Sources page probability recreations

Prob. 12, p. 61. O-(60, 100), D = 100. (Mikami 16 gives English.)

Prob. 13, pp. 62 63. Slower starts 10 ahead. After the faster goes 100, he is 20 ahead. When did he overtake?

Prob. 14, p. 63. Hare starts 100 ahead of hound. Hound runs 250 and is then 30 behind. How much further for hound to overtake? (Swetz; Was Pythagoras Chinese?; p. 21, gives this with numbers 50, 125, 30.)

Prob. 16, p. 64. Slower (the guest) goes 300 per day. Faster (the host) starts after ⅓ day, catches the slower and returns after ¾ of a day. How fast is the faster?

Prob. 20, p. 66. Birds can traverse a distance, M-(7, 9).

Prob. 21, pp. 66 67. M-(5, 7) with first delayed by 2.

Chap. VII.

Prob. 10, p. 74. Melon and gourd growing toward each other, MR (7, 10; 90).

Prob. 11, pp. 74 75. Rush grows 3, 3/2, 3/4, .... Sedge grows 1, 2, 4,. ... When are they equal? Solution of 2 6/13 = 2.4615 is obtained by linear interpolation between 2 and 3 days. (Correct solution is log₂ 6 = 2.5848.) (English in Mikami 18.)

Prob. 12, pp. 75 76. Two rats separated by a wall 5 thick. One gnaws 1, 2, 4, .... The other gnaws 1, 1/2, 1/4, .... When do they meet? Solution of 2 2/17 = 2.1176 obtained by linear interpolation between 2 and 3. (Correct answer is log₂ (2 + 6) = 2.1536, though Vogel has 8 instead of 6 in the radical.

Prob. 19, p. 79. Two horses travel 193, 206, 219, 232, ... and 97, 96 ½, 96, 95 ½, .... They set out for a city 3000 away. The faster gets there and then returns. When does he meet the slower? Solution of 15 135/191 = 15.7068 obtained by linear interpolation between 15 and 16. Correct answer is ( 227 + 147529)/10  =  15.7095.

Zhang Qiujian. Zhang Qiujian Suan Jing. Op. cit. in 7.E. 468, ??NYS. Analogous to above hare and hound with values 37, 145, 23. (English in Sanford 212, Mikami 41 and H&S 74.)

Aryabhata. 499. Chap. II, v. 30-31, pp. 72-74. (Clark edition: pp. 40-42.)

V. 30. One man has a objects and b rupees, another has c objects and d rupees, and they are equally wealthy. What is the value of an object? I.e. solve ax + b = cx + d. Though simple, this is the basis of many of the problems in this section, as discussed by Gupta (op. cit. under Bakhshali MS). Clark gives an example with a, b, c, d  =  6, 100, 8, 60.

V. 31. Shukla translates this as describing simple meeting and overtaking problems, MR-(a, b; D) and O-(a, b) with headstart D. Clark translates it as having two planets separated by D = s₁ - s₂ travelling with velocities v₁ and v₂, so they meet in time t  =  (s₁ - s₂) / (v₁ ± v₂). This is discussed in Gupta, loc. cit.

Bakhshali MS. c7C.

Kaye I 43-44 describes the following types in general terms and indicates that general solutions are stated in the MS. O-(a, b), T = t. O-(a, b; c, d). O-(a, b; c, 0). A problem equivalent to MR(a, b; 2D). Kaye I 49-52 describes problems of earning and spending which are equivalent to the following. O-(a, b), headstart 2D. O (a, b) headstart D. Cistern problems with several inlets -- see 7.H.

Kaye III 171, f. 3r, sutra 15 & Gupta. General rule for O-(a, b), headstart D. (Gupta cites Aryabhata as giving the same rule.)

Kaye III 175, f. 4r & Gupta. Examples of the previous with a, b, D = 5, 9, 7 and 18, 25, 8*18. (The second is really an example with the first starting T = 8 days ahead.)

Kaye III 189, f. 83r. O-(3/2, 2), D = 9.

Kaye III 216, ff. 60r-60v, sutra 52 & Gupta. Same as f. 3r, sutra 15, except that distance is replaced by wealth. Example has a man with wealth 30 earning at rate 5/2 and spending at rate 9/3. This can be considered as O (5/2, 9/3),  D = 30. This is also very like the simple versions of a snail climbing out of a well and could be considered as MR(9/3 - 5/2, 0; 30).

Kaye III 217-218, ff. 60v-61v, sutra 53 & Gupta & Hoernle, 1886, p. 146; 1888, p. 44. Two men earn at rates a, b, but the first gives c to the second at the beginning. When are they equally rich? Examples with a, b, c = 5/3, 6/5, 7; 13/6, 3/2, 10. This is equivalent to O-(a, b), D = 2c. See also: G. R. Kaye; The Bakhshāli manuscript; J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349 361; p. 360, Sutra 53 and example, but he has a + for a - in his formula.

Kaye III 186, f. 31r & Gupta. Another example of the above, with a, b, c  =  7/4, 5/6, 7.

Kaye III 172, f. 8r & Gupta. O-(a, b; c, 0), T = -e, i.e. a + (a+b) + ... + (a+(n 1)b)  =  c(n+e). In general this is a quadratic problem -- see below -- but when e = 0, a factor of n cancels leaving a linear problem. Example: O (2, 3; 10, 0).

Kaye III 174-175, ff. 7v & 4r & Gupta. O-(3, 4; 7, 0); O (1, 2; 5, 0).

Kaye III 173, f. 9r & Gupta. O-(1, 1; 10, 0).

Kaye III 176-177, ff. 4.r-5.r, sutra 18 & Gupta & in: G. R. Kaye, The Bakhshāli manuscript; J. Asiatic Soc. Bengal (2) 8:9 (Sep 1912) 349 361; p. 358. O (4, 3; 6, 1); O-(2, 3; 3, 2) and O-(5, 6; 10, 3). See also 7.AF.

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.

Kaye III 173, f. 9v. Travellers set out at rates a, b to a destination D away. The faster, on his return, meets the slower -- when? This is equivalent to MR (a, b; 2D). Example with a, b, D = 1, 6, 70.

Kaye III 190, f. 53v. MR-(96/18, 27/108; 9).

Kaye I 44-46; III 177-178, ff. 5r-6v. O-(3, 4; 5, 0), T = -6, which leads to a quadratic, but has answer 5. O-(5, 3; 7, 0), T = -5 is also given by Hoernle, 1886, pp. 128 129; 1888, p. 33, and Datta, p. 42. This also leads to a quadratic, but the answer is (7 + 889)/6 and the MS gives an approximation for the root. I have recorded that Kaye also has O-(3, 4; 5, 0) with no delay, but I can't find this.

Bhaskara I. 629. Commentary to Aryabhata, chap. II, v. 30-31. Sanskrit is on pp. 127-132; English version of the examples is on pp. 308-309.

V. 30, Ex. 1: 7x + 100 = 9x + 80.

V. 30, Ex. 2: 8x + 90 = 12x + 30.

V. 30, Ex. 3: 7x + 7 = 2x + 12.

V. 30, Ex. 4: 9x + 7 = 3x + 13.

V. 30, Ex. 5: 9x - 24 = 2 x + 18.

V. 31, Ex. 1: MR-(3/2, 5/4; 18).

V. 31, Ex. 2: O-(3/2, 2/3), D = 24.

Anania Schirakatzi (= Ananias of Shirak). Arithmetical problems. c640. Translated by: P. Sahak Kokian as: Des Anania von Schirak arithmetische Aufgaben; Zeitschrift für d. deutschösterr. Gymnasien 69 (1919) 112-117. See 7.E for description.