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

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

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.

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.