 & 6/5; 7 & 7/6; 8 & 8/7; and says it continues.

Calandri. Arimethrica. 1491. F. 97v. A + B = A * B. He finds 5/2, 5/3.

Riese. Die Coss. 1524.

No. 38, p. 45. A + B = A/B, with A/B = 3/2.

No. 41, p. 46. A   B = A/B, with A/B = 4/3.

Ozanam. 1694. Prob. 5, question 14, 1696: 13; 12. Prob. 8, quest. 14, 1725: 27-28. A + B  =  A * B. Gives (a+b)/a and (a+b)/b, e.g. 5/2 & 5/3.

Eadon. Repository. 1794.

P. 299, no. 8. A + B = A * B. Gives B = A/(A-1).

P. 299, no. 9. A - B = A² - B². Gives B = 1-A.

P. 299, no. 10. A + B = A² - B². Gives B = A-1.

Manuel des Sorciers. 1825. ??NX P. 85. A + B = A * B.

Walter Taylor. The Indian Juvenile Arithmetic .... Op. cit. in 5.B. 1849. Pp. 193-194. Solves A + B = A * B as A = (a+b)/a, B = (a+b)/b for any a, b.

Mittenzwey. 1880. Prob. 80, pp. 15 & 66; 1895?: 88, pp. 20 & 69; 1917: 88, pp. 18 & 66. A + B = 3 (A - B); AB = 3 (A + B). 1917 gives an algebraic solution.

Briggs & Bryan. The Tutorial Arithmetic, Part II. Op. cit. in 7.H. 1898. Exercises X, prob. 16, pp. 124 & 579. A + B  =  A * B  =  A²   B².

C. T. C. Wall. The (σ, x, p) problem. Eureka 20 (Oct 1957) 20-25. This starts with finding x, y, p, q such that x + y = pq, xy = p + q. There are several special solutions: x,  1, 1-x, -1; x, -x, 0, -x² and two sporadic solutions: 3, 2, 5, 1; 2, 2, 2, 2. He then considers more variables such that the symmetric functions of one set are the same as for the other set, but in reverse order.

Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. Prob. e, pp. 189-190. A - B = A/B.

Jerome S. Meyer. Arithmetricks. Scholastic Book Services, NY, 1965. Pp. 35-36. Considers all four cases.

A * B = A + B. (n+1)/n and n+1.

A * B = A - B. n and n/(n+1).

A / B = A + B. n + 1/(n+2) and (n+1)/(n+2).

A / B = A - B. n + 1/(n-2) and n-1.

Bronnie Cunningham. Funny Business. An Amazing Collection of Odd and Curious Facts with Some Jokes and Puzzles Too. Puffin, 1978. Pp. 37 & 142. A - B = A/B = 4.

7.AY. SUM OF POWERS OF DIGITS
A PDI (Perfect Digital Invariant) is an n digit number which is equal to the sum of the k-th powers of its digits. If k = n, it is called a PPDI (PluPerfect Digital Invariant). More generally, one can consider the function f_k(N) = sum of the k-th powers of the digits of N (in base 10 usually) and study its iterates. It is not hard to see that f_k(N) < N from some point on, so the iterates must lead to cycles. E.g., for k = 3, we have:

160  217  352  160.

See Reviews in Number Theory, sections A62 and A64 for related material. See 7.BB for other iterated functions of integers.
F. Hoppenot. Courrier du "Sphinx". Sphinx 7:4 (Apr 1937) 72. Gives 153, 371, 407, 8208, 9474 as PPDIs.

M. Fistié. Courrier du "Sphinx". Sphinx 7:5 (May 1937) 87. Adds 370 to Hoppenot's list and says these are all the PPDIs for k = 3.

Ball. MRE, 11th ed., 1939, p. 13. Gives the four PPDI's for k = 3: 153, 370, 371, 407, citing Sphinx.

G. H. Hardy. A Mathematician's Apology. CUP, 1940. "There are just four numbers (after  1) which are the sums of the cubes of their digits .... These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to a mathematician." He cites Ball for them.

H. Steinhaus. One Hundred Problems in Elementary Mathematics. (As: Sto Zadań, PWN -- Polish Scientific Publishers, Warsaw, 1958.) Pergamon Press, 1963. With a Foreword by M. Gardner, Basic Books, NY, 1964. Prob. 2: An interesting property of numbers, pp. 11-12 & 55-58. Considers iterates of f₂(n). He shows this always leads to a cycle and find there are two cycles, 1 of length 1 starting from 1 and 1 of length 8 starting from 4.

Kiyoshi Iseki and coworkers extended Steinhaus's idea to k = 3, 4, 5 in 5 papers during 1960 1963. See Leveques's Reviews in Number Theory, A64-12 -- A64-16 for details. For k = 3, there are 9 cycles (5 of which have period 1: 1 and the four PPDI's for k = 3). For k = 4, there are 6 cycles (4 of which have length 1: 1, 1634, 8208, 9474). For k = 5, there are 15 cycles, the longest having length 28. There are 5 cycles of length 1: 1, 54748, 93084, 92727, 194979.

Max Rumney. Digital invariants. RMM 12 (Dec 1962) 6-8. Quotes Hardy. General introduction to the ideas. Interested in repeating the process of summing the n-th powers of an n-digit number, so his PDI is what is now called a PPDI. He also considers cases where the process cycles through a few values, e.g. for the sum of the cubes of the digits, we get the cycle: 133, 055, 250, 133. There are such cycles starting with 133, 136, 217, 919.

Gardner. SA (Jan 1963) c= Magic Numbers, chap. 3. Surveys what is known with numerous references up through 1973. In particular, D. St. P. Barnard showed that the number of PPDIs is finite by noting that 61 * 9⁶¹ < 10⁶⁰, so that a PPDI must have < 61 digits.

Harry L. Nelson. More on PDI's. Publication UCRL-7614, Univ. of California. 1 Dec 1963. ??NYS -- cited by Schwarz and others below, who say it proves and improves Barnard's result and that it makes the distinction between PDI and PPDI as now generally used.

Azriel Rosenfeld, proposer; Nathan J. Fine, solver. Problem E1651 -- Sums of squares of digits. AMM 71 (1964) 90 & 1042-1043. "Prove that no multidigit integer is equal to the sum of the squares of its digits."

Benjamin L. Schwarz. Finiteness of a set of self-generating integers. JRM 2:2 (Apr 1969) 79 83. Proves Barnard's result and improves it to < 60 digits.

Benjamin L. Schwarz. Finite bounds on digital invariants -- some conjectures. JRM 3:2 (Apr 1970) 88-92. General discussion of whether the number of PDIs is infinite. He gives a heuristic argument that there are infinitely many.

Joseph S. Madachy. Some new narcissistic numbers. Fibonacci Quarterly 10:3 (Apr 1972) 295-298. "[R]eports on various narcissistic numbers other than digital invariants." Gives various forms and some references.

Benjamin L. Schwarz. Self-generating integers. MM 46:3 (May 1973) 158 160. Gives a simple condition for functions of the digits so that the number of PPDIs (= SGIs in his notation) must be finite. This applies to the sum of the factorials of the digits.

Lionel E. Deimel Jr. & Michael T. Jones. Finding pluperfect digital invariants: techniques, results and observations. JRM 14:2 (1981 82) 87-108. Cites work from 1962. All PPDIs for bases 2 through 10 were found, using months of Data General MV/8000 time, checking primality for bases < 10. There are 89 PPDIs in base 10 (but this includes 0 as an example for exponent 1, though it is not clearly a 1-digit number, so perhaps 88 is a better count).

Mary T. Whalen & Gordon L. Miller. Prime PPDI's. JRM 25:2 (1993) 118 123. Says PPDIs are also known as Armstrong numbers. The present authors have checked the above results and looked for examples of primes in base 10. The base 10 results took a DEC 6310 just over 14 days and three primes (beyond the exponent 1 cases) were found. Results were checked with Mathematica on a NeXT. They give 88 PPDIs for base 10.

Tony Forbes. Recurring digital invariants. M500 165 (Dec 1998) 4-7. After some comments and results on PDIs in base 10 and loops in the iteration process, he decides to examine base 3 and finds there are many examples and gives some very large examples.
7.AZ. DIVINATION OF A PAIR OF CARDS FROM ITS ROWS
A spectator mentally chooses a pair from ten pairs of cards. You then lay them out in a 5 by 4 array and ask him to tell you in which rows his pair appears, whereupon you point out his pair. This is done by laying out the cards in a particular pattern traditionally given by the Latin words: MUTUS DEDIT NOMEN COCIS. In fact, such patterns are easily found for r(r+1)/2 pairs laid out in r rows -- but finding a memorable pattern takes some doing. Since the person's response can be either one row or two rows, the number of responses is r + BC(r, 2) = r(r+1)/2 = BC(r+1, 2).

Unger, Secret Out, Ball, Ahrens and Indoor Tricks and Games deal with triples. With r rows, there are r + BC(r, 2) + BC(r, 3) = r (r² + 5)/6 possible responses. However, this does not always give a possible situation. E.g., for r = 2, there are 3 possible responses, but one cannot layout 3 triples in two rows, and the same problem occurs when r is even. For r = 3, there is an easy pattern: AAADDEG BBBDFFG CCCEEFG and I imagine it can be easily done whenever r is odd. However, we are getting to more responses than we can handle and it might be easier to use fewer of them. See Unger for an approach.

This is actually a combinatorial problem and I will probably shift this section into Chapter 5.
Bachet. Problemes. 1612. Prob. XVI: De plusieurs cartes disposées en divers rangs deviner laquelle on aura pensée: 1612: 87-92. Prob. 18, 1624: 143 151; 1884: 72-83. This problem covers several card divinations. The present topic has been added to "cette seconde [impression de ce livre]". He discusses the problem for r rows and shows the simplest pattern, but gives no mnemonics. For 10 pairs, the simplest pattern has rows: AABCD, BEEFG, CFHHI, DGIJJ.

Wingate/Kersey. 1678?. Prob. 5, pp. 539-542. Does 10 and 15 pairs as in Bachet.

Hooper. Rational Recreations. Op. cit. in 4.A.1. 1774. Recreation XXVII: The ten duplicates, pp. 70-71. MUTUS DEDIT NOMEN COCIS.

Endless Amusement I. c1818. P. 106: The ten duplicates. As in Hooper.

Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 122-123, no. 184: The ten duplicates. c= Endless Amusement I.

Rational Recreations. 1824. Feat 38, p. 168: Cards in couples. MUTUS DEDIT NONEM [sic] COCIS.

Manuel des Sorciers. 1825. Pp. 48-51, art. 21. ??NX MUTUS NOMEN DEDIT COCIS. Extends to 15 pairs, in a different way than Bachet, by bordering the usual array of letters with a left hand column 5, 4, 3, 2, 1 and adding a bottom row which continues from the 1 just added with 1, 2, 3, 4, 5.

Unger. Arithmetische Unterhaltungen. 1838. Pp. 233-235, nos. 897-901. He first gives the 10 pair case, using ABCBD EFAGE HGHIC KFKID, which is isomorphic to MUTUS NOMEN DEDIT COCIS, but he never gives any mnemonics. He then asks which other arrays can be done and sees that r rows can be used with r(r+1)/2 pairs. He gives a scheme for r = 3: ABAC BDDE EFCF, and a scheme for r = 5: ABCDEA BFGHFI KGLLCM ONNDHK EIOPMP. He then considers triples and gives a scheme for 8 triples: AAA123 BBB124 CCC143 DDD423. This has the feature that the responses are either one row or three rows, giving r + BC(r, 3) = r (r²   3r + 8)/6 responses. He shows an easy method for extending a scheme to the next number of rows getting AAA123567 BBB124589 CCC143860 DDD423907 EEE567890 with 15 triples. He notes that one may want to rearrange the columns to make it less obvious when the cards are laid out. He really should have started with AAA1 BBB1 CCC1 and then constructed AAA123 BBB124 CCC134 DDD234 in order to make the construction clearer.

Young Man's Book. 1839. Pp. 205-206. The Ten Duplicates. Identical to Endless Amusement I.

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Parlour magic, no. xxii, pp. 202-203 (1868: 213-214): The ten duplicates. MUTUS DEDIT NOMEN COCIS.

Magician's Own Book. 1857. No. 21: The pairs re-paired, pp. 65-66. MUTUS DEDIT NOMEN COCIS ["Mutus gave a name to the Coci," a people who have yet to be discovered.] = Boy's Own Conjuring Book, 1860, p. 73.

The Secret Out. 1859.