7. arithmetic & number theoretic recreations a. Fibonacci numbers

Prob. 53: Can you do this?, p. 44. Consider 1, 2, 3, 4 and 5, 7, 8, 9. Rearrange the sets so both have the same sum! He phrases it in terms of numbers on jerseys of football players. Cf Morris, 1991.

Prob. 55: A matter of multiplication, p. 45. AB*CDE consists of the same five digits and A = 1. Answer is: 14 * 926 = 12964. He says the only solutions when the condition A = 1 is dropped are: 24 * 651 = 15624; 42 * 678 = 28476; 51 * 246 = 12546; 57 * 834 = 47538; 65 * 281 = 18265; 65 * 983 = 63895; 72 * 936 = 67392; 75 * 231 = 17325; 78 * 624 = 48672; 86 * 251 = 21586; 87 * 435 = 37845. Cf Ripley below.

A. B. Nordmann. One Hundred More Parlour Tricks and Problems. Wells, Gardner, Darton & Co., London, nd [1927 -- BMC]. No. 94: The "100" problem, pp. 88 & 114. Use 1, 2, ..., 9 to make 100. Answer: 15 + 37 + 46  =  98 + 2 = 100.

Perelman. FFF. 1934. 1957: prob. 99 & 101, pp. 137 & 144; 1979: probs. 102 & 104, pp. 166-167 & 174-175. = MCBF, probs. 102 & 104, pp. 167 & 177-178.

102: One. "Write one by using all the ten digits." 148/296 + 35/70. Also (123456789)⁰, etc.

104: Ten Digits. "Write 100 using all the ten digits. How many ways are there of doing it? We know at least four." 70 + 24 9/18 + 5 3/6 and three other similar answers.

See: Meyers in 7.I.1 for the largest integer constructible with various sets of numbers.

Haldeman-Julius. 1937. No. 29: The 700 problem, pp. 6 & 22. Arrange the ten digits so "they'll add up to 700." Answer is: 102 4/8 + 597 3/6.

J. R. Evans. The Junior Week End Book. Op. cit. in 6.AF. 1939. Prob. 32, pp. 264 & 270. Find largest and smallest amounts in pounds, shillings, pence and farthings using the nine positive digits. Solutions as in Dudeney.

The Little Puzzle Book. Op. cit. in 5.D.5. 1955. P. 56: One hundred per cent. Form 100 using all 10 digits, but not in order. Gives just one solution.

Philip E. Bath. Fun with Figures. Op. cit. in 5.C. 1959. No. 27: Another sum of money, pp. 14 & 48. Least amount of English money using all 10 digits. Answer: £20567  18s  9¾d.

Robert Harbin. Party Lines. Op. cit. in 5.B.1. 1963. 1, 2, 3 ... 100, p. 77. Use each of the 10 figures to make a total of 100. Answer:  57 + 23 = 80 + 1 + 4 + 6 + 9 = 100.

Jonathan Always. Puzzles to Puzzle You. Op. cit. in 5.K.2. 1965. No. 91: Monetary matter, pp. 30 & 82. Find smallest amount, as in Dudeney.

Ripley's Puzzles and Games. 1966. P. 14. Says there are five examples of integers A, B such that A + B and A x B have the same digits: 9, 9; 3, 24; 2, 47; 2, 263; 2, 497.

There are infinitely more examples. One can show that one of A, B must be a single digit, say A, and that except for the special case 9, 9, the number of digits in A + B must be the same as in B, say n. There are two special solutions with n = 1, namely 0, 0 and 2, 2. The number of solutions for n = 2, 3, 4, 5, 6, 7 is 2, 2, 8, 29, 184, 1142. I have found solutions for each digit A greater than one, except for A = 4, 7 and I searched up to nine digit numbers.

Ripley's Believe it or Not, 14th series. Pocket Books, NY, 1968. Unpaged -- about 85% of the way through. AB * CDE gives a product of five digits which is a permutation of ABCDE. E.g. 14 * 926 = 12964. Asserts there are just 12 "sets of figures in which the multiplicand and the multiplier reappear in the product." Gives six examples. Cf Wood, 1927.

Ripley's Believe it or Not, 15th series. Pocket Books, NY, 1968. Unpaged -- about 30% of the way through. Gives 7 examples of the above situation, one of which was given above, saying "The original figures reappear in the results ...." Cf Wood, 1927. They do not seem to have considered other forms, e.g. I find 3 * 51 = 153; 6 * 21 = 126; 8 * 86  = 688.

Doubleday - 3. 1972. Sum total, pp. 129-130. Put 1, 2, 3, 5, 6, 7, 8, 9 into two groups of four with the same sum. Answer: 173 + 5 = 86 + 92.

Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 20: Make a century, pp. 20 & 99-100. Express 100 as a mixed fraction using the nine positive digits, e.g. 81 5643/297. There are 10 of this form and one other: 3 69258/714.

Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. Fodder for number crunchers, pp. 69 & 128. Find the 10-digit numbers, using all 10 digits, which are divisible by 1, 2, ..., 18 [i.e. are multiples of 12252240]. There are four such and none is divisible by 19. [I checked and found no smaller multiples of 12252240 with distinct digits. I found that there are 94 multiples of 9! (=  362880) with distinct digits. Of these, there are 2, 3, 16, 73 with 7, 8, 9, 10 digits. (Any multiple of 10! (= 3628800) has its last two digits equal.)]

Scot Morris. The Next Book of Omni Games. Op. cit. in 7.E. 1991. Pp. 55 & 192. Form the digits into two sums: 2 + 6 + 7 + 9 and 1 + 3 + 4 + 5 + 8. Make the sums equal by moving one number! Cf Wood, 1927.

Nob Yoshigahara. Puzzlart. Tokyo, 1992. Pandigital times, pp. 16 & 93. Using two digits for month, day, hour, minute, second, one can use the nine positive digits as in 8:19:23:46:57. This happens 768 times a year -- what are the earliest and latest such times. He gets: 3:26:17:48:59 and 9:28:17:56:43 and I have checked this. [?? -- what if we use all ten digits?]

Ed Barbeau. After Math. Wall & Emerson, Toronto, 1995. Four problems with whole numbers, pp. 127-130.

1. Find all integers such that the number and its square contain all nine positive digits just once. Answers: 567 and 854.

3. Find all integers such that its cube and its fourth power contain all ten digits just once. Answer: 18.

Jamie & Lea Poniachik. Cómo Jugar y Divertirse con su Inteligencia; Juegos & Co. & Zugarto Ediciones, Argentina & Spain, 1978 & 1996. Translated by Natalia M. Tizón as: Hard-to-Solve Brainteasers. Ed. by Peter Gordon. Sterling, NY, 1998. Pp. 15 & 71, prob. 21: John Cash. Find a three digit number abc with abc = 5*bc and bc = 5*c.

7.AC.7. SELF-DESCRIPTIVE NUMBERS, PANGRAMS, ETC.
New section. Are there older examples?
Solomon W. Golomb. Shift Register Sequences. Holden-Day, 1967. ??NYS -- cited by a 1996 article, but I cannot locate the material in the revised ed., Aegean Park Press, Laguna Hills, California, 1982; perhaps it is in some other work of Golomb ??check. Find a non-decreasing sequence of positive integers, (a_i), i = 1, 2, ..., such that i appears a_i times. Unique answer is: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, ....

Douglas Hofstadter. SA (Jan 1982) c= Metamagical Themas, Basic Books, NY, 1985, chap. 2, pp. 25-48. In Jan 1981, he had discussed self-referential sentences, and readers sent in a number of numerical ones.

Jonathan Post:

This sentence contains ten words, eighteen syllables and sixty four letters.

John Atkins:

'Has eighteen letters' does.

Howard Bergerson:

In this sentence, the word and occurs twice, the word eight occurs twice, the word four occurs twice, the word fourteen occurs four times, the word in occurs twice, the word seven occurs twice, the word the occurs fourteen times, the word this occurs twice, the word times occurs seven times, the word twice occurs eight times and the word word occurs fourteen times.

Lee Sallows:

Only the fool would take trouble to verify that his sentence was composed of ten a's, three b's, four c's, four d's, forty-six e's, sixteen f's, four g's, thirteen h's, fifteen i's, two k's, nine l's, four m's, twenty-five n's, twenty-four o's, five p's, sixteen r's, forty-one s's, thirty seven t's, ten u's, eight v's, eight w's, four x's, eleven y's, twenty-seven commas, twenty-three apostrophes, seven hyphens, and last, but not least, a single !

Raphael Robinson asks to fill in the blanks in the following and says there are two

solutions:

In this sentence, the number of occurrences of 0 is __, of 1 is __, of 2 is __, of 3 is __, of 4 is __, of 5 is __, of 6 is __, of 7 is __, of 8 is __, and of 9 is __.

The supplemental material in the book includes

J. K. Aronson:

'T' is the first, fourth, eleventh, sixteenth, twenty-fourth, twenty-ninth, thirty-third, ....

See below for further material.

Douglas Hofstadter. Metamagical Themas, Basic Books, NY, 1985, chap. 3, pp. 49-69. In the Post Scriptum, pp. 68-69, he reports on Sallows' search for a 'pangram'. Apparently the first of the type he wants is in Dutch, by Rudy Kousbroek and Sarah Hart: Dit pangram bevat vijf a's, twee b's, .... After some months search, Sallows' computer found:

This pangram tallies five a's, one b, one c, two d's, twenty-eight e's, eight f's, six g's, eight h's, thirteen i's, one j, one k, three l's, two m's, eighteen n's, fifteen o's, two p's, one q, seven r's, twenty-five s's, twenty-two t's, four u's, four v's, nine w's, two x's, four y's, and one z.

He challenges people to compute a version starting:

This computer-generated pangram contains ....

Douglas Hofstadter. Metamagical Themas, Basic Books, NY, 1985, chap. 16 pp. 364-395. In the Post Scriptum, pp. 387-395, he continues his discussion of the above material. He notes that Robinson's problem is convergent in the sense that if one inserts a random sequence of numbers, then counts the occurrences of the numbers and uses the counts as a new number, etc., then this iterative process usually converges to a solution. There are two solutions, but there is also a two term cycle and Hofstadter conjectures all initial values converge to one of these three situations. Sallows' challenge was given in A. K. Dewdeney's Computer Recreations column (SA, Oct 1984) and Larry Tesler used an iterative program on it. Tesler soon found a loop and modified the program a bit to obtain a solution:

This computer-generated pangram contains six a's, one b, three c, three d's, thirty-seven e's, six f's, three g's, nine h's, twelve i's, one j, one k, two l's, three m's, twenty-two n's, thirteen o's, three p's, one q, fourteen r's, twenty-nine s's, twenty-four t's, five u's, six v's, seven w's, four x's, five y's, and one z.

Lee Sallows. In Quest of a Pangram. Published by the author, Holland, nd [1985?]. Describes his search for a pangram.

Lee Sallows. Reflexicons. Word Ways 25 (1992) 131-141. A 'reflexicon' is a list of numbers and letters which specifies the number of times the letter occurs in the list. There are two in English.

fifteen e's, seven f's, four g's, six h's, eight i's, four n's, five o's, six r's, eighteen s's, eight t's, four u's, three v's, two w's, three x's.

sixteen e's, five f's, three g's, six h's, nine i's, five n's, four o's, six r's, eighteen s's, eight t's, three u's, three v's, two w's, four x's.

He also discusses 'pangrams', which are sentences containing the above kind of information -- e.g. This sentence contains one hundred and ninety-seven letters: four a's, .... The search for these is described in his booklet cited above. He then discusses crosswords using the number names.

Tony Gardiner. Challenge! What is the title of this article? Mathematics Review 4:4 (Apr 1994) 28-29. Following on a previous article in 4:1, he discusses self-describing sequences, where the description arises by reading the sequence. E.g. 22 is read as 'two twos'; 31 12 33 15 is 'three ones, one two, three threes, one five'. He also mentions self-describing lists, e.g. 1210 contains 'one 0, two 1s, one 2, zero 3s'.

Ed Barbeau. After Math. Wall & Emerson, Toronto, 1995. Pp. 117-122. Considers self-describing lists of length n and shows there are only the following: 1210, 2020, 21200 and, for n > 6, (n-4)2100...001000

Lee Sallows. Problem proposal to Puzzle Panel, 18 Jun 1998. "How many letters would this question contain, if the answer wasn't already seventy three?"
7.AD. SELLING, BUYING AND SELLING SAME ITEM
The problem is to determine the profit in a series of transactions involving the same item, but there is usually insufficient information. But see Clark and Sullivan for an unusual answer.
Clark. Mental Nuts. 1897, no. 1; 1904, no. 2; 1916, no. 24. The horse question. Sell a horse for $90, buy back at $80, resell at $100. "What did he make on the transaction?" Answer is $20, but this assumes the horse had no initial cost. If the item has no initial cost and the prices are a, b, c, then the gain is a - b + c. But if the question is asking for the profit, then the data are insufficient as the base cost is not given.

Loyd. The trader's profit. Cyclopedia, 1914, pp. 291 (no solution). (= MPSL1, prob. 13 -- What was the profit?, pp. 12 & 125.) Sells a bicycle at 50, buys back at 40, sells again at 45. Lengthy discussion of various 'solutions' of 15, 5 and 10. He says "the President of the New York Stock Exchange was bold enough to maintain over his own signature that the profit should be $10." Gardner points out that there isn't enough information.

H. E. Licks. Op. cit. in 5.A. 1917. Art. 25, p. 18. Sells at 50, buys back at 45, sells at 60 -- what is the profit? Author labels it impossible.

Smith. Number Stories. 1919. Pp. 126 127 & 146. Buys at 5000, sells at 5000, buys back at 4500, sells again at 5500.

Loyd Jr. SLAHP. 1928. The used car puzzle, pp. 9 10 & 88. Sell a used car at 100, buy back at 80, resell at 90. "This popped into my head one morning ...." Gives arguments for profits of 30, 10 and 20. Solution says the information is insufficient.

Collins. Fun with Figures. 1928. This sticks 'em up, p. 69. Buys at $55, sells at $55, buys back at $50, sells again at $60.

Rudin. 1936. No. 160, pp. 57 & 113. Buy for $70, sell for $80, buy back for $90, sell for $100.

Sullivan. Unusual. 1943. Prob. 2: A business transaction. Sell for $4000, buy back for $3500, sell again for $4500. Says the gain is $5000, composed of the initial $4000 plus the gain on the buying and selling. This is like Clark and markedly different than most approaches, which refer to profit.

Hubert Phillips. Something to Think About. Ptarmigan (Penguin), 1945. Problem 12: Alf's bike, pp. 15 & 89. More complex version with four persons and each person's percentage profit or loss given.

The Little Puzzle Book. Op. cit. in 5.D.5. 1955. P. 40: Multiple choice (B). Sell a cow for $100, buy back at $90, resell at $120. Says the profit is $30.

7.AD.1. PAWNING MONEY
Viscount John Allsebrook Simon. [Memory of Lewis Carroll.] Loc. cit. in 7.S.2. = Carroll-Wakeling II, prob. 33: Going to the theatre, pp. 51 & 73. He says Carroll gave the following. Man pawns 12d for 9d and sells the ticket to a friend for 9d. Who loses and how much? Simon relates that when he said that the friend lost 6d, Carroll pointed out that pawnbrokers charge interest. Mentioned in Carroll-Gardner, p. 80, who gives the full name. The DNB says he entered Wadham College, Oxford, in 1892, and his Memory says he met Carroll then. So this dates from 1892.

Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. He doesn't have the problem, but on p. 155 he gives current pawnbrokers' charges. These would be ½d for the ticket and ½d per 2s or part thereof per month or part thereof. So the above will cost at least ¾d.

Smith. Number Stories. 1919. Pp. 131 & 148. Pawn $1 for $.75 and sell ticket for $.50. Doesn't consider interest.

Lilian & Ashmore Russan. Old London City A Handbook, Partly Alphabetical. Simpkin, Marshall, Hamilton, Kent & Co., London, 1924, p. 222. Says the Bank of England once contemplated setting up a pawn business, to charge 1d per £1 per month.

R. Ripley. Believe It Or Not! Book 2. Op. cit. in 7.J. 1931. P. 143. "A man owed $3.00. He had a $2.00 bill, which he pawned for $1.50, and then sold the pawn ticket to another man for $1.50, who redeemed the $2.00 bill. Who lost?" No answer given.

H. Phillips. The Playtime Omnibus. Op. cit. in 6.AF. 1933. Section XVII, prob. 7, pp. 54 & 236. Jones pawns 6d for 5d, then sells the ticket to Brown for 4d. Who lost? Doesn't consider interest.

John Paul Adams. We Dare You to Solve This!. Op. cit. in 5.C. 1957? Prob. 52: Losers weepers, pp. 32 & 49. Pawn $5 for $3, sell ticket to a friend for $3 and he redeems it. Who lost and how much? Doesn't consider interest.
7.AE. USE OF COUNTERFEIT BILL OR FORGED CHEQUE
These problems give one or several transactions involving a bill or cheque which is then found to be counterfeit or forged. Who loses and how much? In the classic version of the hatter, the straightforward answer is that he loses the value of the counterfeit bill. However, there are two values for the hat -- the sale price and the cost price, whose difference is the profit that would be made in a normal sale. One can argue that one has only lost the cost price of the hat, so the loss is the value of the counterfeit bill less the expected profit.
Magician's Own Book. 1857. The unlucky hatter, p. 245. Man buys $8 hat with counterfeit $50 bill. "... and in almost every case the first impression is, that the hatter lost $50 besides the hat, though it is evident he was paid for the hat...." = Book of 500 Puzzles, 1859, p. 59. = Boy's Own Conjuring Book, 1860, pp. 215 216, but this spells out $ as dollars..

Lemon. 1890. The unlucky hatter, no. 225, pp. 34 & 106. Man pays for $8 hat with counterfeit $50 bill. "In almost every case the first impression ... is that the hatter lost $50 beside [sic] the hat..."

Hoffmann. 1893. Chap. IV, no. 43: What did he lose?, pp. 153 154 & 205 206 = Hoffmann Hordern, p. 129. Man pays for hat with a counterfeit bill. How much does the seller lose? Answer says that "The reply of most people is, almost invariably, that the hatter lost [the change] and the value of the hat, but a little consideration will show that this is incorrect." He then says that the seller loses the amount given in change less his profit on the goods sold; "the nett value of the hat, plus such trade profit, being balanced by the difference ... which he retained out of the proceeds of the note." This is wrong, as the sale price is part of the the refund that he has to make to the person who changed the note. Hordern notes that Hoffmann is wrong and 'the reply of most people' is indeed correct.

Clark. Mental Nuts. 1897, no. 2; 1904, no. 21; 1916, no. 7. The shoe question. Boy pays for $4 pair of shoes with counterfeit $10 bill. Answer says he lost $6 and the pair of shoes.

Dudeney. Some much discussed puzzles. Op. cit. in 2. 1908. Dud cheque used to buy goods and get cash. "Perpetually cropping up in various guises."

H. E. Licks. Op. cit. in 5.A. 1917. Art. 21, p. 17. Use of a counterfeit bill.

Lynn Rohrbough, ed. Mental Games. Handy Series, Kit E, Cooperative Recreation Service, Delaware, Ohio, 1927. Counterfeit Bill, p. 10. Man buys $6 pair of shoes with a phoney $20 bill. How much did seller lose? No solution given.

Ahrens. A&N, 1918, pp. 95 96 gives such a problem.

Dudeney. PCP. 1932. Prob. 34: The banker and the note, pp. 21 & 131. = 536; prob. 31: The banker and the counterfeit bill, pp. 10 11 & 230. Counterfeit bill goes in a circle so no one loses.

Phillips. Week End. 1932. Time tests of intelligence, no. 15, pp. 14 15 & 188. Forged cheque used to settle account and receive cash.

Robert A. Streeter & Robert G. Hoehn. Are You a Genius? Vol. 1, 1932; vol. 2, 1933, Frederick A. Stokes Co., NY. Combined ed., Blue Ribbon Books, NY, 1936. Vol. 1, p. 46, no. 10: "Brain twister". Man owes me 40¢, he gives me a knife worth at least 60¢ and I give him 20¢. I then find the knife was stolen and I pay the owner 75¢, which is its value. How much have I lost? Answer is 60¢. This is based on my payment of 75¢ being a fair purchase, so my loses are the 40¢ debt and the 20¢ change, which are now irrecoverable. However, if the first transaction is considered fair, then I've lost 75¢. Further, I might consider the original debt as a past loss, which would reduce the present loss to 20¢ or 35¢.

Phillips. The Playtime Omnibus. Op. cit. in 6.AF. 1933. Section XVII, prob. 5: Pinchem, pp. 53 & 236. Identical to Week End.

Dr. Th. Wolff. Die lächelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. Prob. 3, pp. 187-188 & 196-197. Sell bracelet worth 60 for phoney 100 bill, which is changed by the neighbouring shopkeeper. Author says he gets many answers, including 140, 200 and even 340, but the right answer is 100.

Depew. Cokesbury Game Book. 1939. Shoe dealer, p. 211. Sell shoes worth $8 for phoney $20 bill. Loss is $12 plus value of shoes.

McKay. Party Night. 1940. No. 7, p. 177. Man buys boots worth 15s with a bad £1 note. Says he gets answers up to "35s and a 15s pair of boots". Notes that the neighbouring grocer who changed the bill is irrelevant and the bootseller is simply £1 out of pocket. This ignores his profit on the boots.

Meyer. Big Fun Book. 1940. No. 10, pp. 171 & 754. Same as Streeter & Hoehn.

Doubleday - 1. 1969. Prob. 8: Cash on delivery, pp. 15 & 157. = Doubleday - 4, pp. 19-20. Forged $5 bill used to settle a circle of debts. Solution claims everything returns to the previous state, except for one stage -- this seems very confused and incorrect.

Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 13: The counterfeit note, pp. 17 & 98. Counterfeit note goes in a circle. She claims all the transactions are invalid -- but the bill has simply allowed a circuit of debts to be cancelled and hence no one has lost and there is no reason to cancel anything.

7.AF. ARITHMETIC PROGRESSIONS
See Tropfke 625.

There are many problems which are based on this. Some occur in 7.H.7 and 10.A. Here I only include the most interesting.

Bakhshali MS. c7C. In: G. R. Kaye, The Bakhshāli manuscript; J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349 361; p. 358 and in Kaye I 43 & III 176-177, ff. 4r-5r, sutra 18 and in Gupta. Consider two APs a, a + b, ..., and c, c + d, ..., and suppose the sums after n terms are equal to S. In the notation of 10.A, this is O-(a, b; c, d). Then n = 2(a   c)/(d - b) + 1. Does examples with (a, b; c, d) = (4, 3; 6, 1); (2, 3; 3, 2) and (5, 6; 10, 3).

Kaye III 174, f. 4v & Gupta. This is a problem of the same type, but most of it is lost and the scribe seems confused. Gupta attempts to explain the confusion as due to using the data a, b; c, d = 3, 4; 1, 2, with the rule n  =  2(c a)/(b-d) + 1, where the scribe takes the absolute values of the differences rather than their signed values. In this way he gets n = 3 rather than n = -1.

Pacioli. Summa. 1494. F. 44v, prob. 32. 1 + 2 + ... + 10½. He gets 10½ x 11½ / 2.

Unger. Arithmetische Unterhaltungen. 1838. Pp. 182 & 263, no. 693. Man digging a well 49 feet deep. First foot costs 15, but each successive foot costs 6 more than the previous. Find cost of last foot and total cost. So this is really an arithmetic progression problem, but I haven't seen others of those using this context.

M. Ph. André. Éléments d'Arithmétique (N^o 3) a l'usage de toutes les institutions .... 3rd ed., Librairie Classique de F.-E. André-Guédon, Paris, 1876. Prob. 550, p. 239. How many edges and diagonals does a convex octagon have?

(Beeton's) Boy's Own Magazine 3:6 (Jun 1889) 255 & 3:8 (Aug 1889) 351. (This is undoubtedly reprinted from Boy's Own magazine 1 (1863).) Mathematical question 59. Seller of 12 acres asks 1 farthing for the first acre, 4 for the second acre, 16 for the third acre, .... Buyer offers £100 for the first acre, £150 for the second acre, £200 for the third acre, .... What is the difference in the prices asked and offered? Also entered in 7.L.

Perelman. 1937. MCBF, A team of diggers, prob. 195, pp. 372-373. A team can dig a ditch in 24 hours, but just one digger begins and then the others join in at equal intervals, with the work finished in one interval after the last man joined. The first man works 11 times as long as the last man. How long did the last man work? Perelman finds this noteworthy (and I agree) because the number of men in the team cannot be determined!
7.AF.1. COLLECTING STONES
Alcuin. 9C. Prob. 42: Propositio de scala habente gradus centum. Computes 1 + 2 + ... + 100 as 100 + (1+99) + (2+98) + ... + 50.

Pacioli. Summa. 1494. F. 44v, prob. 31. Collect 100 oranges.

Pacioli. De Viribus. c1500. Ff. 122v - 124r. C(apitolo) LXXIII. D(e). levare .100. saxa a filo (To pick up 100 stones in a line). Wager on the number of steps to pick up 100 stones (or apples or nuts), one pace apart. Gives the number for 50 and 1000 stones.

Cardan. Practica Arithmetice. 1539. Chap. 66, section 55, f. EE.iiii.r (pp. 150-151). (The 55 is not printed in the Opera Omnia.) Picking up 100 stones in a line. (H&S 56-57 gives Latin with English summary.)

Buteo. Logistica. 1559. Prob. 87, pp. 299-300. Ant collecting 100 grains. (H&S 56.)

H&S 56 says this occurs in Trenchant (1566), ??NYS.

Baker. Well Spring of Sciences. 1562? ??check if this is in the Graves copy of the 1562/1568 ed. Prob. 2, 1580?: f. 36r; 1646: p. 56; 1670: pp. 72-73. 100 stones.

van Etten. 1624. Prob. 87 (84), part IV (8), p. 114 (184). 100 apples, eggs or stones. Henrion's Notte, p. 38, observes that there are many arithmetical errors in prob. 87 which the reader can easily correct.

Ozanam. 1694. Prob. 7, question 6, 1696: 53; 1708: 29. Prob. 10, question 6, 1725: 65 66. Prob. 1, 1778: 64-65; 1803: 66-67; 1814: 59-60; 1840: 32. 100 apples, becoming stones in 1778 et seq. 1778 describes a bet based on this process versus a straight run of the same distance.

Wells. 1698. No. 101, p. 205. 20 stones.

Dilworth. Schoolmaster's Assistant. 1743. P. 94, no. 6. 100 stones 2 yards apart. Answer in miles, furlongs and yards.

Walkingame. Tutor's Assistant. 1751. Arithmetical Progression, prob. 3, 1777: p. 90; 1835: p. 98; 1860: p. 118. 100 eggs a yard apart. Answer in miles and yards.

Edmund Wingate (1596-1656). A Plain and Familiar Method for Attaining the Knowledge and Practice of Common Arithmetic. .... 19th ed., previous ed. by John Kersey (1616-1677) and George Shell(e)y, now by James Dodson. C. Hitch and L. Hawes, et al., 1760. Quest. 44, p. 366. 100 stones a yard apart.

Mair. 1765? P. 483, ex. II. 100 eggs a yard apart. Answer: 5 miles, 1300 yards.

Euler. Algebra. 1770. I.III.IV: Questions for practice, no. 4, p. 139. 100 stones a yard apart. Answer: 5 miles, 1300 yards.

Vyse. Tutor's Guide. 1771?