7. arithmetic & number theoretic recreations a. Fibonacci numbers

Pp. 35-37, art. 12. Salve Certa Anima Semita Vita Quies.

Pp. 38-39, art. 13. Extends to four objects. Start with 88 counters and give 1, 2, 3, 4 to the people, leaving 78. Then tell them to take 1, 4, 16, 0 times more, so this is the same as Bachet's four person version.

Endless Amusement II. 1826?

P. 115: "Three things being privately distributed to three Persons, ...." c= Badcock.

P. 179: "Three Cards being presented to Three Persons, ...." M = (6, 4, 3) using the inverse permutation. No mnemonic.

Young Man's Book. 1839. Pp. 198-199. Identical to Endless Amusement II, p. 179.

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Parlour magic, no. xii, pp. 196-198 (1868: 209-210): How to discover the possessors of any articles taken from the table during your absence. Salve cesta animæ semita vita quies.

Magician's Own Book. 1857.

Prob. 36: The infallible prophet, pp. 23-24. Mnemonics: Salve certa animae semita vita quies; Par fer Cesar jadis devint si grand prince.

The three graces, pp. 218-221. Mnemonics: James Easy admires now reigning with a bride; Anger, fear, pain, may be hid with a smile; Graceful Emma, charming she reigns in all circles; Brave dashing sea, like a giant revives itself; Salve ....

Boy's Own Conjuring Book. 1860.

Prob. 35: The infallible prophet, pp. 35-36. Identical to Magician's Own Book.

The three graces, pp. 188-190. Identical to Magician's Own Book.

Vinot. 1860.

Art. XXXII: Trois cartes déterminées étant prises par trois personnes, deviner celle que chaque personne aura prise, p. 51. Essentially M = (6, 4, 3) set up by giving them 12, 24, 36 and using (1/2, 1/3, 1/4).

Art. XXXIV: Les trois bijoux, pp. 53-54. Usual (1, 2, 4) versions starting with 24 counters, hence forming 18 - P.

Boy's Own Book. Divination by cards. 1868: 637-638. Basically M = (1, 2, 4) but set up as (2, 3, 5). Vowel mnemonic: Par fer  César jadis devint si grand prince.

Magician's Own Book (UK version). 1871. To tell which article each of three persons took, pp. 35-36. Mnemonics: Take her certain anise seedlings Ida quince; Salve certa animæe servita vita quies; Par-fer Cæsar jadis devint si-grand prince.

Hanky Panky. 1872. A new three-card trick, pp. 256-257. M = (6, 4, 3), which is equivalent to (4, 2, 1).

Bachet-Labosne. Problemes. 3rd ed., 1874. Supp. prob. VII, 1884: 192-193. M = (6, 4, 3), which is the same as (4, 2, 1). No mnemonics.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Simple Tricks by Calculation No. I: To tell three persons which card each one has chosen, pp. 32-33. M = (4, 2, 1).

É. Ducret. Récréations Mathématiques. Op. cit. in 4.A.1. 1892? Pp. 107-?? (didn't photocopy the following pages): Les trois Bijoux. Seems to be usual form.

Lucas. L'Arithmétique Amusante. 1895. Prob. VIII-X, pp. 21-25. M = (1, 2, 4) with result subtracted from 18: Par fer César jadis devint si grand prince; Avec éclat l'Aï brillant devint libre. M = (0, 1, 3) with result subtracted from 12: Il a jadis brillé dans ce petit État.

Meyer. Big Fun Book. 1940. The picture card trick, pp. 509-510. Uses K, Q, J. Forms P  =  (2a₁-1)m₁ + (2a₂-1)m₂ + (2a₃-1)m₃ for multipliers (m_i) = (2, 3, 6). Then examines the remainder of the deck, which has 49 - P = R cards. 2 - [R/11] is the person holding the J. A counting process, which is erroneously described, gives the person with the Q as congruent to R + [R/11] (mod 3). In fact 3 - R(mod 3) is the position of the K.

Gardner. MM&M. 1956. The purloined objects, pp. 57-59. Gives several mnemonics using consonants. E.g. when the objects are Toothpick, Lipstick and Ring, use: tailor altar trail alert rattle relate. If the objects are denoted by A, B, C, use: Abie's bank account becomes cash club; or if they are denoted Small, Medium, Large, use: Sam moves slowly (since) mule lost limb.
7.AP. KNOWING SUM VS KNOWING PRODUCT
Two persons are told the sum and product of two integers. They then have some conversation such as: "I don't know what the numbers are." "I knew that." "I now know what they are." "So do I." What were the numbers? This seems to be a recent problem and I have only a few references. There are older versions, often called census-taker problems, where one knows the sum and product of three numbers (usually ages), but needs more information (often whether there is an oldest, which eliminates twins).

Thanks to Leroy F. Meyers for many of these references.

W. T. Williams & G. H. Savage. The Penguin Problems Book. Penguin, 1940. No. 96: The church afloat, pp. 53 & 135. Three ages: product = 840, sum is twice the curate's age. This is insufficient, but whether the eldest is older or younger than the vicar is sufficient to decide.

Lester R. Ford, proposer. Problem E776. AMM 54 (1947) 339 & 55:3 (Mar 1948) 159-160. ??NYS/NX -- proposal is quoted and the location of the solution given in the Otto Dunkel Memorial Problem Book (= AMM 64:7, part II, (Aug-Sep 1957) 61 & 89) and the location is more specifically given by Meyers & See. Four families of different sizes, not enough to form two baseball teams (i.e. the total is < 18), and the product of the numbers is the host's house number. The guest says he needs more information -- does the smallest family have just one child? The host's answer allows him to determine the numbers -- what were they?

Meyers writes that he first heard of the census-taker problem in 1951.

AMM problem E1126, 1954?. ??NYS.

W. A. Hockings, proposer; A. R. Hyde, solver. Problem E1156 -- The dimensions of Jones's ranch. AMM 62 (1955) 181 (??NX) & 63:1 (Jan 1956) 39-42. This is a continuation of E1126 and involves four men and their ranches.

Hubert Phillips. My Best Puzzles in Mathematics. Dover, 1961. Prob. 87: The professor's daughter, pp. 48 & 101. Youngest daughter is at least three. The product of their ages is 1200 and the sum is ten less than the wife's age. Visitor computes and then makes two wrong guesses as to the age of the youngest daughter. How old is the wife? The fact that the visitor made two wrong guesses means there must be at least three sets of ages, all  3, with product 1200 and the same sum, and indeed there is just one such situation, and this has three daughters. Allowing younger children permits some more complicated possibilities since we have

1 + 3 + 20 + 20 = 2 + 2 + 10 + 30 = 3 + 16 + 25 = 4 + 10 + 30

and 2 + 2 + 15 + 20 = 4 + 15 + 20 = 5 + 10 + 24 = 6 + 8 + 25.

(Phillips had most of these problems in his newspaper and magazine columns so it is likely that this will turn up in the period 1930-1950.)

M. H. Greenblatt. Mathematical Entertainments, op. cit. in 6.U.2, 1968. Chap. 1: "Census-taker" problems, pp. 1-7. Says he believes these problems came from some wartime project at MIT. Discusses three similar types.

1: The neighborhood census. Product of three ages is 1296 and the sum is the house number. Census taker asks if any of them are older than than the informant, who says 'no' and then the census taker knows the ages.

2: The priest and the banker. Product of three ages of ladies with the banker is 2450 and the sum is the same as the priest's age. The priest says this is insufficient and asks if any of the ladies is as old as the banker. When he says 'no', the priest knows all the ages. This even determines the banker's age!

3: The three Martians. Product of three ages is 1252 and the sum is the age of the informant Martian's father. The interrogator says this is insufficient and asks if any of the Martians is as old as the informant. When he answers 'no', the interrogator knows the ages. When you start on the problem, you find that 1252 = 2²·313 is not a suitable number. However, the problem shows a drawing of the Martian and one see he only has three fingers on each hand! Interpreting 1252 as a base 6 number gives the decimal 320 and the problem proceeds as before.

Gardner. SA (Nov & Dec 1970) c= Wheels, Chap. 3, prob. 10: The child with the wart. Supplied by Mel Stover. The product of the ages of three children is 36 and the sum is the questioner's house number. When the questioner says the information is insufficient, the father says the oldest child has a wart, which is sufficient to determine the ages. See Meyers & See for a generalization.

A. K. Austin. A calculus for know/don't know problems. MM 49:1 (Jan 1976) 12-14. He develops a set-theoretic calculus for systematically solving problems involving spots on foreheads, etc. (see 9.D), including problems similar to the present section. He gives the following problem considered by Conway and Patterson, but apparently unpublished. Two persons each have a card on their back bearing a positive integer, visible to the other person, but not to the person. They are told that the sum of the two integers is 6 or 7. They are then asked, in turn, to state whether they know what their own integer is. If they both have 3, what is the sequence of responses? Austin finds there are five 'no' answers, then the answers alternate yes, no.

David J. Sprows, proposer; Problem Solving Group, Bern, solver. Prob. 977 -- Mr. P. and Ms. S. MM 49:2 (Mar 1976) 96 & 50:5 (Nov 1977) 268. P & S are given the product and sum of two integers a, b, greater than one, but the sum is  100. P says he doesn't know the numbers. S says she knew that. P replies that he now knows the values and S responds that so does she. The unique answer is a, b = 4, 13. Editor cites the above AMM problems, though they are not quite the same type. See the discussion below.

Gardner. SA 241:6 (Dec 1979) 20-24. Problem 1: The impossible problem. Gives a version of Sprows' problem and says the problem was sent by Mel Stover and had been circulating for a year or two. This version assumes the numbers are greater than 1 but at most 20, which gives the unique solution 4, 13. However Gardner asserts that the solution remains the same if the bound is increased to 100 -- but I think there are further answers, e.g. 4, 61, see the discussion below. Stover says a computer program has checked and found no further solutions up to 2,000,000 and it may be that there is no further solution when the upper bound is removed -- this is a mistake of some sort, see the discussion below. Further, Kiltinen & Young say they had a letter from Gardner conjecturing that there are infinitely many solutions.

John O. Kiltinen & Peter B. Young. Goldbach, Lemoine, and a know/don't know problem. MM 58:4 (Sep 1985) 195-203. This discusses the Sprows problem, without the bound on S and various generalizations involving more stages of conversation. This leads to use of Goldbach's Conjecture and a conjecture of Lemoine that every odd number  7 can be expressed as 2p + q, where p, q are odd primes.

Friedrich Wille. Humor in der Mathematik. Vandenhoek & Ruprecht, Göttingen, (1984), 3rd ed., 1987. Paul und Simon, pp. 62 & 121 122. This has solution 4, 13. Note to the solution says there was much correspondence after the 1st ed., leading to a new solution and the computation of further examples with Sum = 17, 65, 89, 127, 137, 163, 179, 185, 191, 233, 247, 269, 305.
I recently was sent a version of the Sprows problem, without a bound on S, by Adrian Seville and solved it before checking this section. Dr. S (= Simon) and Dr. P (= Paul) are given the sum and product of two integers greater than 1. Dr. S says he doesn't know the numbers, but he can tell that Dr. P cannot determine the numbers. After a bit, Dr. P says he now knows the numbers. After a bit more, Dr. S says he now knows the numbers also. The version I received implies there is a unique solution, but there are more. Wille's discussion notes that the possible values of the sum, after Dr. S's statement are the {odd composites + 2}, which much simplifies the analysis. I found some further values of Sum: 343, 427, 457 and then extended considerably, finding: 547, 569, 583, 613, 637, 667, 673, 697, 733, 757, 779, 787, 817, 821, 853, 929, 967, 977, 989, 997, giving 36 values less than 1000 and there are another 42 solutions between 1000 and 2000. In looking at the earlier solutions, it appeared that one of the two numbers was always a power of 2, with the other number being odd, but for S = 757, P = 111756, one has the numbers being 202 and 556. Five more counterexamples to the initial appearance occur up to 2000.

While doing this investigation, I wondered what would happen if the condition 'greater than 1' was reduced to 'positive'. After some calculation, it became apparent and is provable that all solutions have P = S - 1 with the numbers being 1 and S - 1. Solutions occur for S = 5, 9, 10, 16, 28, 33, 34, 36, 46, 50, 52, 66, 78, 82, 88, 92, 96, giving 17 values less than 100, and there are another 87 values between 100 and 1000, making a total of 104 values less than 1000, which is where I stopped. If one assumes Goldbach's Conjecture, or part of it, one can show there are infinitely many solutions of the form S = pq + 1, where p, q are primes such that p + q = r + 1 for a prime r. However, I have not been able to see if there are infinitely many solutions of the original form of the problem.

I later was referred to the MM problem 977 and was surprised to see that the solution is unique when S  100 is imposed -- offhand, one might expect S = 65 and 89 to also be solutions in this case, but the additional knowledge about S affects the intermediate stages of the deduction. Rerunning an early version of my program with additional printout, I find that one needs S < 107 to prevent S = 65 being a solution. I then found Gardner's 1979 version and examination of the same data indicates that S = 65 is a solution when the numbers are bounded by 100. All in all, I find the presence of a bound has subtle effects and I find it unsatisfying.
Tim Sole. The Ticket to Heaven and Other Superior Puzzles. Penguin, 1988. Standard and Poor, pp. 116 & 126 128. Solution is 2, 15.

Leroy F. Meyers & Richard See. The census-taker problem. MM 63:2 (Apr 1990) 86-88. The product of the three ages in the next house is 1296 and the sum is the house number. The census-taker then asks if any of those people is older than the respondent. When he says 'yes', the census-taker says he knows the ages. This is a slight variant on Greenblatt's type 1. Authors investigate what other values, N, can be used instead of 1296. That is, we want N = ABC = DEF with A + B + C = D + E + F with just two such triples A, B, C; D, E, F. They determine the simple forms of such N and list the 45 examples  1296. The first few are: 36, 40, 72, 96, 126, 176, 200, 234, 252, 280, 297, 320, 408, 520, 550, 576, 588, 600. They give a number of references to similar questions.

Nicolas Guerrero. Problem 184.7 -- Deux nombres. M500 184 (Feb 2002) 25. The problem is given in French, but I will translate and paraphrase. Dr. P is given the product and Dr. S is given the sum of two integers between 2 and 100. (This is ambiguous -- the range might be [2, 100] or [3, 99].) Dr. P says he doesn't know the numbers. Dr. S then says he doesn't know the numbers. Dr. P then says he knows them, and Dr. S then says he knows them.

ADF [Tony Forbes]. Editorial remark on Problem 184.7. M500 188 (Oct 2002) 26. His citation is erroneously to M500 185, p. 25. He says he has not been able to obtain a satisfactory answer, in particular his answer is not unique.

My analysis gives four answers for S, P in the range [2, 100]:

a, b, S, P = 2, 6, 8, 12; 4, 19, 23, 76; 4, 23, 27, 92; 7, 14, 21, 98. However, it is clear that the intention is for just a, b to be in this range and this leads to so many possible values of P that a computer must be used. My program finds two solutions:

a, b, S, P = 2, 6, 8, 12; 84, 88, 172, 7392. Adapting the program to the range [2, 200], the larger solution disappears, but is replaced by six solutions with P larger than 30,000. I suspect that the proposer of the problem may have intended the earlier problem of Sprows, but either misheard or misremembered the exact details.

If one takes range [3, 99] for all the variables, the calculation is a little simpler, and there are three solutions: a, b, S, P = 3, 8, 11, 24; 9, 9, 18, 81; 9, 11, 20, 99. But there is an extra feature -- the values a, b, S, P = 6, 13, 19, 78; 8, 11, 19, 88 lead to the above sequence of statements except that Dr. S's last statement is that he still doesn't know the numbers, but we know his sum!

7.AQ. NUMBERS IN ALPHABETIC ORDER
I recall the problem of figuring out the reason for the sequence: 8, 5, 4, 9, 1, 7, 6, 3, 2 from 1956 or 1957.
?? Alphabetic Number Tables, 0   1000. MIT, Cambridge, Massachusetts, 1972. ??NYS.

"Raja" [= Richard & Josephine Andree]. Puzzle Potpourri #3. Raja Books, Norman, Oklahoma, 1976. No. 17: volumes 1, 2, ..., 12 shelved alphabetically.

Harvey & Robert Dubner. A tabulation of the prime numbers in the range of one to one thousand, in English and in Roman numerals, in alphabetic order. JRM 24 (1992) 89-93.
7.AR. 1089
The basic process is to take a three digit number, find the difference between it and its reversal, then add that result to its reversal and one gets 1089. The items by Meyer and Langford note that 9 * 1089 = 9801, so this is connected to 7.AH.

I now think that this ought to have originated from the fact that a number minus its reversal is divisible by nine, cf Berkeley & Rowland below and Section 7.K.1, but the monetary version seems to have arisen first. I have a note that I have seen a c1881 reference to the monetary version of this problem, but I cannot find it.

Prof. Orchard, proposer; Prof. Anderson, Rev. H. Sewell, et al, solvers. Prob. 10441. Mathematical Questions with Their Solutions from the "Educational Times" [generally known as Educational Times Reprints] 53 (1890) 78-79. Prove and generalise the fact that the process done with old English money gives £12 18s 11d. Solution says that if the multipliers of the units are m and n so that a, b, c denotes amn + bm + c, then the result of the calculation is m, n-2, m-1. The proposal makes no mention of the problems when the number of pounds is 12 or more or when the numbers of pounds and pence are equal and assumes the reversal is less than the original, but the solver assumes that a < m to make the reversal be an ordinary sum of money and then that c < a "to make the subtraction possible". Mr. Davis notes that this trick has been "current in well-informed City-circles for some months." If you perform it more than once, instruct the victim to add some convenient amounts to the result.

Don Lemon. Everybody's Scrap Book of Curious Facts. Saxon, London, 1890. Curious arithmetical puzzle, pp. 302-303. Quotes a letter to the London Globe from a 'constant reader' asking for an explanation for the monetary version, assuming you start with less than £11 19s 11d and the number of pence is less than the number of pounds.

Ball. MRE, 1st ed., 1892.