7. arithmetic & number theoretic recreations a. Fibonacci numbers

Prob. 102, p. 79 & Answer 30, p. 140: One hundred puzzles. Three problems to insert the nine positive digits into formulae to make 100. E.g. + + + - - - = 100 is solved as: 3² + 91 + 7 + 8 - 6 - 5 - 4.

Prob. 140, p. 106 & Answer 104, p. 181: Plus and Minus. Find all ways to insert + and - signs into 1 2 ... 9 to yield 100. Finds 12 ways, one with a leading minus, as given by Gardner, 1962.

7.AC.4. PAN DIGITAL PRODUCTS
Loyd. Problem 36: Juggling with figures. Tit Bits 32 (5 Jun & 3 Jul 1897) 173 & 258. 10 digital product with smallest result is 3907 * 4 = 15628. Solution also gives the largest case: 9403 * 7  =  65821. [I have verified that these are correct.]

Dudeney. The miller's puzzle. The Canterbury puzzles. London Mag. 8 (No. 46) (May 1902) 367-371 & 8 (No. 47) (Jun 1902) 480-482. = CP, prob. 3, pp. 26 & 164-165. Find a solution of A*BC = D*EF = GHI using the 9 positive digits and which is closest to a given pattern. Answer says there are four solutions, the closest is 2 * 78  =  4 * 39  =  156.

Dudeney. CP. 1907.

Prob. 93: The number blocks, pp. 139-140 & 238. Find a solution of AB*CDE  =  FG*HIJ. Answer: 64 * 915 = 80 * 732.

Prob. 101: The three motor-cars, pp. 147-149 & 242-243. Wants a solution of AB*CDE  =  FGHIJ such that AB divides CDE. Answer is 27 * 594  =  16038 and he says it is hard to show this is unique. He says there are many solutions for A*BCDE  = FGHIJ, e.g. 3 * 5694 = 17082. = Wood, 1927, prob. 62.

Anon. Prob. 50. Hobbies 30 (No. 765) (11 Jun 1910) 228 & (No. 768) (2 Jul 1910) 289. Same pattern as in Dudeney, 1902, with different set-up.

M. Thié. ?? Nouv. Ann. Math. (4) 11 (1911) 46. ??NYS -- cited by Dickson I 463, item 62a. Found examples with 9 positive digits like 12 * 483 = 5796.

T. C. Lewis. ?? L'Intermédiaire des Math. 19 (1912) 26-27 & 187. ??NYS -- cited by Dickson I 463, item 66. Examples with the 10 digits, like 7 * 9403 = 65821 and 3 * 1458 = 6 * 0729.

Dudeney. AM. 1917.

Prob. 80: The three groups, pp. 14 & 156. Cites Thié and his example as being an extension of CP, prob. 101. Asks for solutions to Thié's form and to A*BCDE  =  FGHI, with the 9 positive digits, e.g. 4 * 1738 = 6952. Answer gives 7 solutions in the first case and 2 in the second case.

Prob. 81: The nine counters, pp. 14 & 156. Find solution of AB*CDE = FG*HI, with the 9 positive digits, such that the product is maximal. Answer: 32 * 174  =  58 * 96  =  5568.

Prob. 82: The ten counters, pp. 15 & 156. Divide the 10 digits into two equal products giving maximum and minimum products. Answers: 2 * 3485  =  1 * 6970  =  6970 & 64 * 915  =  80 * 732  =  58560. = Wood, 1927, probs. 59 & 60.

Prob. 85: The cab numbers, pp. 15 & 157. Find two numbers, using all 9 positive digits, whose product contains all 9 positive digits and is maximal. He believes the maximum is 96 * 8745231 = 839542176.

Prob. 86: Queer multiplication, pp. 15-16 & 157. Examples of A*BCDEFGHI  =  abcdefghi, where both sides use the 9 positive digits: 3 * 51249876 and 9 * 16583742. Asks for a solution with A = 6. Answer:  6 * 32547891.

Peano. Giochi. 1924. Prob. 34, p. 9. Notes 2 * 78  =  39 * 4  =  156. (Cf Dudeney, 1902.)

Hummerston. Fun, Mirth & Mystery. 1924. Grand-dad's age, Puzzle no. 68, pp. 155 & 182. A product of two 2-digit numbers contains the same digits and is the birth date of the grandfather, namely 21 * 87 = 1827. [There are three other examples, but none has a product in the recent past: 15*93 = 1395; 27*81 = 2187; 35*41 = 1435. After making the first number less than or equal to the second, eliminating leading and trailing zeroes in the factors, I find the following numbers of solutions for factors of M, N digits. 1, 1 : 0; 1, 2 : 3; 1, 3 : 7; 1, 4 : 36; 2, 2 : 4; 2, 3 : 41; 2, 4 : 170; 3, 3 : 119; 3, 4 : 972. Note that 0 * 0 = 00, etc. has been eliminated. The solutions in the 1, 3 case are: 3*51 = 153; 6*21 = 126; 8*86 = 688.]

Wood. Oddities. 1927.

Prob. 14: A problem in multiplication, p. 16. A * BC = D * EF = GHI. Gives four solutions: 2 * 78 = 4 * 39 = 156; 3 * 58 = 6 * 29 = 174 and their reversals (i.e. 4 * 39  = 2 * 78 = 156; 6 * 29 = 3 * 58 = 174) and implies there are no more.

Probs. 59 & 60: Number blocks & More number blocks, p. 47. Same as Dudeney's AM prob. 82.

Prob. 62, pp. 47-48. Same as Dudeney's CP prob. 101.

W. F. Cheney Jr, proposer; Victor Thébault, solver. Problem E13. AMM 39 (1932) 606 & 41 (1934) 265 266. Two factor products using all the digits just once. Gives all solutions without 0: 2 of form A*BCED = FGHI (confirming results of Buker in AMM 40 (1933) 559 ??NYS); 7 of form AB*CDE = FGHI. Gives some solutions with 0: 4  of form A*BCDE  = FGHIJ; 3 of form AB*CDE = FGHIJ.

Perelman. FFF. 1934. Tricky multiplication. 1957: prob. 45, pp. 56 & 61; 1979: prob. 48, pp. 71 & 77. = MCBF: prob. 48, pp. 69 & 74. Gives all 9 solutions without 0 to Dudeney's AM prob. 80.

Victor Thébault, proposer; G. H. Biucliu & L. Tits, solvers. Mathesis 44 (1935) 205 207. ??NYS -- described in CM 9 (1983) 89. All 94 solutions of n*ABCDE = FGHIJ.

Jerome S. Meyer. Arithmetricks. Scholastic Book Services, NY, 1965. No repeating digits, pp. 16-17. Says A*BCDEFGHI = abcdefghi with A = 9 has four solutions such that we also have 2*abcdefghi  =  abcdefghij. 81274365; 72645831;  58132764; 76125483 (where the last 3 is misprinted as 4). I find seven other examples: 58463721; 57624831; 85372461; 72534861; 83257614; 82164735; 71465328. However this doesn't include Dudeney's example in AM 86, because it doesn't satisfy the extra condition.

Charles L. Baker. (Presumably in RMM. ??NYS) Reported in Madachy's Mathematical Recreations, op. cit. in 5.O, (1966), 1979, pp. 183 185. Confirms Thébault's (1934) and Perelman's results without 0 and presents all two factor products with 0: 13 of form A*BCED  =  FGHIJ; 9 of form AB*CDE = FGHIJ.

Charles W. Trigg, proposer; Edward Moylan, solver; David Daykin, commenter. Problem 691 -- A product of integers. MM 41:3 (1968) 158; 42:1 (Jan 1969) 44-45 & 42:2 (Mar 1969) 102-103. Solve a = 8b, where a and b together use the 9 positive digits once each. Must have the form ABCDE = 8*FGHI and there are 46 solutions, all listed. D. Sumner assumed that both a and b contained all nine positive digits and found a unique solution with b = 123456789. Daykin gives the number of solutions of the first problem in base β, with multiplier m, for 2  m < β  15, and also considering the use of 0 as a digit and the use of just odd or just even digits. The tables show surprising irregularity. [Is this really surprising??]

Stewart Metchette. A note on digital products. JRM 10 (1977 78) 270 271. Extends Thébault and Baker to three factor products and gives all of the following forms: 12 of form A*BC*DE = FGHI; 10  of form A*BC*DE = FGHIJ; 2 of form A*BC*DEF = GHIJ.

Birtwistle. Calculator Puzzle Book. 1978.