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Ten rows to form -- in each row three --

Tell me, ye wits, how this can be?

Robina.

Answer has

Good-tempered Friends! here nine stars see:

Ten rows there are, in each row three!

W. S. B. Woolhouse. Problem 39. The Mathematician 1 (1855) 272. Solution: ibid. 2 (1856) 278 280. ??NYS -- cited in Burr, et al., below, who say he does (15, 26, 3).

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Mechanical puzzles.

No. 1, p. 176 (1868: 187). (9, 10, 3).

Ingenious artist pray disclose,

How I nine trees can so dispose,

That these ten rows shall formed be,

And every row consist of three?

No. 12, p. 182 (1868: 192-193). (24, 28, 3), but with a central pond breaking 4 rows of 6 into 8 rows of 3.

Magician's Own Book. 1857.

Prob. 33: The puzzle of the stars, pp. 277 & 300. (9, 10, 3),

Friends one and all, I pray you show

How you nine stars would so bestow,

Ten rows to form -- in each row three --

Tell me, ye wits, how this can be?

Prob. 41: The tree puzzle, pp. 279 & 301. (21, 9, 5), unequally spaced on each row. Identical to Book of 500 Puzzles, prob. 41.

The Sociable. 1858. = Book of 500 Puzzles, 1859, with same problem numbers, but page numbers decreased by 282.

Prob. 3: The practicable orchard, pp. 286 & 302. (16, 10, 4).

Prob. 8: The florist's puzzle, pp. 289 & 303-304. (31, 6, 6) with 7 circles of 6.

Prob. 9: The farmer's puzzle, pp. 289 & 304. (11, 11, 3).

Prob. 12: The geometrical orchard, p. 291 & 306. (27, 9, 6).

Prob. 17: The apple-tree puzzle, pp. 292 & 308. (10, 5, 4).

Prob. 22: The peach orchard puzzle, pp. 294 & 309. (27, 10, 6).

Prob. 26: The gardener's puzzle, pp. 295 & 311. (12, 6, 4) two ways.

Prob. 27: The circle puzzle, pp. 295 & 311. (37, 20, 5) equally spaced along each row.

Prob. 29: The tree puzzle, pp. 296 & 312. (15, 16, 3) with some bigger rows. Solution is a 3 x 4 array with three extra trees halfway between the points of the middle line of four.

Prob. 32: The tulip puzzle, pp. 296 & 314. (19, 9, 5).

Prob. 36: The plum tree puzzle, pp. 297 & 315. (9, 10, 3).

Family Friend (Dec 1858) 359. Practical puzzles -- 2. "Make a square with twelve counters, having five on each side." (12, 4, 5). I haven't got the answer, but presumably it is the trick version of a hollow square with doubled corners, as in 7.Q. See Secret Out, 1859 & Illustrated Boy's Own Treasury, 1860.

Book of 500 Puzzles. 1859. Prob. 3, 9, 12, 17, 22, 26, 27, 29, 32, 36 are identical to those in The Sociable, with page numbers decreased by 282.

Prob. 33: The puzzle of the stars, pp. 91 & 114. (9, 10, 3), identical to Magician's Own Book, prob. 33.

Prob. 41: The tree puzzle, pp. 93 & 115. (21, 9, 5), identical to Magician's Own Book, prob. 41. See Illustrated Boy's Own Treasury.

The Secret Out. 1859.

To place twelve Cards in such a manner that you can count Four in every direction, p. 90. (12, 7, 4) trick of a 3 x 3 array with doubling along a diagonal. 'Every direction' must refer to just the rows and columns, but one diagonal also works.

The magical arrangement, pp. 381-382 = The square of counters, (UK) p. 9. (12, 4, 5) -- trick version. Same as Family Friend & Illustrated Boy's Own Treasury, prob. 13.

The Sphynx, pp. 385-386. (21, 30, 3). = Hoffmann, no. 15.

Charades, Enigmas, and Riddles. 1860: prob. 13, pp. 58 & 61; 1862: prob. 13, pp. 133 & 139; 1865: prob. 557, pp. 105 & 152. (9, 10, 3). (The 1862 and 1865 have slightly different typography.)

Sir Isaac Newton's Puzzle (versified).

Ingenious Artist, pray disclose

How I, nine Trees may so dispose,

That just Ten Rows shall planted be,

And every Row contain just Three.

Boy's Own Conjuring Book. 1860.

Prob. 40: The tree puzzle, pp. 242 & 266. (21, 9, 5), identical to Magician's Own Book, prob. 41.

Prob. 42: The puzzle of the stars, pp. 243 & 267. (9, 10, 3), identical to Magician's Own Book, prob. 33, with commas omitted.

Illustrated Boy's Own Treasury. 1860.

Prob. 2, pp. 395 & 436. (37, 20, 5), equally spaced on each row, identical to The Sociable, prob. 27.

Prob. 13, pp. 397 & 438. "Make a square with twelve counters, having five on each side." (12, 4, 5). Trick version of a hollow square with doubled corners. Presumably identical to Family Friend, 1858. Same as Secret Out.

J. J. Sylvester. Problem 2473. Math. Quest. from the Educ. Times 8 (1867) 106 107. ??NYS -- Burr, et al. say he gives (10, 10, 3), (81, 800, 3) and (a, (a 1)²/8, 3).

Magician's Own Book (UK version). 1871. The solution to The florist's puzzle (The Sociable, prob. 8) is given at the bottom of p. 284, apparently to fill out the page as there is no relevant text anywhere.

Hanky Panky. 1872.

To place nine cards in ten rows of three each, p. 291. I.e. (9, 10, 3).

Diagram with no text, p. 128. (37, 20, 5), equally spaced on each line as in The Sociable, prob. 27.

Hoffmann. Modern Magic. (George Routledge, London, 1876); reprinted by Dover, 1978. To place twelve cards in rows, in such a manner that they will count four in every direction, p. 58. Trick version of a 3 x 3 square with extras on a diagonal, giving a form of (12, 7, 4). Same as Secret Out.

Lewis Carroll. MS of 1876. ??NYS -- described in: David Shulman; The Lewis Carroll problem; SM 6 (1939) 238-240.

Given two rows of five dots, move four to make 5 rows of 4. Shulman describes this case, following Dudeney, AM, 1917, then observes that since Dudeney is using coins, there are further solutions by putting a coin on top of another. He refers to Hoffmann and Loyd. The same problem is in Carroll-Wakeling, prob. 1: Cakes in a row, pp. 1-2 & 63, but undated and the answer mentions the possibility of stacking the counters.

(9, 10, 3). Shulman quotes from Robert T. Philip; Family Pastime; London, 1852, p. 30, ??NYS, but this must refer to the item in Family Friend, which was edited by Robert Kemp Philp. BMC indicates Family Pastime which may be another periodical. Shulman then cites Jackson and Dudeney. Carroll-Wakeling, prob. 2: More cakes in a row, pp. 3 & 63, gives the problems (9, 8, 3), (9, 9, 3), (9, 10, 3), undated.

Mittenzwey. 1880.

Prob. 151, pp. 31 & 83; 1895?: 174, pp. 36 & 85; 1917: 174, pp. 33 & 82. (6, 3, 3) in three ways.

Prob. 152, pp. 31 & 83; 1895?: 175, pp. 36 & 85; 1917: 175, pp. 33 & 82. Arrange 16 pennies as a 3 x 3 square so each row and column has four in it. Solution shows a 3 x 3 square with extras on the diagonal -- but this only uses 12 pennies! So this the trick version of (12, 7, 4) as in Secret Out & Hoffmann (1876).

Prob. 153, pp. 31 & 83; 1895?: 176, pp. 36 & 85; 1917: 176, pp. 33 & 82. (21, 10, 5).

Cassell's. 1881. P. 92: The six rows puzzle. = Manson, 1911, p. 146.

J. J. Sylvester. Problem 2572. Math. Quest. from the Educ. Times 45 (1886) 127 128. ??NYS -- cited in Burr, below. Obtains good examples of (a, b, 3) for each a. In most cases, this is still the best known.

Richard A. Proctor. Some puzzles; Knowledge 9 (Aug 1886) 305-306 & Three puzzles; Knowledge 9 (Sep 1886) 336-337. (19, 9, 5). Generalises to (6n+1, 3n, 5).

Richard A. Proctor. Our puzzles. Knowledge 10 (Nov 1886) 9 & (Dec 1886) 39-40. Gives several solutions of (19, 9, 5) and asks for (19, 10, 5). Gossip column, (Feb 1887) 92, gives another solution

William Crompton. The odd half-hour. The Boy's Own Paper 13 (No. 657) (15 Aug 1891) 731-732. Sir Isaac Newton's puzzle (versified). (9, 10, 3).

Ingenious artist pray disclose

How I nine trees may so dispose

That just ten rows shall planted be

And every row contain just three.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Card Puzzles No. IV, p. 3. (9, 10, 3).

Hoffmann. 1893. Chap. VI, pp. 265 268 & 275 281 = Hoffmann-Hordern, pp. 174-182, with photo.

No. 1: (11, 12, 3).

No. 2: (9, 10, 3).

No. 3: (27, 9, 6).

No. 4: (10, 5, 4).

No. 5: (12, 6, 4). Photo on p. 177 shows L'Embarras du Brigadier, by Mauclair-Dacier, 1891 1900, which has a board with a 7 x 6 array of holes and 12 pegs. The horizontal spacing seems closer than the vertical spacing.

No. 6: (19, 9, 5).

No. 7: (16, 10, 4).

No. 8: (12, 7, 4) -- Trick version of a 3 x 3 square with extras on a diagonal as in Secret Out, Hoffmann (1876) & Mittenzwey.

No. 9: 9 red + 9 white, form 10 + 8 lines of 3 each. Puts a red and a white point at the same place, so this is a trick version.

No. 11: (10, 8, 4) -- counts in 8 'directions', so he counts each line twice!

No. 12: (13, 12, 5) -- with double counting as in no. 11.

No. 15: (21, 30, 3) -- but points must lie on a given figure, which is the same as in The Secret Out.

Clark. Mental Nuts. 1897, no. 19: The apple orchard; 1904, no. 91: The lovers' grove. (19, 9, 5). 1897 just has "Place an orchard of nineteen trees so as to have nine rows of five trees each." 1904 gives a poem.

I am required to plant a grove

To please the lady whom I love.

This simple grove to be composed

Of nineteen trees in nine straight rows;

Five trees in each row I must place,

Or I shall never see her face.

Cf Ripley, below.

Dudeney. A batch of puzzles. Royal Magazine 1:3 (Jan 1899) & 1:4 (Feb 1899) 368-372. (22, 20, 4) with trees at lattice points of a 7 x 10 lattice. Compare with AM, prob. 212.

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. (16, 15, 4). Cf 1902.

Dudeney. "The Captain" puzzle corner. The Captain 3:2 (May 1900) 179. This gives a solution of a problem called Joubert's guns, but I haven't seen the proposal. (10, 5, 4) but wants the maximum number of castles to be inside the walls joining the castles. Manages to get two inside. = Dudeney; The puzzle realm; Cassell's Magazine ?? (May 1908) 713-716; no. 6: The king and the castles. = AM, 1917, prob. 206: The king and the castles, pp. 56 & 189.

H. D. Northrop. Popular Pastimes. 1901. No. 11: The tree puzzle, pp. 68 & 73. = The Sociable, no. 29.

Dudeney. The ploughman's puzzle. In: The Canterbury Puzzles, London Magazine 9 (No. 49) (Aug 1902) 88 92 & (No. 50) (Sep 1902) 219. = CP; 1907; no. 21, pp. 43 44 & 175 176. (16, 15, 4). Cf 1899.

A. Héraud. Jeux et Récréations Scientifiques -- Chimie, Histoire Naturelle, Mathématiques. Baillière et Fils, Paris, 1903. P. 307: Un paradoxe mathématique. (24, 28, 4). I haven't checked for this problem in the 1884 ed.

Pearson. 1907.

Part I, no. 77: Lines on an old sampler, pp. 77 & 167. (17, 28, 3).

Part II, no. 83: For the children, pp. 83 & 177. Trick version of (12, 4, 5), as in Family Friend (1858).

Dudeney. The world's best puzzles. Op. cit. in 2. 1908. He says (9, 10, 3) "is attributed to Sir Isaac Newton, but the earliest collection of such puzzles is, I believe, in a rare little book that I possess -- published in 1821." [This must refer to Jackson.] Says Rev. Mr. Wilkinson gave (11, 16, 3) "some quarter of a century ago" and that he, Dudeney, published (16, 15, 4) in 1897 (cf under 1902 above). He leaves these as problems but doesn't give their solutions in the next issue.

Wehman. New Book of 200 Puzzles. 1908.

P. 4: The practicable orchard. (16, 10, 4). = The Sociable, prob. 3.

P. 7: The puzzle of the stars. (9, 10, 3). = Magician's Own Book, prob. 33.

P. 8: The apple-tree puzzle. (10, 5, 4). = The Sociable, prob. 17.

P. 8: The peach orchard puzzle. (27, 10, 6). = The Sociable, prob. 22.

P. 8: The plum tree puzzle. (9, 10, 3). = The Sociable, prob. 36.

P. 12: The farmer's puzzle. (11, 11, 3). = The Sociable, prob. 9.

P. 19: The gardener's puzzle. (12, 6, 4) two ways. = The Sociable, prob. 26.

P. 26: The circle puzzle. (37, 20, 5) equally spaced along each row. = The Sociable, prob. 27.

P. 30: The tree puzzle. (15, 16, 3) with some bigger rows. = The Sociable, prob. 29.

P. 31: The geometrical orchard. (27, 9, 6). = The Sociable, prob. 12.

P. 31: The tulip puzzle. (19, 9, 5). = The Sociable, prob. 32.

P. 41: The florist's puzzle. (31, 6, 6) with seven circles of six. = The Sociable, prob. 8.

J. K. Benson, ed. The Pearson Puzzle Book. C. Arthur Pearson, London, nd [c1910, not in BMC or NUC]. [This is almost identical with the puzzle section of Benson, but has 13 pages of different material.] A symmetrical plantation, p. 99. (24, 28, 4).

Williams. Home Entertainments. 1914. Competitions with counters, p. 115. (11, 12, 3); (9, 10, 3); (10, 5, 4).

Dudeney. AM. 1917. Points and lines problems, pp. 56-58 & 189-193.

Prob. 206: The king and the castles. See The Captain, 1900.

Prob. 207: Cherries and plums. Two (10, 5, 4) patterns among 55 of the points of an 8 x 8 array.

Prob. 208: A plantation puzzle. (10, 5, 4) among 45 of the points of a 7 x 7 array.

Prob. 209: The twenty-one trees. (21, 12, 5).

Prob. 210: The ten coins. Two rows of five. Move four to make (10, 5, 4). Cf Carroll, 1876. Shows there are 2400 ways to do this. He shows that there are six basic solutions of the (10, 5, 4) which he calls: star, dart, compasses, funnel, scissors, nail and he describes the smallest arrays on which they can fit.

Prob. 211: The twelve mince-pies. 12 points at the vertices and intersections of a Star of David. Move four to make (12, 7, 4).

Prob. 212: The Burmese plantation. (22, x, 4) among the points of a 7 x 7 array. Finds x = 21. Cf 1899.

Prob. 213: Turks and Russians, pp. 58 & 191 193. Complicated problem leading to (11, 16, 3) -- cites his Afridi problem in Tit-Bits and attributes the pattern to Wilkinson 'some twenty years ago', cf 1908.

Blyth. Match-Stick Magic. 1921.

Four in line, p. 48. (10, 5, 4).

Three in line, p. 77. (9, 10, 3).

Five-line game, pp. 78-79. (21, 9, 5).

King. Best 100. 1927. No. 62, pp. 26 & 54. = Foulsham's no. 21, pp. 9 & 13. (10, 5, 4).

Loyd Jr. SLAHP. 1928. Points and lines puzzle, pp. 20 & 90. Says Newton proposed (9, 10, 3). Asks for (20, 18, 4) on a 7 x 7 array.

R. Ripley. Believe It or Not! Book 2. Op. cit. in 5.E, 1931. The planter's puzzle, p. 197, asks for (19, 9, 5) but no solution is given. See Clark, above, for a better version of the verse.

"I am constrained to plant a grove

For a lady that I love.

This ample grove is too composed;