7. arithmetic & number theoretic recreations a. Fibonacci numbers

Seven and eight, p. 149. **7** / ** = 8***, full skeleton shown, but no other values given. Unique solution is 99708 / 12 = 8309.

Long division, pp. 150, 185 & 197. Skeleton of 123195 / 215 = 573 with four values given.

Letter division, pp. 151, 186 & 198. Complete layout of a division with letters denoting the digits, as in a cryptarithm. Two possible answers: 6420 / 20 = 321 or 6930 / 30 = 231.

On all fours, pp. 151 & 198. Skeleton of 31666 / 142 = 223 with all fours given.

7.AC.3. PAN DIGITAL SUMS
These are generally of the form ABC + DEF = GHI. The digits may be positive or 9 of the 10 digits. One can also have a 10 digital form, e.g. ABC + DEF = GHIJ.
Dudeney. Problem 64: The lockers puzzle. Tit Bits 33 (18 Dec 1897 & 5 Feb 1898) 220 & 355. = AM, prob. 79, pp. 14 & 156. Find ABC + DEF = GHI using 9 of the 10 digits which have the least result, the greatest result and a result whose digits are distinct from the first two. Answers: 107 + 249 = 356; 235 + 746 or 324 + 657 = 981; 134 + 586 = 720 or 134 + 568 = 702 or 138 + 269 = 407.

Dudeney. AM. 1917. Prob. 77: Digits and squares, pp. 14 & 155. For ABC + DEF = GHI, he wants DEF  =  2 * ABC, so GHI  =  3 * ABC, using the 9 positive digits. Says there are four solutions, the tops being 192, 219, 273, 327.

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 89, pp. 38 & 142: Doubles and trebles. Same as Dudeney, AM, prob. 77.

Morley Adams. Puzzle Parade. Op. cit. in 7.AC.1. 1948. No. 12: Figure square, pp. 146 & 150. As in Dudeney, AM, prob. 77. Says there are four solutions, but wants the one with minimal E. Solution: 219 + 438 = 657.

Philip E. Bath. Fun with Figures. Op. cit. in 5.C. 1959. No. 15: Skeleton addition, pp. 10 & 41. Complete ABC + DE7 = GH8 using all nine positive digits. Gets four forms by reversing A and D and/or B and E.

Jonathan Always. Puzzles to Puzzle You. Op. cit. in 5.K.2. 1965. No. 135: All the numbers again, pp. 42 & 90. As in Dudeney, AM, prob. 77. Gives one solution: 192 + 384  =  576.

Ripley's Puzzles and Games. 1966. Arrange the nine positive digits in two columns with the same sum. He forms an X shape with 5, 9, 1, 6, 2 down one line and 4, 8, 1, 7, 3 down the other. Both 'columns' add to 23. [Another solution is to invert the 6 or the 9.]

Wickelgren. How to Solve Problems. Op. cit. in 5.O. 1974. Integer path addition

problem, pp. 130 132. Wants the 9 positive digits in a pan-digital sum, so the 1 2 9

resulting 3 x 3 array has each digit i horizontally or vertically adjacent + 4 3 8

to the digit i+1. Says there is one solution, shown at the right. = 5 6 7

Shakuntala Devi. Puzzles to Puzzle You. Op. cit. in 5.D.1. 1976. Prob. 41: The

number and the square, pp. 31 and 107. As in Dudeney, AM, prob. 77. Gives all solutions.

Birtwistle. Calculator Puzzle Book. 1978. Prob. 87: Three by three -- part two, pp. 61 & 120 121. Which distributions of the nine positive digits as ABC, DEF, GHI have the lowest sum and product? 147 + 258 + 369 = 774 gives the lowest sum and the digits in each position can be permuted -- e.g. 348 + 257 + 169 gives the same sum. The lowest product is uniquely given by 147 x 258 x 369 = 13994694.

Johannes Lehmann. Kurzweil durch Mathe. Urania Verlag, Leipzig, 1980. No. 14, pp. 39 & 139-140. A + B + C = D + E + F = G + H + I has just two solutions using the positive digits. [Interestingly, one gets no more solutions using the ten digits.]

David Singmaster. Determination of all pan-digital sums with two summands. JRM 27:3 (1995) 183-190. AB + CDE = FGHI has no solutions with the nine positive digits and ten basic solutions using nine of the ten digits. AB + CDEF = GHIJ has nine basic solutions. (Each basic solution gives four or eight equivalent solutions.) ABC + DEF = 1GHI has 12 basic solutions, which can be paired. ABC + DEF = GHI has 216 basic solutions, but 80 have A = 0. 42 cases use the positive nine digits. The 216 can be grouped into 72 triples and a canonical example is given for each triple. The cases where the three terms form a simple proportion are listed.

7.AC.3.a INSERTION OF SIGNS TO MAKE 100, ETC.
I include here problems like inserting + and - (and perhaps  and ) signs into 12...9 to yield 100, which I call 'insertion to make 100'. This has two quite different sets of answers depending on whether the operations are carried out sequentially (as on an old calculator) or in algebraic order of precedence (as on a computer or modern calculator). See also 7.AC.6 for similar problems.
Leske. Illustriertes Spielbuch für Mädchen. 1864? Prob. 564-16, pp. 253 & 395. Make 100 from 1, 2, ..., 9, 0. Answer: 9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. (0 isn't used, but could be added.)

Mittenzwey. 1880. Prob. 139, pp. 30 & 79; 1895?: 159, pp. 33 & 82; 1917: 158, pp. 31 & 79. Make 100 from 1, 2, ..., 9 using only multiplication and addition. Same answer as Leske.

Anon. & Dudeney. A chat with the Puzzle King. The Captain 2 (Dec? 1899) 314-320; 2:6 (Mar 1900) 598-599 & 3:1 (Apr 1900) 89. Insert as few signs as possible in 12...9 to make 100. Usual answer is 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8x9) (cf Leske), but he gives 123   45   67 + 89. Cf. AM, 1917.

Dudeney. AM. 1917. Prob. 94: The digital century, pp. 16-17 & 159-160. Insert signs into 12...9 to make 100, using: (1) as few signs as possible; (2) as few strokes as possible, with - counting as 1 stroke; +,  () &  counting as 2;  counting as 3. He finds his 1899 result is best under both criteria. Cf. Anon & Dudeney, 1899.

Hummerston. Fun, Mirth & Mystery. 1924. Century making, p. 66. 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8x9) (cf Leske). 12 + 3 - 4 + 5 + 67 + 8 + 9. 123 + 45   67 + 8 - 9. 9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 (cf Leske; Hummerston notes this is the reverse of his first example). 98 - 76 + 54 + 3 + 21. Also the trick version: 15 + 36 + 47 = 98 + 2 = 100 (cf 7.A.6).

M. Adams. Puzzles That Everyone Can Do. 1931. Prob. 262, pp. 98 & 169: A number puzzle. Insert 'mathematical signs' into 4 3 2 1 to make 100. Answer:  4 * [(3 + 2) / .1] .

Perelman. 1934. See in 7.AC.6 for pandigital sum yielding 1 and 100.

McKay. At Home Tonight. 1940. Prob. 19: Centuries, pp. 66 & 80. Insertion to make 100. -1x2 - 3 - 4 - 5 + 6x7 + 8x9. -1x2 -3 - 4 + 5x6 +7 + 8x9. 1x2x3x4 + 5 + 6 + 7x8 + 9.

Anonymous. The problems drive. Eureka 12 (Oct 1949) 7-8 & 15. No. 6. Insert symbols into 1 2 ... 9 to make 1000; 1001; 100. Answers: 1234 - (5+6+7+8)x9; (12 x 34)   (5 x 6) + (7 x 89); 123 - 45 - 67 + 89. "These solutions are not unique."

Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 14-16 & 30. No. 10. Use 1, 2, 3, 4, 5, in order to form 100; 3 1/7; 32769.

Philip E. Bath. Fun with Figures. Op. cit. in 5.C. 1959. No. 48: A tricky problem, pp. 20 & 48. "Can you replace the asterisks by the digits of the number 216345879 in this order so the resulting total is 100? * + * + * + * + * + *" Answer:  2/1 + 6/3 + 4 + 5 + 8 + 79. His fraction bars are horizontal, but this problem seems a bit unreasonable to me.

Richard E. Bellman. On some mathematical recreations. AMM 69:7 (Aug/Sep 1962) 640 643. Develops a general theory for the number of ways n can be obtained by inserting + or x into a₁a₂...a_N and for determining the minimum number of + signs occurring. He computes the example of inserting into 12...9 to make 100 by use of recursion, finding that 1x2x3x4 + 5 + 6 + 7x8 + 9 = 100 has the minimal number of + signs, cf McKay.

Gardner. SA (Oct 1962) c= Unexpected, chap. 15. Insertion to make 100. Cites Dudeney. Asks for minimum number of insertions into 98...1 to make 100. Answer with four signs.

Gardner. SA (Jan 1965) c= Magic Numbers, chap. 6. Considers inserting + and - signs in 12...9 or 98..1 to yield 100. Says he posed this in SA (Oct 1962) c= Unexpected, chap. 15 and many solutions for both the ascending and descending series were printed in Letters in SA (Jan 1963). He gives a table of all the answers for both cases: 11  solutions for the ascending series and 15 solutions for the descending series. He extends the problem slightly by allowing a - in front of the first term and finds 1 and 3 new solutions.

Jonathan Always. Puzzles for Puzzlers. Tandem, London, 1971. Prob. 10: Very simple arithmetic, pp. 14 & 62. Insert signs into 1 2 3 4 5 6 to form an equation. He gives 12  3  4 + 5 = 6. I find 1 + 2 x 3 + 4 = 5 + 6. which seems more satisfactory.

Birtwistle. Calculator Puzzle Book. 1978. Prob. 65: Key to the problem, pp. 47-48 & 104. Using a calculator, insert operations in 012...9 and 98...10 to produce 100. Gives one example of each: 0 + 1/2 + 3x4x5 + 6 + 7 + 8 + 9; 9x8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 (cf Leske).

Steven Kahane. Sign in, please! JRM 23:1 (1991) 19-25. Considers inserting + and - signs in 12...n and in n...21 to produce various results, e.g. 0, n+1, n, among others.

Ken Russell & Philip Carter. Intelligent Puzzles. Foulsham, Slough, 1992.