DH/dccmt

[{"id_text": "P010721", "project_name": "dccmt", "raw_text": "1 granary of grain. [1] man receives 7 sila. Its men are 45 (\u0161ar), 42 (sixties), 51. 3 sila of grain remain."}, {"id_text": "P010737", "project_name": "dccmt", "raw_text": "40 sons of builders (each) received 2 ban as a flour gift. (Total) 3 lidga, 1 barig, 2 ban of flour."}, {"id_text": "P010882", "project_name": "dccmt", "raw_text": "45 (\u0161ar), 36 (sixties). 1 man recieved 7 sila of grain from the granary."}, {"id_text": "P020479", "project_name": "dccmt", "raw_text": "The long side is 1 (sixty) 7 1/2 rods. (What is) the short side of 1 iku area? (Its short side is) 1 (sixty) rods, 5 cubits, 2 double-hands, 3 fingers (and) 1/3 finger. (random signs)"}, {"id_text": "P128042", "project_name": "dccmt", "raw_text": "The length is 3 22 rods, its breadth 6 cubits. 4 bur of <baked> bricks. Brought with the brick wall. Its 5th part ... there is.1 9 bur, 1 e\u0161e, 1 02 1/2 sar of unbaked bricks for a footing of height 10 cubits. 1 e\u0161e, 3 iku, 10 sar of baked bricks for a footing of height 10 cubits. First time. 1 bur, 4 1/2 iku, 6 sar of unbaked bricks for a footing of height 3 cubits. 2 iku, 8 1/3 sar of baked bricks for a footing of height 3 cubits. Second time. Total 4 bur, 1 e\u0161e, 5 iku, 18 1/3 sar of baked bricks. Total 10 bur, 1 e\u0161e, 5 iku, 18 1/2 sar of unbaked bricks. Temple of \u0160ara. 1 bur, 3 1/2 iku, 10 sar of unbaked bricks for a footing of height 3 cubits. 1 1/2 iku, 30 sar of baked bricks for a footing of height 3 cubits. First time. 1 e\u0161e, 3 iku of unbaked bricks for a footing of height 3 cubits. 1 iku, 4 1/2 sar of baked bricks for a footing of height 2 cubits. Second time. 52 rods, 2 cubits of brick wall. Its breadth is 6 cubits, its height 22 cubits. Its baked bricks are 1 bur, 1 00 sar. 1 08 rods, 2 cubits of brick wall. Its breadth is 5 cubits, its height 22 cubits. Its baked bricks are 1 bur, 2 iku, 22 sar. The length is 12 rods, 4 cubits, its breadth 6 cubits, its height 22 cubits. Its unbaked bricks are 2 e\u0161e, for the wall of the temple walkway. Total: 2 bur, 5 1/2 iku, 16 1/2 sar of baked bricks. Total: 2 bur 1 e\u0161e, [35] sar of unbaked bricks. Temple of Ninurra. Grand total: 6 bur, 2 e\u0161e, 4 1/2 iku 35 sar of baked bricks. Grand total: 12 bur, 2 e\u0161e, 5 1/2 iku, 28 1/2 sar of unbaked bricks. Bricks for the temple of \u0160ara and for the temple of Ninurra."}, {"id_text": "P128566", "project_name": "dccmt", "raw_text": "6 rods, 4 1/3 cubits length. 1/2 rod height. 2 cubits, 4 fingers width. Its volume is 3 1/2 sar, 2 1/2 shekels.1 Its bricks are 25 1/2 sar.2 6;31 503 34 0;10 505 "}, {"id_text": "P129726", "project_name": "dccmt", "raw_text": "1 05 rods length, at 2/3 sar volume for 1 rod - piling up an embankment. Its earth is 43 1/3 sar; its workers' wages are 4 20 for 1 day. 1 05 rods length, reed bundles on an embankment. A worker did a length of 1 rod in 1 day. Its workers' wages are 1 05 for 1 day. 5 iku - a worker hoed 2 1/2 sar in 1 day. Its workers' wages are 3 20 for 1 day in the first year. 5 iku - a worker hoed 3 sar in 1 day. Its workers' wages are 2 53 for 1 day in the second year. Ploughing for 1 day (unfinished)."}, {"id_text": "P212456", "project_name": "dccmt", "raw_text": "[1(?) mina of silver: <PN>] 5/6 [mina of silver]: [Dingir]-bani. 2/3 (mina) of silver, two-thirds: Dingir-ku 1/2 mina of silver: Dingir-azu 1/3 (mina) of silver, one-third: Dingir-muda 15 shekels of silver: Dingir-aha 1 shekel of silver: Dingir-aba [1(?)] mina of silver: <PN> [1(?) mina of silver <PN>, 5/6 mina of silver]: [Dingir]-bani. 2/3 (mina) of silver, two-thirds: Dingir-ku 1/2 mina of silver: Dingir-azu 1/3 (mina) of silver, one-third: Dingir-muda 15 shekels of silver: Dingir-aha 1 shekel of silver: Dingir-aba"}, {"id_text": "P212534", "project_name": "dccmt", "raw_text": "The short side is 13 1/2 rods width; its length is 1 (sixty) 30 rods.  <Its> area is 2 e\u0161e, 15 sar, in which the reaping (workrate) is 2 iku (a day). The short side is 13 1/2 rods; its length is 2 (sixties) rods. Its area is 1 bur (?) ..., in which the reaping (workrate) is 1 iku (a day)."}, {"id_text": "P212605", "project_name": "dccmt", "raw_text": "The short sides are 30 1/2 <rods> (and) 22 <rods>; the long sides are 5 (sixties) 30 <rods> (and) 5 (sixties) 30 <rods>.  Its area is 4 bur less 2 iku. Per 1 iku of area (there are) 1 gur 2 barig of grain. Its grain is 1 (sixty) 27 gur 2 barig. Amar-ezem (PN?)"}, {"id_text": "P213160", "project_name": "dccmt", "raw_text": "36 <rods> north, 1 (sixty) 50 <rods> east; 1 (sixty)-less-2 <rods> south, 50-less-2 <rods> west. Its area is 2 bur, 1 e\u0161e, 3 iku. Ur-I\u0161kur, Vinegar Meadow."}, {"id_text": "P213161", "project_name": "dccmt", "raw_text": "(The square side is) 11 rods, 1 seed-cubit, 1 cubit, 1 half-cubit. Its area of 1 1/4 iku, 2 1/2 sar, 6 shekels, 15 small shekels was found."}, {"id_text": "P213162", "project_name": "dccmt", "raw_text": "The square side is 1 (\u0161ar) 5 (sixties) <rods> less 1 seed cubit . Its area of 2 20 49 bur, 5 iku, 5 1/2 sar, 19 2/3 shekels is found."}, {"id_text": "P213163", "project_name": "dccmt", "raw_text": "The average long side is 2 (sixties) 40 <rods>. (What is) the short side of 1 iku area? Its short side is 3 seed-cubits, 1 cubit, 1 half-cubit."}, {"id_text": "P213165", "project_name": "dccmt", "raw_text": "The long side is 4 (sixties) 3 <rods>. (What is) the short side of 1 iku area? Its short side is to be found."}, {"id_text": "P214319", "project_name": "dccmt", "raw_text": "The length is 4 rods, the breadth 1 1/2 rods, the height 6 cubits. The length is 2 rods, [the breadth] 1 1/2 rods, [the height] 5 cubits. Mouth of the Thrown Throwstick River"}, {"id_text": "P214402", "project_name": "dccmt", "raw_text": "30 talents of silver; 10 talents of gold. 1 \u0161ar wild cattle; 1 \u0161ar donkeys; 1 \u0161ar ...; 1 \u0161ar ... Ur-... (PN)."}, {"id_text": "P215323", "project_name": "dccmt", "raw_text": "The lower long side is 1 (sixty) 55 rods; the upper long side is 1 (sixty) 10 rods. The upper short side is 1 (sixty) 6 rods; the lower <short side> is 56 rods. 2 e\u0161e, 1 1/2 iku area."}, {"id_text": "P215324", "project_name": "dccmt", "raw_text": "The lower <long side> is 8 (sixties) 25 <rods>; the upper <short side> is 1 (sixty) 25 rods, 2 seed-cubits; the long side is 9 (sixties)-less-7 (?) <rods>; the short side is 52 <rods>."}, {"id_text": "P215326", "project_name": "dccmt", "raw_text": "The long side is 1 (sixty) 3 1/2 rods; the short side by the watercourse is 22 1/2 <rods>; the lower long side is 1 (sixty) 30 <rods>; the short side (reached by) irrigation is 22 1/2 <rods>. The harrowed area is 2 e\u0161e, 1 1/2 iku. Foxy the felter."}, {"id_text": "P215434", "project_name": "dccmt", "raw_text": "The square side is 1 (\u0161argal), 4 (sixties) rods, 4 seed-cubits. (blank line) The square side is 1 (\u0161ar) 1 (sixty), 32 rods, 1 seed-cubit. Ur-I\u0161taran. The area of 7 47 17 bur, 1 e\u0161e, 3 1/2 iku, 10 sar, 16 2/3 shekels is found."}, {"id_text": "P215972", "project_name": "dccmt", "raw_text": "The short sides are 1 (sixty) <rods> (and) 44 1/2 <rods>; the average long side is 5 (sixties) 50 <rods>. Its area is 10 bur. Sarruturi, Field of the Grand Vizier."}, {"id_text": "P217125", "project_name": "dccmt", "raw_text": "East 16 <rods> (and) 20-less-3 <rods>; north 37 <rods> (and) 31 1/2 <rods>. Its area is (blank). Barran the \u0161abra. North 40 (and) 25 1/2; east 30 (and) 25 1/2. Its area is (blank). Uruna the \u0161abra. 4 1/2 iku area ... Dudu ...."}, {"id_text": "P217673", "project_name": "dccmt", "raw_text": "The square side is 33 (sixties) 33 rods, 1 half-reed. The area of 1 11 27 bur, 5 1/2 iku, 1/2 sar, 3 2/3 shekels, 5 <small> shekels is found."}, {"id_text": "P217674", "project_name": "dccmt", "raw_text": "The square side is 28 (\u0161ar) <rods> less 1 double-hand. (Assigned to) Meluhha. The square side is 1 (\u0161ar) 5 (sixties) <rods> less 1 seed-cubit. (Assigned to) Ur-I\u0161taran."}, {"id_text": "P218053", "project_name": "dccmt", "raw_text": "The long side is 9 (sixties) <rods>. (What is) the short side of 1 iku area? 2 cubits, 6 2/3 fingers."}, {"id_text": "P222256", "project_name": "dccmt", "raw_text": "[1] cubit square side: 1 small mina, 15 shekels (area). 2 cubits square side: 2 shekels, less 1 <small> mina. [3 cubits] square side: [4] shekels, less a 4th part. 4 cubits square side: 6 shekels, 2 <small> minas. 5 cubits square side: 10 [1/3] shekels, 15 <small minas> 6 cubits square side: 1[5] shekels. 7 cubits square side: 1/3 sar, 1/3 <shekel>, 5 <small minas>. 8 cubits square side: 1/2 sar less 3 shekels, 1 small mina. 10-less-1 cubits square side: 1/2 sar, 4 shekels, less a 4th part. 10 cubits square side: 2/3 sar, 2 shekels, less 1 small mina. 11 cubits square side: 1 sar less 10 shekels, (plus) 1 <small mina>, 15 small shekels. 3 reeds square side: 2 sar, 15 shekels. Nammah the accountant wrote it."}, {"id_text": "P240964", "project_name": "dccmt", "raw_text": "3 gubar, 4 anzam (of grain): 4 hundred gubar. 30 gubar 6 sila, 4 anzam: 1 thousand men. 3 hundred, 3 gubar, 4 anzam: 1 myriad men. 3 thousand, 30 gubar, 6 sila, 4 anzam: 1 hundred thousand men. 6 thousand 6 1/2 gubar, 2 sila, 2 anzam: 2 hundred thousand men. 1 thousand 8 hundred 20-less-2 gubar, 4 sila: 6 myriad men. Total: 7 thousand, 8 hundred, 1 (sixty), 20-less-1 gubar of grain. 2 hundred thousand, 6 myriad men."}, {"id_text": "P241144", "project_name": "dccmt", "raw_text": "1 (sixty) less 1 minas, 1/2 shekel of copper, 7 minas, 1/2 shekel of tin. Steward of the assembly. 1 hundred helmets. 1/2 shekel, 5 1/2 dilmun shekels of copper, 4 1/2 dilmun shekels of tin. Steward of the assembly. 1 helmet."}, {"id_text": "P250375", "project_name": "dccmt", "raw_text": "The long side is 2 rods. <What is> the short side? <The area is> 2 less 1/4 iku. In it you put a double-hand, a 6th part of it (a seed-cubit). You put a 4th part of it (a seed-cubit). It was found. Its short side is 5 seed cubits, 1 double-hand, [5] fingers. (random signs?)"}, {"id_text": "P254390", "project_name": "dccmt", "raw_text": "O Nisaba! Length and width. I combined length and width and then I made an area.  I turned around. I added as much as the length exceeded the width to the middle of the area so that (it was) 3 03.  I returned. I summed the length and width and (it was) 27.  What are the length, width, and area? 27 and 3 03, sums 15,\tlength 3 00, area 12, width You, when you proceed: add 27, the sum of the length and width, to the middle of [3 03] so that (it is) 3 30.  Add 2 to 27 so that (it is) 29.  You break off half of 29 and then 14;30 times 14;30 is 3 30;15.  You take away 3 30 from the middle of 3 30;25 so that the remainder is 0;15. 0;15 squares 0;30.  Add 0;30 to one 14;30 so that the length is 15.  You take away 0;30 from the second 14;30 and the width is 14.  You take away the 2 that you added to 27 from 14, the width, so that the true width is 12. I combined 15, the length, and 12, the width and then 15 times 12 is 3 00, the area.  By what does 15, the length exceed 12, the width?  It exceeds by 3. Add 3 to the middle of 3 00, the area. The area is 3 03.  Length and width. I combined length and width and then I built an area.  I turned around. I added half of the length and a third of the width to the middle of my area so that (it was) 15.  I returned. I summed the length and width and (it was) 7.  What are the length and width? You, when you proceed: you write down [2], the writing of a half, [and] 3, the writing of a third, and then you solve the reciprocal of 2 and 0;30 times 7 is 3;30. I multiply <0;30> by 7, the sums (sic) of length and width, and then take away 3;30 from 15, my sums, so that the remainder is 11;30.  Go no further. I combine 2 and 3 \u2014 3 times 2 is 6 \u2014 and then the reciprocal of 6 gives you 0;10.  I take away 0;10 from [7], your sums of the length and width, so that the remainder is 6;50.  You break off half of 6;50 so that it gives you 3;25.  You write down 3;25 twice and then \u2014 3;25 times 3;25 \u2014 I take away 11;40 15 from the middle of 11;30 (sic) so that the remainder is 0;10 25. <10;25 squares 0;25.> You add 0;25 to one 3;25 so that (it is) 3;50.  And you add that which I took away from the sum of the length and width to 3;[50] so that the length is 4.  I take away 0;25 from the second 3;25 so that the width is 3. <15,> 7, sums 4, length 12, area 3, width Length and width. I combined length and width and then I built an area.  I turned around and then I combined as much as the length exceeded the width with the sum of the length and my [width] and then I added it to the middle of my area so that (it was) 1;13 20.  I returned. I summed the length and width [so that] it was 1;40. 1;40,\t1;13 20, sums 1, length 0;40, area 0;40, width You, when you proceed: <combine> 1;40, the sum of the length and width \u2014 1;40 times 1;40 is 2;46 40.  You take away 1;13 20, the area, from 2;46 40 so that (it is) 1;33 20.  Go no further. You break off half of 1;40 and then \u2014 0;50 times 0;50 \u2014 you add 0;41 40 to 1;33 20 and then 2;15 squares 1;30.  By what does 1;40 exceed 1;30? It exceeds by 0;10.  Add 0;10 to 0;50. The length is 1.  Take away 0;10 from 0;50 so that the width is 0;40. Length and width. I combined the length and width \u00abthe length and width\u00bb and then I built an area.  I returned. I summed the length and width so that it was square with the area.  I summed the length, width, and area so that (it was) 9.  What are the length, width and area? For a length of three ropes one man carried 9 sixties of bricks here and then I gave him 2 s\u016btu of grain.  Now, the builder has had me provide (for them) and so I called for 5 labourers.  And then one carried one part of it to me, the second twice it, the third three times it, the fourth four times it, the fifth five times it.  How many of the bricks the first one carried to me did he entrust to me, and then how much grain did I give him? 30 , \t36 1 20, \t9 1 ,\t1 12 , 2 40 2 ,\t1 48\t, 4 3 ,\t2 24\t, 5 20 4 , \t3\t, 6 40 5 ,\t<3 36>\t, <8> 20 grains For three ropes one man carried 9 sixties of bricks here and then I gave him 2 s\u016btu grain.  Now, the builder has had me provide (for them) and so I called for 4 labourers.  And then the first one carried a seventh part of it, the second an eleventh of it, the third a thirteenth of it, the fourth a fourteenth of it.  How many bricks did he entrust to me, and then how much grain did I give him? 30 ,\t9 7 ,\t1 24 ,\t3 06 40 11 ,\t2 12 , <4> 53 20 13 ,\t2 36 ,\t5 46 40 14 ,\t2 48 ,\t6 [1]3 20 1 30 ,\t20 ,\tgrain If (they ask you) \u2018For a length of three ropes I added the bricks, the labourers [and] my days, so that (it was) 2 20.  My days were (equal to) two-thirds of my workers.  Select the bricks, the labourers, and my days for me.' [6]       ,             2\t, 2 20 work rate of a day 1       ,              30\t, \t6 one labourer, 30 work rates, 30 labourers 1 30 bricks 40      ,             20\t, \t20 days. I summed the areas, my volume, [and] the brick, so that (it was) 12;30 02.  I converted [...] into a depth. [...] my earth."}, {"id_text": "P254399", "project_name": "dccmt", "raw_text": "(A field.) ... \u00ablength\u00bb. The upper <length> is 2 15, the lower length 1 21, [the upper] width 3 33, the lower width 51. There are 6 brothers. The oldest and the following are equal; 3 and 4 are equal; 5 and 6 are equal. What are the boundaries: the diagonals and the descenders? You, when you proceed: sum 3 33, the upper width, and 51, the lower width, so that 4 24 results. Return, and then solve the reciprocal of 2 15, the <upper> length, so that 0;00 26 40 results. Multiply 0;00 26 40 by 1 21, the <lower> length, so that 0;36 results. Add 0;36 to 4 24 \u00ablower\u00bb, so that 4 24;36 results. Return, and then sum 2 15, the upper length and 1 21, the lower length, so that 3 36 results. Break 3 36, so that 1 48 results. Solve the reciprocal of 1 48, so that 0;00 33 20 results. Multiply 0;00 33 20 by 4 24;36, so that 2;27 results. The 2nd diagonal is 2 27 (sic). Return, and then by what does 2 15, the upper length, exceed 1 31, the lower length? It exceeds by 54. Take away 54 from 2 27, the 2nd diagonal, so that the remainder is 1 33. \u00abThe remainder\u00bb The 4th diagonal is 1 33. Return, and then combine 3 33, the upper width. 12 36 09 results. Return, and then combine 2 27, the second diagonal, so that 6 00 09 results. Sum 6 00 09 and 12 36 09, so that 18 36 18 results. Break 18 36 18, so that 9 18 09 results. Make its square-side come up, so that 3 03 results. The upper diagonal is \u00ab2\u00bb <3> 03.  Return, and then [combine] 2 27, [the 2nd] diagonal, [so that] 6 00 09 results. Return, and then [combine] 1 33, the 4[th] diagonal, [so that] 2 24 09 results. [Sum] 2 24 09 and 6 00 09 [so that] 8 24 18 results. Break 8 24 18, so that 4 12 09 results. Make its square-side come up, so that 2 03 results. The 3rd diagonal is 2 03.  Return, and then combine 1 33, the 4th diagonal, so that 2 24 09 results. Return, and then combine 51, the lower width, so that 43 21 results. sum 43 21 and 2 24 09, so that 3 07 30  <results>. Break 3 07 30, so that 1 33 45 results. Make its square-side come up, so that 1 15 results. The 5th diagonal is 1 15.  Return, and then by what does 3 33, the <upper> width, exceed 51, the lower width? It exceeds by 2 42. Solve the reciprocal of 2 42, so that 0;00 22 13 20 results. Return, and then by what does 3 33 exceed 3 03, the <upper> diagonal? It exceeds by 30. Multiply 30 by 0;00 22 13 20, so that 0;11 06 40 results. Multiply 0;11 06 40 by 2 15, the upper length, and 1 21, the lower length, so that 25 and 15 result. The upper desc<ender> is 25, the lower desc<ender> is 15.  Return, and by how much does 3 33, the upper diagonal, exceed 2 27, the 2nd diagonal? it exceeds by 36. Multiply 36 by 0;00 22 13 20, so that 0;13 20 results. Multiply 0;13 20 by 2 15, the <upper> length, and 1 21 the <lower> length, so that 30 and 18 result. The 2nd upper desc<ender> is 30, the 2nd lower desc<ender> is 18. Return, and by how much does 2 27, the second diagonal, exceed 2 03, the 3rd diagonal? [It exceeds by] 24. Multiply 24 by 0;00 22 13 20, so that 0;08 53 20 [results]. Multiply [0;08 53 20] by 2 15, the <upper> length and 1 [21, the <lower> length], so that 20 and 12 result. The [upper] desc <ender> is 20, the lower [descender is 12]. Return. You make the 3 remaining desc<enders> like [the previous ones]. That is the procedure."}, {"id_text": "P254406", "project_name": "dccmt", "raw_text": "I summed the area and my square-side and it was 0;45.  You put down 1, the projection. You break off half of 1. You combine 0;30 and 0;30.  You add 0;15 to 0;45. 1 squares 1. You take away 0;30 which you combined from inside 1 so that the square-side is 0;30. I took away my square-side from inside the area and it was 14 30.  You put down 1, the projection. You break off half of 1. You combine 0;30 and 0;30.  You add 0;15 to 14 30. 14 30;15 squares 29;30. You add 0;30 which you combined to 29;30 so that the square-side is 30. I took away a third of the area. I added a third of the square-side to inside the area and it was 0;20.  You put down 1, the projection. [You take away] a third of 1, the projection, [and] you multiply 0;40 by 0;20.  You write down 0;13 20. You break [half of 0;20], the third which you added. You combine 0;10 and 0;10, You add 0;01 40 to 0;13 20.  0;15 [squares] 0;30. You take away [0;10 which you combined from inside 0;30] and (it is) 0;20.  The reciprocal of 0;40 [is 1;30. You multiply by 0;20 and] the square-side is [0;30]. [I took away] a third [of the area]. I summed [the area and] my square-side: it was 4 46;40.  You put down [1, the projection]. You take away 0;20, a third of 1, the projection, and you multiply 0;40 by 4 46;40, and  [you write down] 3 11;[06 40]. You break [in half] 1, the projection. You [combine] 0;30 and 0;30. You add [0;15 to 3 11;06 40].  3 11;21 40 squares 13;50. [You take away 0;30] which you combined from [inside 13;40 and <(it is) 13;20. The reciprocal of 0;40 is> 1;30. You multiply [by] 13;20 and the square-side is 20. [I summed the area and my square-side and a third] of my square-side [and it was 0;55].  You put down [1, the projection]. You add a third of [1, the projection to 1]: 1;20. [You combine] its half, 0;40, [and 0;40]. You add 0;26 40 to 0;55 and  [1;21 40 squares 1;10. Take away 0;40 that you] combined from the middle of 1;10 and the square-side is 0;30]. [I summed the area and two-thirds] of my square-side [and it was 0;35].  You put down [1, the projection]. Two-thirds of [1, the projection] (is) 0;40. You combine [its half, 0;20 and] 0;20.  [You add 0;06 40 to 0;35 and] 0;41 40 squares 0;50. You] take away 0;20 that you combined from the middle of 0;50] and the square-side is 0;30. [I summed my square-side seven times and the area] eleven times [and it was 6;15].  You write down [7 and 11. You multiply] 11 by 6;15 [and (it is) 1 08;45]. You break off [half of 7. You combine] 3;30 and 3;30.  You add [12;15 to 1 08;45] and [1 21 squares 9. You take away 3;30 that you] combined from the middle of 9 [and you write down 5;30. The reciprocal of] 11 cannot be solved.  [What should I put down by 11 that] will give me [5;30]? [Its quotient is 0;30. The square side is 0;30.] [I summed the areas of my two square-sides and] (it was) 0;21 40. [And I summed my square-sides and it was 0;50].  You break [off half of 0;21] 40. [You write down 0;10 50. You break off half of 0;50.] You combine [0;25 and 0;25].  [You take away 0;10 25 from the middle of 0;10 50 and 0;00 25] squares [0;05].  You add 0;05 to the first 0;25 [and the first square-side is 0;30].  You take away 0;05 from the second 0;25 and [the second square-side is 0;20]. I summed the areas of my two square-sides and it was 0;21 40.  A square-side exceeds the (other) square-side by 0;10.  You break off half of 0;21 40 and you write down 0;10 50.  You break off half of 0;10 and you combine 0;05 and 0;05.  You take away 0;00 25 from the middle of 0;10 50 and 0;10 25 squares 0;25.  You write down 0;25 twice. You add 0;05 that you combined to the first 0;25 and the square-side is 0;30.  You take away 0;05 from the middle of the second 0;25 and the second square-side is 0;20. I summed the areas of my two square-sides and (it was) 21;15.  A square-side is less than the (other) square-side by a seventh.  You put down 7 and 6. You combine 7 and 7: 49.  You combine 6 and 6. You sum 36 and 49 and (it is) 1 25. The reciprocal of 1 25 cannot be solved. What should I put down by 1 25 that will give me 21;15? 0;15 squares 0;30.  You multiply 0;30 by 7 and the first square-side is 3;30.  You multiply 0;30 by 6 and the second square-side is 3. I summed the areas of my two square-sides and (it was) 28;15.  A square-side exceeds the (other) square-side by a seventh.  You write down 8 and 7. You combine 8 and 8: 1 04.  You combine 7 and 7. You sum 49 and 1 04: 1 53.  The reciprocal of 1 53 cannot be solved. What should I put down by 1 53 that will give me 28;15? 0;15 squares 0;30.  You multiply 0;30 by 8 and the first square-side is 4.  You multiply 0;30 by 7 and the second square-side is 3;30. I summed the areas of my two square-sides and (it was) 0;21 40.  I combined my two square-sides and (it was) 0;10.  You break off half of 0;21 40 and you combine 0;10 50 and 0;10 50.  It is 0;01 57 \u00ab46\u00bb <21> 40. You combine 0;10 and 0;10. You take away 0;01 40 from 0;01 57 \u00ab46\u00bb <21> 40 and 0;00 17 \u00ab46\u00bb <21> 40 squares 0;04 10.  You add 0;04 10 to the first 0;10 50 and 0;15 squares 0;30. The first square-side is 0;30.  You take away 0;04 10 from the middle of the second 0;10 50 and 0;06 40 squares 0;20. The second square-side is 0;20. I summed the areas of my two square-sides and (it was) 0;28 20.  A square-side was a quarter of the (other) square-side.  You write down 4 and 1. You combine 4 and 4: 16.  You combine 1 and 1. You sum 1 and 16 and (it is) \u00ab16\u00bb <17>.  The reciprocal of 17 cannot be solved. What should I put down by 17 that will give me 0;28 20? 0;01 40 squares 0;10.  You multiply 0;10 by 4 and the first square-side is 0;40.  You multiply 0;10 by 1 and the second square-side is 0;10. I summed the areas of my two square-sides and (it was) 0;25 25.  A square-side was two-thirds of the (other) square-side [and 0;05] rods.  You write down 1 and 0;40 and 0;05 [over] 0;40.  [You combine] 0;05 and 0;05.  [You take away 0;00 25 from the middle of 0;25 25 and you write down 0;25. You combine 1 and 1: 1. You combine 0;40 and 0;40. You sum 0;26 40 and 1 and you multiply 1;26 40 by 0;25 and you write down 0;36 06 40.] You [multiply 0;05 by] 0;40 and you multiply 0;03 20 by 0;03 20. You add 0;00 11 06 40] to 0;36 06 40 [and 0;36 17 46 40 squares 0;46 40. You take away 0;03] 20 that you combined [from the middle of 0;46 40] and you write down 0;43 20. [The reciprocal of 1;26 40 cannot] be solved. What [should I put down] by 1;26 40 [that] will give me [0;43 20]? Its quotient (?) is 0;30. [You multiply 0;30 by 1 and] the first square-side is [0;30. You multiply 0;30 by 0;40 and] you sum [0;20] and [0;05] and the second square-side [is 0;25]. I summed [the areas of my four] square-sides and (it was) 0;28 05. [A square-side was two-thirds], a half, a third of the (other) square-side(s). You write down [1 and 0;40 and 0;30 and 0;20]. You combine 1 and 1: <it is> 1. [You combine 0;40 and 0;40]: it is [0;26] 40. You combine 0;30 and 0;30: it is 0;15. [You combine 0;20 and 0;20. You sum 0;06] 40 and 0;15 and 026 40 and 1. [The reciprocal of 1;48] 20 cannot be solved. [What] should I put down [by 1;48 20] that will give me 0;27 05? [0;15 squares 0;30.] You multiply [0;30 by 1] and the first square-side is 0;30. [You multiply 0;30 by 0;40] and the second square side is 0;20. [You multiply 0;30 by 0;30] and the third square-side is 0;15. [You multiply 0;30 by 0;20 and] the fourth square-side is 0;10. I took away [a third of the square-side] from inside the area and (it was) 0;05. [You put down 1, the projection.] A third of 1, the projection is 0;20. You break off [half of 1, the projection]. You multiply 0;30 by 0;20 and (it is) 0;10. [You] combine [0;10 and 0;10]. You add 0;01 40 to 0;05 and [0;06 40 squares 0;20]. You add 0;10 that you combined to 0;20 and the square-side is 0;30. I summed [the areas of] my three square-sides and (it was) 10 12;45. A square-side was a seventh of the (other) square-side. You write down 49 and 7 and 1.[You] combine 49 and 49: <it is> 40 01. You combine 7 and 7: it is 49. You combine 1 and 1: <it is> 1. You sum 10 01 and 49 and 1 and (it is) 40 51. The reciprocal of 40 51 cannot be solved. What should I put down by 40 51 that will give me 10 12;45? Its half is 0;15. 0;15 squares 0;30. You multiply 0;30 by 49 and the first square-side is 24;30. You multiply 0;30 by 7 and the second square-side is 3;30. You multiply 0;30 by 1 and the third square-side is 0;30. I summed the areas of my three square-sides and (it was) 0;23 20. A square-side exceeds the (other) square-side by 0;10. You multiply the 0;10 which exceeds by 1. You multiply 0;10 by 2. <You combine> 0;20 and 0;20: it is 0;06 40. You combine 0;10 and 0;10. You add 0;01 40 to 0;06 40. You take away 0;08 20 from the middle of 0;23 20 and you multiply 0;15 by 3, the square-sides. You write down 0;45. You sum 0;10 and 0;20 and you combine 0;30 and 0;30 You add 0;15 to 0;45 then 1 squares 1. You take away 0;30 that you combined and you write down 0;30. The reciprocal of 3, the square-sides, is 0;20. You multiply by 0;30. The square-side is 0;10. You add 0;10 to 0;10 and the second square-side is 0;20. You add 0;10 to 0;20, then the third square-side is 0;30. I combined the square-sides then I summed (it and) the area. I combined as much as a square-side exceeds the (other) square-side with itself. I took it away from [the middle of the area] and (it was) 0;23 20. I [summed] my square-sides [and (it was) 0;50]. You copy 0;23 20 twice. You write down 0;46 40. [You combine 0;50 and 0;50. You take away 0;41 40 from the middle of 0;46 40 and the reciprocal of 0;05 is 12. You multiply 0;05 by 0;05. 0;00 25 squares 0;05. You break off half of 0;50. You add 0;25 to 0;05, then the first square-side is 0;30. You take away 0;05 from the middle of 0;25, then the second square-side is 0;20.] An area. I added four widths and the area and (it was) 0;41 40. You write down 4, the four sides. The reciprocal of 4 is 0;15. You raise 0;15 by 0;41 40, and (it is) 0;10 25. You write it down. You add 1, the projection and 1;10 25 squares 1;05. You take away 1, the projection, which you added and you copy 0;05 twice and 0;10 rods squares itself. I summed the areas of my three square-sides and (it was) 0;29 10. A square-side is (equal to) two-thirds of the (other) square-side and 0;05 rods, (which is equal to) half the (third) square-side and 0;02 [30] rods. You write down 1 and 0;40 and 0;20 <and> 0;05 over 0;40. You write down 0;02 30 over 0;20. You break off half of 0;05. You add 0;02 30 to 0;02 30. You combine 0;05 and 0;05. You write down 0;00 25. You combine 0;05 and 0;05. You add 0;00 25 to 0;00 25 and you take away \u00ab0;25 25\u00bb <0;00 50> from the middle of 0;29 10. You write down 0;03 45. You combine 1 and 1: 1. You combine 0;40 and 0;40: 0;26 40. You combine 0;20 and 0;20. You sum 0;06 40 and 0;26 40 and 1 and you multiply 1;22 30 by 0;03 45 and (it is) 0;05 50. You multiply 0;40 by 0;05: 0;03 20. You multiply [0;20] by 0;02 30 <added to half of 0;05>: 0;00 50. You sum 0;03 20 and 0;50 and you combine 0;04 10 and 0;04 10. You add 0;00 17 21 40 to 0;05 50 and 0;06 07 21 40 squares 0;19 10. You take away 0;04 10 from the middle of 0;19 10 and you \u00abcopy 0;15 twice\u00bb <multiply by the reciprocal of 1;33 40>. You multiply 0;30 by 1 and the first square-side is 0;30. You multiply 0;30 by 0;40 and you add 0;20 to 0;05 and the second square-side is 0;25. You break off half of 0;25 and you add 0;12 30 to 0;02 30 and the third [square-side] is 0;15."}, {"id_text": "P254407", "project_name": "dccmt", "raw_text": "The square-side is 1 cable. I extended a border on each side and then I drew a second square-side. Inside the square-side I drew a circle. What are their areas? The square-side is 1 cable. I extended a border on each side and I drew a circle. What is its area? The square-side is 1 cable. Inside it I drew a square-side and a circle. The circle that I drew touched the square-side. What are their areas? [The square-side is 1 cable.] <I drew> a second [square-side. Inside the second square-side I drew] 4 wedges and 1 circle. What are their areas? [The square-side is 1 cable. <I drew> a second square-side. Inside the second square-side] I drew [4 squares and 1 circle.] What are their areas? The square-side is 1 cable. Inside it I drew a second square-side. The square-side that I drew touches the outer square-side. What is its area? The square-side is 1 cable. Inside it <I drew> 4 wedges and 1 square-side. The square-side that I drew touches the second square-side. What is its area? The square-side is 1 cable. Inside it I drew a square-side. The square-side that I drew touches the <first> square-side. Inside the second square-side I drew a third square-side. <The square-side> that I drew touches the <second> square-side. What is its area? The square-side is 1 cable. Inside it I drew 8 wedges. What are their [areas]? The square-side is 1 cable. Inside it I drew a square-side. The square-side that I drew touches the <first> square-side. Inside the <second> square-side I drew 4 wedges. <What are their areas?> The square-side is 1 cable. Inside it [I drew] 16 wedges. What are their areas? The square-side is 1 cable. Inside it I drew 4 ox-brows and 2 wedges. What are their areas? The square-side is 1 cable. I extended half on each side and I drew a square-side. Inside the second square-side I drew a third square-side. What is its area? The square-side is 1 cable. (Inside it) I drew 12 wedges and 4 squares. What are their areas? The square-side is 1 cable. Inside it I drew 4 wedges. What are their areas? The square-side is 1 cable. Inside it <I drew> 4 squares, 4 rectangles and 4 wedges. What are their areas? The square-side [is 1 cable]. Inside it I drew 16 squares. What are their areas? The ... is 1 cable. [...] ... the width. The square-side is 1 [cable]. I extended a border on each side and I drew a square-side. Inside the square-side that [I drew is] 1 concave square. What is [its area]? The square-side is 1 cable. I extended a border on each side and I drew the shape of a lyre. What is its area? The square-side is 1 cable. [Inside it] are 2 crescent moons, [1] wedge, 1 cone, 1 rectangle and 4 squares. What are their areas? The square-side is [1] cable. Inside [it are 2] rectangles, [1] oval and 4 squares. What are their areas? The square-side is 1 cable. Inside it are 3 bows and 1 rectangle. What are their areas? The square-side is 1 cable. Inside it are 2 bows, 1 barge, and 4 ox-brows. [...] What are their [areas]? The square-side is 1 cable. Inside it are 1 circle and 6 crescent moons. What are their areas? The square-side is 1 cable. <Inside it are> 2 circles, 2 crescent moons and 4 squares. What are their areas? The square-side is 1 cable. <Inside it are> 4 wedges, 16 barges, 5 concave squares. What are their areas?"}, {"id_text": "P254440", "project_name": "dccmt", "raw_text": "[...] ... [...] ... [...] the procedure.  [...] what? [...] ... [...] you will see. [...] Square 2. [...] Multily [3 by] 2. The height is 6. [That is] the procedure. [...] what? [...] h1 Problem (i') [...]. What are [the length and width]? [...] You will see 3. Break 3 in half. You will see 1;30. [... Release the reciprocal of 1;30.] You will see 0;40. The ratio of the width. Release the reciprocal of 12, the ratio of the depth. [You will see 0;05.] Multiply 0;05 by 1. You will see 0;05. Multiply by 0;40. You will see 0;03 20. Multiply 0;03 20 by 0;05. You will see 0;00 16 40. Find the reciprocal of 0;00 16 40. You will see 3 36. Multiply 3 36 by 1;10. You will see 4 12. The square-side is 6. Multiply 6 by 0;05. The length is 0;30. Multipy 6 by 0;03 20. The width is 0;20. Multiply 6 by 1. You will see 6, the depth. That is the procedure. h1 Problem (ii') An excavation. The length is equal to the depth. I removed 1, the volume. I summed my ground-area and the volume: 1;10. The (sum of the) length and width is 0;50. What are the length and width? You: Multiply 0;50 by 1, the ratio. You will see 0;50. Multiply 0;50 by 12. You will see 10. Square 0;50. You will see 0;41 40. Multiply by 10. You will see 6;56 40. Find its reciprocal. You will see 0;08 38 44. Multiply by 1;10. You will see 0;10 04 48. The square-sides are 0;36, 0;24, 0;42. Multiply 0;36 by 0;50. The length is 0;30. Multiply 0;24 by 0;50. The width is 0;20. <Multiply> 0;36 by 10. The depth is 6. The procedure. h1 Problem (iii') An excavation. The depth is as much as the length. I removed 1, the volume. I summed my ground-area and volume: 1;10. The length exceeded the width by 0;10. You: Put down 1 and 12, the ratios. Multiply 0;10, the excess, by 1. You will see 0;10. Multiply by 12. You will see 2. Square 0;10. You will see 0;01 40. Multiply by 2. You will see 0;03 20. Find the reciprocal of 0;03 20. You will see 18. Multiply by 1;10. You will see 21. The square-sides are 3, 2, 21. Multiply [0;10 by 3]. The length is 0;30. Multiply 0;10 by 2. The width is 0;20. Multiply 3 by 2. You will see 6. The depth is [6]. The procedure. h1 Problem (iv') An excavation. The length is as much as the depth. I removed a volume.I summed my ground-area and the volume: 1;10. The length is 0;30. What is the width? You: multiply 0;30, the length, by 12. You will see 6, the depth. Add 1 to 6. You will see 7. The reciprocal of 7 cannot be found. What should I put to 7 that will give me 1;10? Put 0;10. Solve the reciprocal of 0;30, the length. You will see 2. Multiply 0;10 by 2. You will see 0;20, the width. The procedure. h1 Problem (v') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume: 1;10. The width is 0;20. <What is> the length? You: multiply 0;20 by 12. You will see 4. Multiply 4 by 1;10. You will see 4;40. Break in 1/2 0;20, the width. You will see 0;10. Square 0;10. You will see 0;01 40. Add to 4;40. You will see 4;41 40. The square-side is 2;10. Subtract 0;10 that you squared and you will see 2. Find the reciprocal of 4. you will 0;15. Multiply by 2. You will see 30, the length. The procedure. h1 Problem (vi') [...] add [...] 4;30, 53;20, 1;45 ... [...] The length is 0;30. Multiply 0;22 30 by 0;53 20. The width is 0;20. [...] The procedure. An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume. I took a 7th. I added (it) to my ground-area and it was 0;20. [The length is] 0;30. [What is the width?] You: multiply 0;30 by 12. You will see 6, the depth. [Add] 1 to 6. You will see 7. Take a 7th. You will see 1. Sum 1 and 1. You will see 2. Find the reciprocal of 2. You will see 0;30. Multiply 0;30 by 0;20, the sum.  You will see 0;10. Find the reciprocal of 0;30, the length. You will see 2. Multiply 2 by 0;10. [The width is 0;20.] The procedure. h1 Problem (vii') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume: 1;10 I took a 7th of it. I added (it) to my ground-area: 0;20. The width was 20. You: multiply 0;20 by 7. You will see 2;20. Multiply 0;20, the width, by 12.  You will see 4. Multiply 4 by 2;20. You will see 9;20. Add 1 to 7. You will see 8. Multiply 8 by 0;20. You will see 2;40. Break 1/2 of 2;40. [Square (it).] You will see 1;46 10. Add to 9;20. You will see 11;06 40. The square-side is 3;20. Subtract the 1;20 that you squared. You will see 2. Find the riciprocal of 4. You will see 0;15. Multiply 0;15 by 2. [The length is] 0;20. The procedure. h1 Problem (viii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as that by which the the reciprocal exceeds its reciprocal pair. I removed [a volume] of 16. What are the length, width, and depth? You: find the reciprocal of 12. You will see [0;05]. Multiply 0;05 by 16. You will see [1];20. The reciprocal is 1;20. [Find] the reciprocal of 1;20. You will see [0;45]. Its recipocal pair is 0;45. The depth is [16]. The procedure. h1 Problem (ix') An excavation. The length is as much as a reciprocal. The width is as much as [its reciprocal pair]. <The depth is> as much as that by which the reciprocal exceeds its reciprocal pair. I removed a volume of 26. What are the reciprocal, [its reciprocal pair, and the depth]? You: solve the reciprocal of 12. You will see 0;05. Multiply 36 by 0;05 <<by 36>>. You will see 3. Break 3 in 1/2. [You will see 1;30.] The recriprocal is 1;30. Its reciprocal pair is 0;40. The depth is 36. The procedure. h1 Problem (x') An excavation. The length is as much as a reciprocal. [The width is as much as its reciprocal pair]. The depth is as much as the total of the reciprocal and its reciprocal pair.  I removed a volume of 26. What are the reciprocal, its reciprocal pair, and the depth? You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 26.  You will see 2;10. Break 2;10 in 1/2. Square (it). You wil see 1;10 25. <Subtract 1 from 1;10 25. You will see 0;10 25.> The square-side is 0;25. Add and subtract (it) to <1>;05. You will see 1;30 and 0;40. The reciprocal is 1;30. Its reciprocal pair is 0;40. The depth is 26. The procedure. h1 Problem (xi') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as that by which the reciprocal exceeds its reciprocal pair, subtracted from the reciprocal. I removed a volume of 6. [What are] the reciprocal and its reciprocal pair? You: find the reciprocal of 12. You will see 0;05. Multiply by 6. You will see 0;30. Find the reciprocal of 0;30. You will see 2. The reciprocal is 0;30, its reciprocal pair 2, the depth 6. The procedure. h1 Problem (xii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The [depth] is as much as the total of the reciprocal and its recriproal pair. [I removed] a volume of 30. You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 30, the volume. You will see 2;30. Break 2;30 in 1/2. Square (it). You will see [1;33] 45. Subtract 1 from 1;33 45. You will see 0;33 45. The square-side is 0;45. Add and subtract from 1;45. You will see 2 and 0;30. The procedure. h1 Problem (xiii') An excavation. The length is as much as a reciprocal. The width is as much as its reciprocal pair. The depth is as much as the reciprocal pair. I removed a volume of 20. What are the reciprocal, its reciprocal pair, and the depth? You: find the reciprocal of 12. Multiply by 20. You will see 1;40. The reciprocal is 1;40. Its reciprocal pair is 0;36. The depth is 20. The procedure. h1 Problem (xiv') An excavation. The depth is as much as I squared and 7 cubits. I removed 3 20, the volume.  What are the length, width, and depth? You: take a 7th of 7. You will see 1. Find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 1. You will see 0;05. Multiply 0;05 by 12. You will see 1. Square 0;05. Multiply 0;00 25 by 1. You will see 0;00 25. Find the reciprocal of 0;00 25. You will see 2 24. Multiply 2 24 by 3 20, the volume. You will see 8 00 00. (With)what does it square itself? It squares itself (with sides of) 1 00, 1 00, 8. Multiply 0;05 by 1 00. You will see 5. The length is 5 cubits. Multiply 8 by 1. The depth is 8 cubits. The procedure. h1 Problem (xv') An excavation. The depth is as much as I squared and 7 cubits. I removed a volume of <3 15> <<13>>. What are the length, width, and depth? You: do as before. (W)ith what does 7 48 square itself? Its square-sides are 6, 6, and 13. It squares itself (with a side of) 6. The depth is 13. The procedure. h1 Problem (xvi') An excavation. The depth is as much as I made square. I removed a volume of 1;30. <What are> the length, width, and depth? You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 1;30. <You will see> 0;07 30. The square-side is 0;30. Multiply 0;30 by 1. It squares itself (with a side of) 0;30. Multilpy 0;30 by 12. The depth is 6. The procedure. h1 Problem (xvii') An excavation. The depth is as much as I made square and 1 cubit excess. I removed a volume of 1;45. You: Multiply 0;05, the excess by 1, the ratio. You will see 0;05. Muliply by 12. You will see 1. Square 0;05. You will se 0;00 25. Multiply 0;00 25 by 1. You will see 0;00 25. [Find] the reciprocal of 0;00 25. You will see 2 24. Multiply 2 24 by 1;45. [You will see] 4 12. In the square-side, 1 added, the square-sides are 6 (and) 1. Multiply 6 by 0;05. You will see 0;30. It squares itself (with a side of) <0;30>. The depth is 7!(6). The procedure. h1 Problem (xviii') An excavation. The depth is 3;20. I removed a volume of 27;46 40. The length exceeds the width by 0;50. You: find the reciprocal of 3;20, the depth. You will see 0;18. Multiply by 27;46 40, the volume. You will see 8;20. Break 50 in 1/2. Square (it). You will see 0;10 25. Add to 8;20. You will see 8;30 25. The square-side is 2;55. [Put it down] twice. Add to 1, subtract from 1. You will see 3;20, the length, (and) 2;30, the width. The procedure. h1 Problem (xix') An excavation. The depth is 3;20. [I removed a volume of] 27;46 40. [I summed the length and width:] 5;50. You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by [27;46 40]. You will see 8;20. Break 1/2 of 5;50. Square (it). [You will see] 8;[30 25]. Subtract 8;20 from inside it. [You will see] 0;10 25. [The square-side is 0;25.] Add and subtract from 2;55. [The length is] 3;20. [The width is 2;30.] The procedure. h1 Problem (xx') An excavation. The depth is 3;20. [I removed] a volume of 27;46 40. [The width exceeds the depth by as much as 2/3 of the length.] You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by 27;[46 40]. You will see 8;20. Multiply 8;20 by 0;40. [You will see] 5;33 [20. Multiply 3;20, the depth, by 0;05: 0;16 40.] Start again. Break 1/2 of 0;16 40. You will see 0;08 20. Square (it). [Add 0;01 09 26 40 to 5;33 20.] (With) what does it square itself? Put down 2;21 40 twice. [Add and subtract] 0;08 20. You will see 2;30, the width (and) 2;13 20. Release the reciprocal of 0;40. You will see 1;30. [Multiply by 2;13 20.] you will see 3;20, the length. The procedure. You: multiply 1;40, the length, by 12, the ratio of the depth. You will see 20. Release the reciprocal of 20. You will see 0;03. [Multiply] 0;03 by 1;40, the volume. [You will see 0;05.] Multiply 7 by 0;05. You will see 0;35. [Break in half 1;40. Square (it). 0;41 40.] [Subtract] 0;35 from it. [You will see 0;06 40. The square-side is 0;20.] [Add and subtract] to [0;50. 1;10 and 0;30, the width. ... Take one-seventh of 1;10. The depth is 0;10. The procedure.] You: multiply 1;40, the length, by 12, the ratio of the depth. You will see 20. Release the reciprocal of 20. You will see 0;03. Multiply 0;03 by 3;20. You will see 0;00 10. Multiply 0;10, the excess, by 7. You will see 1;10. Add 1;40, the length, to 1;10. You will see 2;50. Break in half 2;50. Square (it). You will see 2;00 25. Subtract 1;10 from 2;00 25. You will see 0;50 25. Add and subtract 0;55, the square-side, to 1;25 and you will see 2;20 and 0;30, the width. Take one-seventh of 2;20. The depth is 0;20. The procedure. You: Multiply 1;40, the length, by 12, the ratio of the depth. You will see 20. Release the reciprocal of 20. You will see 0;03. Multiply 0;03 by 0;50. You will see 0;02 30. Multiply 0;02 30 by 7. You will see 0;17 30. Multiply 7 by 0;05, 1 cubit. You will see 0;35. Subtract 0;35 from 1;40, the length. You will see 1;05. Break in half 1;05. Square 0;32 20. You will see 0;17 36 15. Subtract 0;17 30 from it. You will see 0;00 06 15. Its square-side is 0;02 30. Add and subtract to 0;32 30. You will see 0;35 and 0;30, the width. One-seventh of 0;35 is 0;05, the depth. The procedure. 30 procedures."}, {"id_text": "P254450", "project_name": "dccmt", "raw_text": "A grain-pile. The length is 10, the width 6, the top 4, the <grain capacity> 28 48 00, the height 48. It went down 24. What are the cross-width and the grain? You: take the reciprocal of 48, the height. You will see 0;01 15. Multiply 0;01 15 by [6] by which the length exceeds the top. You will see 0;07 30. Multiply [0;07 30 by] 24. You will see 3. [Take] 3 from 10, the length. [You will see 7. The cross-width is 7]. Take 3 from 6, the width. [You will see 3. ...] multiply. You will see 1. ... [A grain-pile. The width is 6, the height 48, the grain capacity 28 48 00. The sum of] the length and top [is 14.] What are [the length and top]? You: [take] the reciprocal of 48. [You will see 0;01 15. Multiply] 0;01 15 by 28 48 00, the grain. You will see 36 00. Take the reciprocal of 1 30, the coefficient. Multiply [by] 36 00. You will see 24. [Take] the reciprocal of 6, the width. You will see 0;10. Multiply 24 by 0;10. You will see 4. Take [4] from the sum. You will see 10. The length is 10; the top is 4. This is the method. [A grain-pile.] The width is 6, the height 48, the grain capacity 28 48 00. [1/2] the top is like a fifth of the length. What are the length and the top? <You:> put down 5, the ratio of the length; 2, the ratio of the top. Return. Take the reciprocal of 48. You will see 0;01 15. Multiply 0;01 15 by 28 48 00. You will see 36 00. Take the reciprocal of 1 30, the coefficient. You will see 0;00 40. Multiply 0;00 40 by 36 00. You will see 24. Take the reciprocal of 6, the width. You will see 0;10. Multiply 0;10 by 24. You will see 4. Break 4 in half. You will see 2. Put it down. Break 2, the ratio of the top, in half. You will see 1. Add to 5, the ratio of the length. You will see 6. Take a third. You will see 2. Multiply 5 by 2. You will see 10. Multiply 2 by 2, the ratio of the top. You will see 4, the top. The method. A grain-pile. The height is 48, the grain capacity 28 48 00, the length 10. 2/3 of the width [is like the top]. What are the width and top? You: [take] the reciprocal of 48. [You will see 0;01 15. Multiply 0;01 15 by 28 48 00. You will see 36 00. Take the reciprocal of 1 30, the coefficient. You will see 0;00 40. Multiply 0;00 40 by 36 00. You will see 24. Take a third of 24. You will see 8.] You will see 3;20. Break 3;20 in half. You will see 1;40. Square 1;40. You will see 2;46 40. Add 2;46 40 to 2;40. You will see 5;26 40. What is the square root? 2;20 is the square root. Take 1;40 from 2;20. You will see 0;40. Break 0;40 in half. You will see 0;20. Square 0;20. You will see 0;06 40. Take the reciprocal of 0;06 40. You will see 9. Multiply 0;40 by 9. You will see 6, the width. Multiply 0;40 by 6. You will see 4. The top is 4. This is the method. A grain-pile. [...] The square sides are each [...]. The height is 6. What is the grain? [You: ...] You will see (...) 5?. Multiply 10 by 5. You will see 50. [...] You will see (...) 3 45 [...] You will see (...) 30, the volume. [...] You will see [...] [A grain-pile. The height is 48, the top 4], the grain capacity 28 48 00. [The sum of the length and width is] 16. What are the length and width? [You:] take [the reciprocal of] 48, the height. You will see 0;01 15. Multiply [0;01 15] by 28 48 00. You will see 36 00. Take the reciprocal of 1 30, the coefficient. You will see 0;00 40. Multiply 0;00 40 by 36 00. You will see 24. Take a third of 24. You will see 8. Break 4, the top, in half. You will see 2. Take 1/3 of 2. You will see 0;40. Take 1/3 of 16. You will see 5;<20>. Add 0;40 to 5;20. You will see 6. Break 6 in half. You will see 3. Square. You will see 9. Take 8 from 9. You will see <1>. What is the square root of \u00c2\u00ab1\u00c2\u00bb? The square root is 1. Add 1 to 3. You will see 4. Take 1 from 3. You will see 2. Triple 4. You will see 12. Take 2 from 12. You will see 10, the length. Take from 16, the sum. You will see 6. The width is 6. The method.  A grain-pile. The top is 4, <the grain capacity> 28 48, the height 48. 1/2 the length plus 1 equals the width. What are the length and width? You: take the reciprocal of 48, the height. You will see 0;01 15. Multiply by 28 48 00. You will see 36 00. Take the reciprocal of 1 30, the coefficient. You will see 0;00 40. Multiply 0;00 40 by 36 00. You will see 24. Because they said 1/2 the length plus 1 equals the ratio of the width, put down 1 and 0;30. Return. Break in half 4, the top. You will see 2. Take 1/3 of 2. You will see 0;40. Take 1/3 of 1. You will see 0;20. Multiply 0;20 by 0;30. You will see 0;10. Multiply 24 by 0;10. You will see 4. What is the square root of 4? The square root is 2. Take the reciprocal of 0;30. You will see 2. Multiply 2 by 2. You will see 4. Triple 4. You will see 12. Take 2 from 12. You will see 10, the length. Take 4, the top, from 10, the length. You will see 6, the width. This is the method. A grain-pile. The top is 4, the grain capacity 28 48 00, the height 48. In 1 cubit the slope is 1;30 cubits. What are the length and width? You: multiply [0;07 30, the slope,] by 48, the height. You will see 6. Add to 4, the top. You will see 10, the length. Take the reciprocal of 48. You will see 0;01 15. Multiply by 28 48 00. You will see 36 00. Multiply 0;00 40 by 36 00. You will see 24. Break 24 in half. You will see 12. <Take> a third. You will see [4]. Take the reciprocal of 4. You will see 0;15. Multiply 24 by 0;15. You will see 6, the width. This is the method. A grain-pile. The length is 10, the height 48, the grain capacity 28 48 00. The sum of all the top and 1/3 of the width is 6. What are the top and width? You: take the reciprocal of 48, the height. You will see 0;01 15. Multiply 0;01 15 by 28 48 00. You will see 36 00. Multiply 0;00 40 by 36 00. You will see 24. Put down 1;30 and 0;30, the ratios. Take a third of 10, the length. You will see 3;20. Take 1/3 of 1;30. You will see 0;30. Take a third of 0;30. You will see 0;10. Multiply 1;30 by 0;10. You will see 0;15. Multiply 1;30 by 3;20. You will see 5. Multiply 24 by 0;15. You will see 6. Take 5 from 6. You will see 1. Take the reciprocal of 0;10. You will see 6. Multiply by 1. You will see 6, the width. Multiply 0;40 by 6. The top is 4. This is the method. A triangular grain-pile. The length is 30, the width 10, the height 48. What is the grain? You: <multiply> 30, the length, by 10, the width. You will see 5 00. Multiply by 48, the height. You will see 4 00 00. Multiply 1 30 by 4 00 00. You will see 6 00 00 00. The grain capacity is 6 00 00 00 gur. This is the method. A grain-pile. The length is 30, the upper width 20, lower width 10, the height 48. <What are> the volume and grain? You: add the upper width and the lower width. You will see 30. Break 30 in half. You will see 15. Multiply 15 by 30. You will see 7 30. Multiply 7 30 by 48, the height. You will see 6 00 00, the volume. Multiply [1] 30 by 6 00 00. You will see 9 00 00 00, the grain. This is the method. [A semicircular grain-pile.] The semicircumference is 30, the diameter 20, the height 48. [What are the volume and grain? You: multiply] 30, the length, by 20, the diameter. [You will see 10 00. Multiply by 0;15,] the coefficient of a semicircle. [You will see 2 30. Multiply by 48,] the height. [You will see 2 00 00, the volume. Multiply 2 00 00 by 1 30. You will see 3 00 00 00, the grain. This is the method.] [A mound. The circumference is 0;30, the height 1.] In 1 cubit, [what is the slope? You:] Take the reciprocal of 1, the height. You will see 1. [Multiply 1 by 0;30, the circumference.] You will see 0;30. Break 0;30 in half. You will see [0;15]. The slope is 0;15 in 1 [cubit. This is] the method. A mound. The circumference is 30. In 1 cubit the slope is 0;15. What is the height? You: Double 0;15, the slope. You will see 0;30. Take the reciprocal of 0;30. You will see 2. Multiply 0;30, the circumference, by 2. You will see 1, the height. The method. [A mound. ...] you will see [...] Put down [0;01 15. Take the reciprocal of 0;05, the coefficient. You will see 12]. Multiply [by 0;01] 15. You will see 0;15. [What is the square root of 0;15?] The square root is [0;30]. The circumference is 0;30 rods. [This is] the method. [A mound.] The volume is 25. What are the mas,s,arum-vessel and dik\u0161um? You: triple 25. You will see 1 15. Take the reciprocal of 0;05, the coefficient. You will see 12. Multiply 12 by 1 15. You will see 15 00. Put it down. Return. Take the reciprocal of 0;15, the slope. You will see 4. Multiply 15 00 by 4. You will see 1 00 00. What is the square root of 1 00 00? The square root is 1 00, the height. Double 0;15, the slope. You will see 0;30. Multiply 0;30 by 1 00, the height. You will see 30, the mas,s,arum-vessel. The method. [A mound.] The mas,s,arum-vessel is 30, the height 1 00, the volume 25. [It went down] 21/2 rods. [...] What is the volume? You: take the reciprocal of 1 00, the height. [You will see 0;01.] Multiply [0;01 by] 30, the mas,s,arum-vessel. You will see 0;30. Multiply 0;30 by [30, by which it went down]. You will see 15. [Take] 15 from 30, the mas,s,arum-vessel. [You will see 15, the upper mas,s,arum-vessel.] Tell the volume. 30, The lower mas,s,arum-vessel is 30, [the upper mas,s,arum-vessel\u00c2\u00b1 is 15.] Multiply [15 by 0;05] the coefficient. You will see 1 15. [... Triple 1 15. You will see 3 4]5. [Multiply] 3 45 by [...] 15 and 18 45 [...] You will see 12 30 [...] Break [...] 45 [in half. ...] 6 15 ... [... You will see] 29 51 40, [the area]. Multiply [29 51 40] by 1, the height. You will see 29 51 40, [the volume]. [...] 29 51 40, the volume [...] 12 30 [...]"}, {"id_text": "P254451", "project_name": "dccmt", "raw_text": "[A wall.] The width is [2 cubits], the length 2 1/2 rods, the height 1 1/2 rods. [How many bricks? You]: multiply 2 cubits, the width, by 2 1/2 rods, the length. You will see 0;25 (the ground). Multiply [0;25] by 18, the height. You will see 7;30 (7 1/2 sar, the volume). Multiply 7;30 by 6, the constant of a wall. You will see 45. The bricks are 45 sar. The procedure. A wall. The length is 2 1/2 rods, the height 1 1/2 rods, the bricks 45 sar. What is the width of my wall? You: multiply 2;30, the length, by 18, the height, so that you will see [45]. Keep (it). Solve (the reciprocal) the reciprocal of 6, the constant of a wall. [Multiply] by [45 sar, the bricks]. You will see 7;30 (the volume). Keep (it). [Solve] the reciprocal of the 45 that you are keeping. [You will see 0;01 20.] Multiply 0;01 20 by the 7;30 (the volume) that you are keeping and [you will see 0;10. The width of the wall is 2 cubits.] The procedure. [...] You will see 1;21 40. Add the 0;25 that you are keeping to 1;21 40. You will see 1;46 40. What is the square-side of 1;46 40? 1;20 is the square-side. Put (it) down twice. Add 1;10 to 1;20. You will see 2;30. The length of the wall is 2;30. Take away 1;10 from 1;20. The remainder is 0;10. The width of the wall is 2 cubits. The procedure. The <area of a> house is 5 sar. For a height of 2 1/2 rods how many bricks should I get made? You: take a 3rd part of 5 sar. You will see 1;40 the walls. Multiply 1;40 by 21/2 rods, the height. You will see 4;10. You will get 2 1/2 iku of bricks made and then you will pile up (a) 5 sar (area of) house to a height of 2 1/2 rods. The procedure. If the bricks are 2 1/2 iku, the height 2 1/2 rods, what (area of) house should I build? You: solve the reciprocal of 2 1/2 rods, the height. Multiply by 4;10, the 2 1/2 iku of bricks. You will see 1;40. [Keep (it).] Solve [the reciprocal of 0;20, the constant of a built house]. You will see 3. [Multiply] 3 by [the 1;40 that you are keeping. You will see 5.] The (area of the) house is 5 sar. [The procedure.] A wall of baked bricks. The width is 2 cubits, the height 1 rod, the baked bricks 9 sar. What is the length of my wall? You: solve the reciprocal of 2;15, the constant of a wall. You will see 0;26 40. Multiply 0;26 40 by 9 sar, the baked bricks. You will see 4 (the volume). Keep (it). Solve the reciprocal of 12, 1 rod, the height of the wall. You will see 0;05. Multiply 0;05 by the 4 that you are keeping. You will see 0;20 (the position). Solve the reciprocal of 0;10, the width; multiply by the 0;20 that you saw. You will see 2. The length is 2 rods. The procedure. The length of a wall of baked bricks is 2 rods, the height 1 rod, the baked bricks 9 sar. What is the width of my wall? You: take the reciprocal of 2;15, the constant of a wall. You will see 0;26 40. Multiply 0;26 40 by 9 sar, the baked bricks <that> you spread out for your wall. You will see 4 (the volume). Solve the reciprocal of 1 rod, the height of your wall. Multiply by 4 that you saw. You will see 0;20. The area of your wall is 1/3 sar. Solve the reciprocal of 2 rods, the length. You will see 0;30. Multiply 0;30 by 0;20. You will see 0;10. The width of the wall is 2 cubits. The procedure. The length of a wall is 2 rods, the width of the wall 2 cubits. 9 sar of baked bricks are put down for the wall. How high can it be with 9 sar of baked bricks? You: solve the reciprocal of 2;15, the constant of baked [bricks]. You will see 0;26 40. Multiply 0;26 40 by 9 [sar, the baked bricks that] are put down [for] the wall. You will see 4 (the volume). [Keep (it). Multiply 2 rods, the length, by 0;10. You will see 0;20.] Solve the reciprocal of 0;20. You will see 3. Multiply 3 by the 4 [that] you are keeping. [You will see 12.] You will elevate (it) [to a height of 1 rod] with 9 sar of baked bricks. The procedure. [A wall of] baked [bricks]. The height of the wall is 1 rod, the baked bricks 9 sar. I summed the length and thickness [of the wall] so that (it was) 2;10. What are the length and thickness of my wall? [You:] solve the [reciprocal of 2;15], the constant of baked bricks. You will see 0;26 40. Multiply 0;26 40 [by] 9 sar, the baked bricks. You will see 4 (the volume). Keep 4, the volume of your baked bricks. Solve [the reciprocal of 12], 1 rod, the height of your baked bricks. You will see 0;05. Multiply 0;05 by the 4 that you are keeping. You will see 0;20. The area is 0;20 (the position). Keep (it). Break off 1/2 of 2;10, [the sum] of the length and thickness of the wall. You will see 1;05. Square 1;05. [You will see] 1;10 25. Take away [0;20], the area [that] you are keeping, from 1;10 25. You will see 0;50 25. What is the square-side of 0;50 25? [The square-side is 0;55.] Add [0;55] to the 1;05 that you squared. You will see 2. The length is 2 rods. Take away [0;55 from 1;05 that] you squared. The remainder is 0;10. The thickness is 2 cubits. The procedure. [A wall of baked bricks.] I laid 9 sar of [baked] bricks. The length [exceeds the thickness of the wall] by [1;50]. The vertical is 1 rod. [What are] the length of the wall and the thickness of my wall? [You:] solve [the reciprocal of 2];15, the constant of a wall of baked bricks. You will see 0;26 40. Multiply 0;26 40 [by 9 sar], the baked [bricks]. You will see 4 (the volume). The volume of the baked bricks is 4 sar. Solve [the reciprocal] of 12, the height of the wall. You will see 0;05. Multiply 0;05 by 4, the volume of baked bricks. You will see 0;20. Keep 0;20, the area. Break off 1/2 of 1;50, by which the length exceeds the thickness of the wall. You will see 0;55. Put down [0;55 twice]. Square 0;55. You will see 0;50 25. Add 0;20, the area, to 0;50 25. You will see 1;10 25. What is the square-side of 1;10 25? [The square-side] is 1;05. Put (it) down twice. Add the [0;55 that] you squared to 1;05. You will see 2. [The length] of the wall is [2 rods. Take away 0;55 from 1;05.] The remainder is 0;10. The thickness [of the wall] is 2 cubits. [The procedure.] A wall of baked bricks. [The length of the wall is 2 rods.] It is [2 cubits] thick at the bottom, [1 cubit thick at the top]. The height is 2 rods. What heaping (?) does my wall [heap up, and] in 1 cubit how much does it slope? You: sum 2 cubits, the lower base, [and 1 cubit], that it tapers to at the top. You will see 0;15. Break off 1/2 of 0;15. You will see 0;07 30 (the position). Multiply 0;07 30 by 2 rods, the length. You will see 0;15. Multiply [0;15 by] 24, the height. You will see 6. Multiply 6 by [2;1]5, the constant of baked bricks. You will see 13;30. The baked bricks are 13 1/2 sar. Break off 1/2 of 13;30. You will see 6;45. A heaping (?) of 6 2/3 <sar> 5 shekels will be heaped up for you with your wall. Return. See how much it slopes in 1 cubit. By how much does 2 cubits, the lower base, exceed 0;05 that is on the top? It exceeds by 0;05. Keep (it). Because they said, \"What slope does it slope in 1 cubit?\", [put down 0;05]. Solve the reciprocal of 0;05. You will see 12. [Multiply] by 2 rods, the height. [You will see 24.] Solve the reciprocal of 24. You will see 0;02 30. Multiply [0;02 30 by] 0;05, (which is) 1 cubit, the ratio. You will see 0;00 12 30. [Multiply 0;00 12 30 by 6 00.] You will see [1;15]. In 1 cubit the wall slopes a slope of 1 finger and a quarter of 1 finger. The procedure. A gate. The height is 1/2 <rod>, 2 cubits, the breadth 2 cubits. What is its diagonal? You: square 0;10, the breadth. You will see 0;01 40, the ground. Take the reciprocal of 0;40 (rods, or 8) cubits, the height. Multiply by 0;01 40, the ground. You will see 0;02 30. Break off 1/2 of 0;02 30. You will see 0;01 15. [Add] 0;01 15 [to 0;40, the height]. You will see 0;41 15. [The diagonal] is 0;41 15. The procedure. If the gate has height 0;40 (rods, or 8) cubits and diagonal 0;41 15, what is the breadth? You: take away 0;40, the height, from 0;41 15, the diagonal. The remainder is 0;01 15. Copy 0;01 15. You will see 0;02 30. Multiply 0;40, the length, by 0;02 30, the product (?) that you saw. You will see 0;01 40. What is the square-side? 0;10 is the square-side. The breadth is 0;10. The procedure. The breadth is 2 cubits, the diagonal 0;41 15. What is the height? You: no (solution). One. Its diagram. The breadth is 2 cubits, the height 0;40 (rods, which is 8) cubits. What is its diagonal? You: square 0;10, the width. You will see 0;01 40, the ground. Multiply 0;01 40 by 0;40 (rods, or 8) cubits, the height, and you will see 0;01 06 40. Copy (it). You will see 0;02 13 20. Add it to 0;40 (rods, which is 8) cubits, the height. You will see 0;42 13 20, the diagonal. The procedure. [...] The procedure. [The height is 0;40 (rods, which is 8) cubits, the diagonal 0;41 13 20.] What is the breadth? You: [square 0;41 13 20.] You will see [0;28] 20. [Square 0;40, the height. You will see 0;26 40. Take away 0;26 40] from 0;28 20. [You will see 0;01 40. What is the square-side? The square-side is 0;10, the breadth.] The procedure. [The breadth is 2 cubits, the diagonal 0;41 13 20. What is the height? You:] square [0;41 13 20)], the diagonal. The squared number is 0;28 20. [Square 0;10, the breadth.] You will see 0;01 40. [Take away] 0;01 40 from 0;28 20. [The remainder is 0;26 40.] What is the square-side? The square-side is 0;40. The procedure. One.  Total 25 trails. I\u0161kur-mansum, son of Sin-iqi\u0161am."}, {"id_text": "P254557", "project_name": "dccmt", "raw_text": "If they ask you about a rectangle, as follows, \u2018The diagonal is 1;15, the area 0;45. How much are my length and the width?\u2019:  You, when you proceed, draw 1;15, the diagonal, (and) its counterpart and then combine them so that 1;33 45 will come up. Its base is 1;33 45.  Copy your area twice so that 1;30 will come up. Take (it) away from 1;33 45 so that \u00ab1 30\u00bb 0;03 45 is the remainder.  Take the square-side of 0;03 45 so that 0;15 will come up. <Break off> its half <so that> 0;07 30 will come up. Multiply by 0;07 30 so that 0;00 56 15 will come up. Your \u0160U is 0;00 56 15.  <Sum> 0;45, your area, over your \u0160U <so that> 0;45 56 15 will come up. Take the square-side of 0;45 56 15 so that 0;52 30 will come up.  Draw 0;52 30 (and) its counterpart and then add the 0;07 30 that you combined to one, take away from one. Your length is 1, the width 0;45. If the length is 1, the width 0;45, how much are the area and my diagonal?  [You, when] you [proceed], combine the lengths so that [1 will come up]. Let your head hold [1]. Return and then combine 0;45, the widths so that 0;33 45 will come up. Add it to your length so that 1;33 45 will come up.  Take the square-side of 1;33 45 so that 1;15 will come up. Your diagonal is 1;15.  Multiply your length by the width so that 0;45, your area, <will come up>. That is the procedure."}, {"id_text": "P254567", "project_name": "dccmt", "raw_text": "The procedure of a log. Its dividing line is 0;05 (rods, or) a cubit. How much is it suitable for storing?  When you proceed, put down the depth equal to the dividing line. Make 0;05 into a depth so that 1 will come up.  Triple 0;05, the dividing line, and the 0;15 will come up. The circle of the log is 0;15.  Combine 0;15 so that 0;03 45 will come up. Multiply 0;03 45 by 0;05, the constant of a circle, so that 0;00 18 45, the area, will come up.  Multiply 0;00 18 45 by 1, the depth, so that 0;00 18 45, the volume, will come up.  Multiply 0;00 18 45 by 6 00 00, (the constant) of storage, so that 1 52;30 will come up.  The log contains 1(barig) 5(ban) 2 1/2 sila of grain. That is the procedure. If a log, whose bottom is 0;05, its top 0;01 40, is 0;30 (rods, or) a reed, long, how much grain does it contain?  When you proceed, make 0;30, the length of the log, into a depth so that 6 will come up. Return.  Sum and break 0;05, the bottom, and 0;01 40, the top, so that 0;03 20 will come up. Triple 0;03 20 so that 0;10, the circle of the log, will come up.  Combine 0;10 so that 0;01 40 will come up. Multiply 0;01 40 by 0;05, the constant, so that 0;00 08 20, the area, will come up.  Multiply 0;00 08 20 by 6, the length of the log, so that 0;00 50, the volume, will come up. Multiply by 0;00 50 by 6 00 00, (the constant) of storage and 5 00 will come up. The log contains 1 gur of grain.  If a log, whose bottom is 0;05, its top 0;01 40, is 5 (rods or), a half-rope, long, its price is 1 talent of silver.  Now, I am carrying 1 mina of silver. Trim (grain to the value) of 1 mina of silver from the log, either from its base or from its top, and sell it to me.  When you proceed, make 5, the length of the log, into a depth so that 1 00 will come up.  Return. Sum (and) break 0;05, the bottom, and 0;01 40, the top, so that 0;03 20, the thickness of the log, will come up.  Triple 0;03 20 so that 0;10, the circle of the log, will come up.  Combine 0;10 so that 0;01 40 will come up. Multiply 0;01 40 by 0;05, the constant, so that 0;00 08 20, the area, will come up.  Multiply 0;00 08 20 by 1 00, the length of the log, so that 0;08 20, the volume, will come up.  Multiply 0;08 20 by 6 00 00, (the constant) of storage, and 50 00 (sila) will come up. The log contains 10 gur of grain.  10 gur grain is the storage (equivalent) of 1 talent of silver. If 50 00 (sila) of grain is the storage of 1 talent of silver, how much is the storage of 1 mina of silver?  Solve the reciprocal of 1 mina of silver so that 1 will come up. Multiply 1 by 1 00, a talent of silver, so that 1 00 will come up. Solve the reciprocal of 1 00 so that 0;01 will come up.  Multiply 0;01 by 50 00, the grain, so that 50 will come up. 5 (ban) of grain is the storage of 1 mina silver.  Return. Ask: \u2018if a log whose bottom is 0;05 (contains) 5 (ban) of grain, how much is its length,?\u2019  Solve the reciprocal of 6 00 00, of storage, so that 0;00 10 will come up. Multiply 0;00 10 by 50, the grain, so that 0;08 20 will come up. Let your head keep 0;08 20.  Return. Triple 0;05, the bottom, so that 0;15, the circle of the log, will come up. Combine 0;15 so that 0;03 45 will come up.  Solve the reciprocal of 0;03 45 so that 16 will come up. Multiply 16 by 0;08 20 that your head is keeping, so that 0;02 13 20 will come up.  You go up 0; 02 13 20, thirteen fingers and 1/3 finger, the length of the log, and then you trim it off and then you give it for 1 mina of silver.  Return. Work out its storage.  Make 0;02 13 20 into a depth so that 0; 26 40 will come up. Triple 0;05, the bottom, so that 0;15, the circle of the log, will come up.  Combine 0;15 so that 0;03 45 will come up. Multiply 0;03 45 by 0;05, the constant of a circle, so that 0;00 18 45, the area, will come up.  Multiply 0;00 18 45 by 0;26 40, the length of the log, so that 0;00 08 20, the volume, will come up.  Multiply 0;00 08 20 by 6 00 00, (the constant) of storage, so that 50 will come up. The storage of 1 mina of silver is 5(ban) of grain.  Return. Trim it from its top. Solve the reciprocal of 1 mina of silver so that 1 will come up.  Multiply 1 by 1 00, a talent of silver, so that 1 00 will come up.  Solve the reciprocal of 1 00 so that 0;01 will come up. Multiply 0;01 by 50 00, the grain of the whole log, and 50 will come up. The storage of 1 mina of silver is 5 (ban) of grain. If 50 (sila of) grain is the storage of 1 mina of silver and 0;01 40 the top of the log, how much should I descend so that I may trim (grain to the value of) 1 mina of silver?  Solve the reciprocal of 6 00 00, (the constant) of storage, so that 0;00 00 10 will come up. Multiply 0;00 00 10 by 50, the grain, so that 0;00 08 20 will come up. Let your head hold 0;00 08 20.  Return. Triple 0;01 40, the top of the log. 0;05, the circle of the log, will come up.  Combine 0;05 so that 0;00 25 will come up. Solve the reciprocal of 0;00 25 so that 2 24 will come up.  Multiply 2 24 by 0;00 08 20 that your head is keeping so that 0;20 will come up. You descend 0;20, four cubits, and then you trim it off and then you sell (it) for 1 mina of silver.  Return. Work out its storage capacity.  Make 0;20, the length of the log, into a depth, so that 4 will come up. Triple 0;01 40, the top of the log, so that 0;05, the circle of the log, will come up.  Combine 0;05 so that 0;00 25 will come up. Multiply 0;00 25 by 0;05, the constant, so that 0;00 02 05, the area, will come up. Multiply 0;00 02 05 by 4, the length of the log, so that 0;00 08 20, the volume, will come up.  Multiply 0;00 08 20 by 6 00 00, (the constant) of storage, so that 50 will come up. The log holds 5 (ban) of grain.  That is the procedure. The procedure of a sila (measuring vessel). The dividing line of my sila (vessel) is 0;01 (rods, or) six fingers.  What should I make deep so that it suffices for 1 sila?  When you proceed: solve the reciprocal of 6 00 00, (the constant) of storage, so that 0;00 00 10 comes up. Multiply 0;00 00 10 by 1 sila, the grain, so that 0;00 00 10 comes up.  Return. Triple 0;01, the dividing line, so that 0;03, the circle of the sila (vessel), comes up.  Combine 0;03, so that 0;00 09 comes up. Solve the reciprocal of 0;00 09, so that 6 40 comes up.  Multiply 6 40 by 0;00 00 10, so that 0;01 06 40 comes up. The depth is 0;01 06 40 (rods, or) six fingers and two-thirds of a finger. If the depth is 0;01 06 40, the dividing line 0;01, how much grain does my sila (vessel) contain?  Make 0;01 06 40 into a depth, so that 0;13 20 (cubits) comes up. Triple 0;01, the dividing line, so that 0;03, the circle of the sila (vessel) comes up.  Combine 0;03, so that 0;00 09 comes up. Multiply 0;00 09 by 0;05 (the constant) of a circle, so that 0;00 00 45, the area, comes up.  Multiply 0;00 00 45 by 0;13 20, the depth, so that 0;00 00 10, the volume, comes up. Multiply 0;00 00 10 by 6 00 00 (the constant) of storage, so that 1 sila of grain comes up. If the grain is 1 sila, my depth 0;01 06 40, what are my diameter and my circle?  Make 0;01 06 40 into a depth, so that 0;13 20 (cubits) comes up. Solve the reciprocal of 0;13 20, so that 4;30 comes up.  Return. Solve the reciprocal of 0;05 (the constant) of a circle, so that 12 comes up. Solve the reciprocal of 6 00 00 (the constant) of storage, so that 0;00 00 10 comes up.  Multiply 0;00 00 10 by 12, so that 0;00 02 comes up. Multiply 0;00 02 by 1 sila, the grain, so that 0;00 02 comes up.  Multiply 0;00 02 by 4;30 so that 0;00 09 comes up. Have its square-side come up so that 0;03 comes up. The circle of the sila (vessel) is 0;03.  Take a third of 0;03 so that 0;01, the dividing line, comes up. That is the procedure. If the dividing line of a ban (measuring vessel) is 0;02, (and) it is filled by a ban and 2/3 sila of grain, what should trim off I so that it amounts to 1(ban)?  When you proceed, have the depth come up. Solve the reciprocal of 6 00 00 (the constant) of storage, so that 0;00 00 10 comes up. Multiply 0;00 00 10 by 1(ban) 2/3 sila of grain, so that 0;00 01 06 40 comes up. Return, and then triple 0;02, the dividing line, so that 0;06, the circle of the ban (vessel), comes up.  Combine 0;06, so that 0;00 36 comes up. Solve the reciprocal of 0;00 36, so that 1 40 comes up.  Multiply 1 40 by 0;00 01 06 40, so that 0;02 57 46 40, the depth, comes up.  Return. The dividing line of 2/3 sila of grain is 0;02. How deep should I make it so that it amounts to 2/3 sila?  Solve the reciprocal of 6 00 00 (the constant) of storage, so that 0;00 00 10 comes up. Multiply 0;00 00 10 by 0;40, the grain, so that 0;00 00 06 40 comes up.  Return. Triple 0;02, the dividing line, so that 0;06, the circle, comes up. Combine 0;06, so that 0;00 36 comes up.  Solve the reciprocal of 0;00 36, so that 1 40 comes up. Multiply 1 40 by the 0;00 00 06 40 that your head was holding, so that 0;00 11 06 40, the depth, comes up. You do down and then you trim off 0;00 11 06 40. Take away 0;00 11 06 40 from the middle of 0;02 57 46 40, the former depth, <so that> 0;02 46 40, the depth of 1 (ban) of grain, comes up.  Make 0;02 46 40 into a depth, so that 0;33 20 (cubits) comes up.  Triple 0;02, the dividing line, so that 0;06, the circle of the ban (vessel), comes up.  Combine 0;06, so that 0;00 36 comes up. Multiply 0;00 36 [by] 0;05 (the constant) of a circle, so that 0;00 03, the area, comes up.  Multiply 0;00 03 by 0;33 20, the depth, so that 0;00 01 40, the volume, comes up. Multiply 0;00 01 40 by 6 00 00 (the constant) of storage, so that 10 comes up. It is sufficient for 1 (ban) of grain. That is the procedure. Three bariga (measuring vessels) and 2 (gur, or 10) sixties (sila) of grain. The first is a bariga, the second 5(ban), the third 4(ban). What grain do they issue?  When you proceed: sum 1 bariga, 5(ban), and 4(ban), so that 2 30 (sila) comes up.  Solve the reciprocal of 2 30, so that 0;00 24 comes up. Multiply 0;00 24 by 10 00, the grain, so that 4 comes up. Multiply 4 by 1 00, so that the big one issues 4 sixties. Multiply 4 by 50 so that the second one issues 3;20 sixties. Multiply 4 by 40, so that the third issues 2;40 sixties. <That is the procedure.> The work rate of plastering. He puts down 1 (rod, or) two reeds square and a thickness of 0;00 10 (rods, or) a finger and then he plasters for the whole day.  If it is 1 (rod, or) two reeds square and he makes it 0;00 10 (rods, or) a finger thick, how much is my clay?  When you proceed, make the thickness of the plastering into a depth so that 0;02 will come up.  Return and combine 1, the square-side of the plastering, so that 1, the area, will come up.  Multiply 1 by 0;02, the thickness, so that 0;02, the volume, will come up.  Multiply 0;02 by 5 00 00, (the constant) of the measured amount, so that 10 00 will come up.  1 man plasters 2 gur of clay for a whole day. That is the procedure. If the height of the plastering is 0;40 (rods, or) eight cubits, the thickness of my plastering 0;00 10 (rods, or) a finger, what are the length I plaster and my clay?  When you proceed, write down 1 sar, the mud that 1 man plasters in a whole day.  Solve the reciprocal of 0;40, the height, so that 1;30 will come up.  Multiply 1;30 by 1 sar of mud so that 1;30 will come up. For a whole day you plaster a length of 1;30 (rods, or) three reeds.  Multiply 1;30, the length, by 0;40, the height of the plastering, so that 1 sar, the area, will come up.  Multiply 1 by 0;02, the thickness of the plastering, so that 0;02, the volume, will come up.  Multiply 2 by 5 00 00, (the constant) of the measured amount (?), so that 10 00 will come up. 1 man plasters 10 00 (sila, or) <2> gur of clay in a whole day. Brickage, combined constant. What is the work rate of brick-making and what is the output of 1 man?  When you proceed, write down 0;20, the work rate of destroying (?), 0; 20, the work rate of brick-making, 0; 10, the work rate of mixing.  Solve the reciprocal of 0;20 so that 3 will come up. Solve the reciprocal of 0;20 so that 3 will come up. Solve the reciprocal of 0;10 so that 6 will come up. Sum them so that 12 will come up.  Solve the reciprocal of 12 so that 0;05 will come up. The combined work rate of is 0;05.  Return. Combine 0;03 20, the square-side of a brick, so that 0;00 11 06 40, the area of a brick, will come up.  Return. Make 0;01, the thickness of the brick, a depth so that 0;12 will come up. Multiply 0;12 by 0;00 11 06 40 so that 0;00 02 13 20, the volume of a brick, will come up.  Solve the reciprocal of 0;00 02 13 20 so that 27 00 will come up. Multiply 27 00 by 0;05, the work rate, so that 2;15 sixties of bricks, the output of 1 man, will come up.  Triple 2;15. 6;45 sixties, the output of the work rate, will come up. The work rate of brick-making. I carry for 5 (rods, or) a half-rope and then I make bricks. What is the output of 1 man?  When you proceed, write down 0;20, the destroying, 0;20, the brick-making, 0;10, the mixing.  Return and then solve the reciprocal of 5, the distance, so that 0;12 will come up. Multiply 0;12 by 45 00, the going, so that 9 00 will come up.  Multiply 9 00 by 0;00 02 13 20, the basket, so that 0;20 is the volume you will carry here for 5 (rods, or) a half-rope.  Write down 0;20, the volume, by the side of 0;10, the mixing.  Return. Solve the reciprocal of 0;20 and 3 then will come up. Solve the reciprocal of 0;20 so that 3 will come up.  Solve the reciprocal of 0;10 so that \u00absolve so that\u00bb 6 will come up. Solve the reciprocal of 0;20, the volume, so that 3 will come up.  Sum (them) so that 15 will come up. Solve the reciprocal of 15 so that 0;04 will come up.  Multiply 0;04 by 1, the day, so that 0;04, the work rate, will come up.  Return. Combine 0;03 20, the square-side of a brick, so that 0;00 11 06 40, the area of a brick, will come up.  Multiply 0;00 11 06 40 by 0;12, the thickness of a brick, so that 0; 00 02 13 20, the volume of a brick, will come up.  Solve the reciprocal of 0;00 02 13 20, the volume of a brick, so that 27 00 will come up. Multiply by 0;04, the work rate, so that 1 48 bricks, the output of 1 man, will come up.  Triple 1 48 so that 5 24, the output of the work rate, will come up. That is the procedure."}, {"id_text": "P254594", "project_name": "dccmt", "raw_text": "19 from the Moon (to) the Bristle. 17 from the Bristle (to) Orion. 14 from Orion to the Arrow. 11 from the Arrow (to) the Bow. 9 from the Bow (to) \u0160UPA. 7 from \u0160UPA to the Scorpion. 4 from the Scorpion to ANTAGUB: I summed and it was 2 sixties leaagues. How distant is god above god? You in your working: sum 19, 17 14, 11, 9, 7, 4 and you will see 1 21. Its reciprocal is 0;00 44 26 40.1"}, {"id_text": "P254600", "project_name": "dccmt", "raw_text": "If the upper length is 1;40, its counterpart is missing, the upper width exceeds the lower width by 0;20, its area is 0;40, what is my (other) length? You in your working: put down 1;30 and break (it in half). Combine (it) so that you see 0;45. Solve the reciproal of 0;45 so that you see 1;20. Multiply [1;20] by 0;40, the area, so that you see 0;53 20. Double 0;53 20 so that you see 1;46 40. Let your head hold 1;46 40. Return and sum 1;40, the upper length, and the 0;20 by which the upper width exceeds the lower width, so that you see 2. Halve 2 and combine (the two halves) so that you see 1. Add 1 to 1;46 40 so that you see 2;46 40. Work out my square-side of 2;46.40 so that you see 1;40. Add to 1;40, your square-side, the 1 that you combined so that you see 2;40. Subtract from the 2;40 that you saw 1;40, the upper length. The remainder is 1, the missing length.  Halve 1 so that you see 0;30. Draw 0;30 (and) the counterpart, and halve the 0;20 by which width exceeds width, so that you see 0;10. Add 0;10 to the first 0;30 so that you see 0;40. Subtract (it) from the second 0;30. You will see 0;20. The lower width is 0;20. That is the procedure. If I added two-thirds of the sum of the upper and lower widths 0;10 to my hand and I built 0;20, the length; the upper width exceeds the lower by 0;05; (and) the area is 0;02 30, what is my length?  You, in your working: my constants are the 0;05 by which it exceeded, the 0;10 that you you added, 0;40 the two-thirds, are my factors of both.1 Solve the reciprocal of 0;40, the two-thirds, so that you see 1;30. Halve 1;30 so that you see 0;45.2 Multiply 0;45 (sic, for 1;30) by 0;02 30, the area, so that you see 0;03 45. Double 0;03 45 so that you see 0;07 30. Let your head hold 0;07 30. Turn back and <solve> the reciprocal of 0;40, the two-thirds, so that you see 1;30. Halve 1;30 so that you see 0;45. Multiply (it) by the 0;10 that you added, so that you see 0;07 30. Let your head hold 0;07 30. Turn back and solve the reciprocal of 0;40 so that you see 1;30. Halve 1;30 so that you see 0;45. Multiply (it) by the 0;10 that you added so that you see 0;07 30.3 Draw 0;07 30 (and) the counterpart, and multiply (them) so that you see 0;00 56 15.  Add 0;00 56 15 to the 0;07 30 that your head is holding, so that you see 0;08 26 15. Work out the square-side of 0;08 26 15 so that its square-side is 0;22 30. Subtract 0;07 30, your takiltum, from 0;22 30, the square-side.  The remainder is 0;15.  Halve 0;15, so that you see 0;07 30. Draw 0;07 30 (and) the counterpart. Halve the 0;05 by which width exceeds width, so that you see 0;02 30. Add 0;02 30 to the first 0;07 30 so that you see 0;10. Subtract (it) from the second 0;07 30. The upper width is 0;10, the lower width 0;05. Turn back and sum 0;20 and 0;05. You will see 0;15. Take two-thirds of 0;15 so that you see 0;10. Add 0;10 to 0;10 and your upper length is 0;20. Halve 0;15 so that you see 0;07 30. Multiply 0;07 30 by 0;20 so that you see 0;02 30, the area. That is the procedure. 0;06 40, the constant of a (storage) box. (r 21) 0;06, of the capacity constant. 0;03 45, a mud wall. (r 22) 0;07 30, a grain-heap. 0;04 10, of a pile of bricks. (r 23) 0;05, of a circle. (r 24) 0;30, of a triangle. If an area (whose) lengths are not equal: you, solve the reciprocal of 4 and then let them say to you the total length(s) and then multiply (it) by the total of your lengths and then inscribe 4, the four cardinal points, and you multiply as much as they took, and then you subtract the area from (it)."}, {"id_text": "P254606", "project_name": "dccmt", "raw_text": "If someone asks, saying this: \"A triangle. The lower length is two-thirds of the upper length; the upper width is half of the lower width; the area is 2;05. What are the length and width?\" You in your working: sum 1 and two-thirds. Halve (it) so that 0;50 comes up. Put down the 0;50 that came up for you. Halve the width so that 0;10 comes up. Multiply the 0;10 that came up for you by 0;05 so that 0;00 08 20 comes up. Solve the reciprocal of the 0;00 08 20 that came up for you so that 7 12 comes up. Multiply the 7 12 that came up for you by 0;02 05, the area, so that 0;15 comes up. What does the 0;15 that came up for you make square? It squares 0;30. Turn back. The upper length is 0;30. The lower width is 0;20. The upper width is 0;10."}, {"id_text": "P254607", "project_name": "dccmt", "raw_text": "If [someone asks you, saying] this: \"I added two-thirds of my two-thirds to a hundred sila of barley <<and two-thirds>>1 and thus the barley was fully used up. How much was my original quantity?\" You, in your working: multiply two-thirds and two-thirds so that you see 0;26 40. Subtract 0;26 40 from [1] so that 0;33 20 is the remainder. Solve the reciprocal of 0;33 20 so that you see [1];48.  Multiply 1;48 by [1 40] so that you see 3 00. The original quantity was 3 00."}, {"id_text": "P254608", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"A mud wall. The breadth is two cubits, the height one cubit. \"What is the daily work-rate of one man?\" You, in your working: multiply 2 by <0;05>, your height, so that 0;10 comes up. The 0;10 that come up for you: solve the reciprocal of 0;10 so that 6 comes up. Multiply the 6 that came up for you by 0;03 45, your constant, so that 0;22 30 comes up.  The 0;22 30 that came up for you is the daily work-rate of one man."}, {"id_text": "P254609", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"A triangle. The width is two-thirds of the length. \"The area is 0;05. What are my length and my width?\" You, in your working: because it has been said to you that the width is two-thirds fo the length, put down 1, the length and 0;40, the width. Turn back. The 0;40 that [you put down] for the width [...]. Multiply 0;30 by 1, the length, so that you see 0;30. The length is 0;30.  Multiply 0;30 by 0;40, the two-thirds, so that you see 0;20. The width is 0;20."}, {"id_text": "P254610", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"I took a reed but I did not know its length.  \"I went sixty lengths and I broke off a cubit and then I went thirty (along) the width. \"The area is 4 0. How much are my length and my width?\" You, in your working: Solve the reciprocal of 0;30, your width, so that you see 2. Multiply 2 by 4 10, your area, so that you see 8 20. Let your head hold 8 20. Multiply the cubit that you broke off by 0;30, your width, so that you see 2;30. Multiply 2;30 so that you see 6;15. Multiply1 6;15 by 8 20 so that you see 8 26;15. What does 8 26;15 square? It squares 22;30. Put down 22;30 (and) the counterpart. Add 2;30 to one. Subtract from one. One is 25, one is 20. 25 to ... [...] 20 [...] ... [...] The length is 25. [(...)] The width is 20. [(...)] [...] ... [...]"}, {"id_text": "P254611", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"Four in a reed, my ..., I harvested 2 sila. \"In a bur area how much did I harvest?\" You in your working: Solve the reciprocal of four, its ..., so that you see 0;15. Multiply 0;15 by a half of your reed, so that you see 0;07 30. Solve the reciprocal of 0;07 30 so that you see 8. Turn back. [...] 2, your reed ... [...] ... what? ... you see. Multiply [...] second (one) so that you see 16. Multiply 16 by 1 annd then multiply 16 by the bur area of your field so that you see 8.  Your ... is 8. How much ... 8 ... and ...."}, {"id_text": "P254612", "project_name": "dccmt", "raw_text": "[If] (someone) asks you, [saying as follows]:  \"An earth wall. The length is a rope; the breadth is two cubits.  \"It is half a cubit thick at the upper part. The height is half a reed. \"What are your volume and [your workers] for a whole day?\"  You, in [your working], sum two cubits, the width, and half a cubit.  Halve (the result) so that [you see] 0;06 15, its halves.  Multiply [0;06] 15 by half a reed, the height, [so that] you see 0;18 45.  Multiply 0;18 45 by a rope, [the length], so that you see 3;[07] 30. []Your volume is] 3;07 30.  Go back, and 0;03 45 is the constant [of an earth wall].  Solve the reciprocal of 0;03 45 and multiply (it) by [3;07 30], your volume, so that you see 50. [Your workers are 50.] If (someone) asks you, saying as follows:  \u2018[How] much is the daily output of one man?\u2019  You, in your working: [add] two [cubits, the width], and half a cubit.  Halve (it) so that [you see] 0;06 15, its halves. Multiply [0;06 15] by [3], the height, so that you see [0;18] 45.  Solve [the reciprocal of 0;18] 45 and [multiply] by 0;03\u00a045 [so that you see 0;12. 0;12] is your [dailu output]."}, {"id_text": "P254616", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"In the exchange rate of 15 sila lard (to) 10 sila oil, two-thirds of the exchange rate of lard is the ulli\u0101nu of the excess. \"I am carrying 1 shekel of silver. Oil and lard ... [...].\" You, in your working: solve the reciprocal of 15, so that [0;04 comes up]. Multiply 0;04 by 1, so that 0;04 comes up for you. Go back, and solve the reciprocal of 10, so that 0;06 comes up. Multiply 0;06 by 0;40, so that 0;04 comes up. Go back, and the ulli\u0101num of the excess. Subtract 6 grains of silver from 1 shekel of silver so that 0;58 is the remainder. Go back, and sum 0;04 and 0;04 so that 0;08 comes up. The 0;08 that came up for you -- find its reciprocal, [so that] 7;30 comes up. Multiply the 7;30 that came up for you by 0;58, the silver, so that 7;15 [comes up]. Multiply the 7;15 that came up for you by 0;04 so that it gives you 0;29. Go back, and multiply 7;15 by 0;04 so that 0;29 comes up. Add the 6 grains of silver that you subtracted from the silver to one. One is 0;29, one is 0;31. The (cost of) buying of one is 7;15, one is 5;10."}, {"id_text": "P254618", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \u201cAs much as I made square I made deep and I removed a sar and half a sar of earth. How much did I make square;  how much did I make deep?\u201d You, in your working: Put down [1;30] and 12, and then solve the reciprocal of 12, so that [you see 0;05]. Multiply [0;05 by 1];30, your earth, so that you see 0;07 30. What does 0;07 30 square? It squares 0;30 Multiply 0;30 by 1, so that you see 0;30. Multiply 0;30 by the second 1, so that you see 0;30. Multiply 0;30 by 12, so that you see 6. Your side-of-square is 0;30. Your depth is 6."}, {"id_text": "P254621", "project_name": "dccmt", "raw_text": "If someone asks you, saying this: \"Ten sar of bricks1 are located over a cable's distance away in the countryside. How much workforce for a day should I assign so that they may complete (it)?\" You, in your working: put down 1 30, the constant (of carrying) and then solve the reciprocal of 1 30, your constant. You will see 0;00 40. Multiply 0;00 40 by the cable distance, so that you see 0;06 40. Multiply the 0;06 40 that you saw with 54 00, your ten sar of bricks, so that you see 6. Your men for a day, who will complete it for you in a day, are 6."}, {"id_text": "P254622", "project_name": "dccmt", "raw_text": "If someone asks, saying this: Two-thirds of the rectangle by [...] I added 0;10 to the width. The area is 0;20. What are the length and the width? You, in your working: Multiply two-thirds by 0;10 so that you see 0;06 40. Turn back. Multiply 0;10 by 1 so that you see 0;10. By what is 0;10 in excess of 0;06 40? It is in excess by 0;03 20. Turn back, and halve 0;03 20, so that you see 0;01 40. Multiply 0;01 40, so that 0;00 02 46 40 comes up. Add 0;00 02 46 40 to 0;13 20 so that 0;13 22 46 40 results. What does 0;13 33 46 40 square? It squares 0;28 20. [Draw] 0;28 20 (and) the counterpart [and] add the 0;01 40 that you multipled to one. Subtract (it) from one. One is 0;30; one is 0;26 40. [Turn back] and the length is 0;40, the width 0;30."}, {"id_text": "P254625", "project_name": "dccmt", "raw_text": "A triangle. The length is 1, the long length 1;15, the upper width 0;45, the complete area 0;22 30. Within 0;22 30, the complete area, the upper area is 0;08 06, the next area 0;05 11 02 24, the third area 0;03 19 03 56 09 36, the lower area 0;05 53 53 39 50 24.  What are the upper length, the middle length, the lower length, and the vertical? You, when you proceed: solve the reciprocal of 1, the length. Multiply by 0;45.  You will see 0;45. Multiply 0;45 by 2. You will see 1;30. Multiply 1;30 by 0;08 06, the upper area.  You will see 0;12 09. What squares 0;12 09? 0;27 squares (it).  The <length of the upper> wedge is 0;27. Break off <half of> 0;27. You will see 0;13 30. Solve the reciprocal of 0;13 30.  Multiply by 0;08 06, the upper <area>. You will see 0;36, the dividing length of 0;45, the width. Turn back. Take away 0;27, the length of the upper wedge, from 1;15.  The remainder is 0;48. Solve the reciprocal of 0;48. You will see 1;15. Multiply 1;15 by 0;36.  You will see 0;45. Multiply 0;45 by 2. You will see 1;30. Multiply 1;30 by 0;05 11 02 24.  You will see 0;07 46 33 36. What squares 0;07 46 33 36?  0;21 36 squares (it). The width of the second triangle is 0;21 35.  Break off half of 0;21 36. You will see 0;10 48. Solve the reciprocal of 0;10 48.  <Multiply> by ..."}, {"id_text": "P254856", "project_name": "dccmt", "raw_text": "A lion [caught] a wild boar.  He roared:  \u2018Your flesh has not yet filled my mouth,  but your squeals have deafened my ears!\u2019 16;40 \u00d7 16;40 = 4 37;46 40"}, {"id_text": "P254859", "project_name": "dccmt", "raw_text": "When a dog snarls,  throw a morsel into his mouth. 1;03 45 \u00d7 1;03 45 = 1;07 44 03 45 \u00d7 16 = 18;03 45 \u00d7 16 = 4 49 = 17 \u00d7 17 0;15 \u00d7 0;15 = 0;03 45 \u00d7 17 = 1;03 45"}, {"id_text": "P254878", "project_name": "dccmt", "raw_text": "He who can say \u2018Let him hurry, let him run, let him be strong, and he will carry it!\u2019 is a lucky man. 2;05 , 12 25 , 2;24 28;48 , 1;15 0;36 , 1;40 2;05"}, {"id_text": "P254936", "project_name": "dccmt", "raw_text": "[...] : 30 by 30 (is) 15. [...] sum: 51 45. : 51 45 rods (?) [...] Sum [15] and 51 45. 52 is remaining. 1 from 1;20 [...] :  the area is 26 00. Multiply 26 00 [by 0;00 00 21 36] so that (you see) 0;09 21 45, or 5 q\u00fb, 4 akalu, 5 grains. [A triangle/rectangle. The long side is 1, the short side 0;45. 1] by 1 is 1. 0;45 by 0;45 is 0;33 45. Sum so that (you see) 1;33 45. How much by how much should I multiply so that [it would be 1;33 45?] 1;15 by 1;15, the diagonal.  [An ox's brow] of long side 30, second long side 30. The upper short side is 50, The lower short side 14.  30 by 30 is 15 00. You lift [14] from 50 so that the remainder is 36. Its 1/2 is 18. 18 (by) 18 is 5 24. You lift 5 24 from 15 00 so that the remainder is 9 36. How much by how much should I multiply so that it would be 9 36? 24 by 24 is 9 36. The transversal of the ox's brow is 24. You sum [50] and 14, the short sides, so that (you see) 1 04. Its 1/2 is 32. You multiply 24, the transversal, by 32, so that (you see) 12 48. You multiply [12] 48 by 0;00 00 21 36, so that (you see) 0;04 26 28 48. The seed-measure (?) is 2 1/2 q\u00fb, 2 1/2 akalu, 10 [...]. [A circle] of circle 1 00 00. 1 00 00 by 1 00 00 is 1 00 00 00 00. 1 00 00 00 00 by 0;05 is 5 00 00 00. The area is 5 00 00 00. You multiply 5 00 00 00 by 0;00 00 21 36, so that (you see) 1 48. 1 hundred 8. [...] a guard-house is 10 cubits by 10 cubits and 10 cubits ... that 1 cubit ... to the top ... [...] ... may you ... from the base of the guard-house so hat the man who is in the guard-house may .... [... The reciprocal of 1] is 1. 1 by 10 is 10. 10 by 10 is 1 40. 1 hundred cubits. [...] 10 cubits by 10 cubits and its weight was 1 mina. From its middle ... 1 finger lower [...] weight of the lower. 10 by 30 is 5 00. 3 hundred fingers. 5 00 by 5 00 is 25 00 00, 1 30 thousand. [... The reciprocal of 25 00 00] is 0;00 00 02 24. You multiply 0;00 00 02 24 by 20 00 so that (you see) 0;00 48. 4 fifths akalu of barley. [...] ... together ... 4 40, the lengths ... lift so that the remainder is 9. 9 by 9 is 1 21. (Rest of tablet missing)"}, {"id_text": "P254982", "project_name": "dccmt", "raw_text": "A trench. [The length is 5 rods], the width is [1 1/2] rods, its depth is 1/2 rod. The work rate is 10 shekels of earth, its wages are 6 grains for a hired man. What are the area, volume, [labourers], and silver? The area is 7 1/2, the volume 45, [the labourers] 4 30, the silver 9 shekels. The silver for a trench is [9 shekels], the width 1 1/2 rods, <its depth> 1/2 rod. The work rate is 10 shekels of earth, its wages are 6 grains for a hired man. [What is] its length? The length is 5 rods \u00abrods\u00bb. The silver for a trench is [9 shekels], the length 5 rods, its depth 1/2 rod. The work rate is 10 shekels of earth, its wages are 6 grains for a hired man. [What is] its width? The width is 1 1/2 rods. The silver for a trench is [9 shekels], its length [5] rods, the width 1 1/2 rods. The work rate is 10 shekels of earth, its wages are 6 grains for a hired man. What is its depth? Its depth is [1/2 rod]. The silver for a trench is 9 shekels, the length 5 rods, the width 1 1/2 rods, its [depth 1/2 rod]. Its wages are 6 grains for a hired man. What is the earth of the work rate? The work rate is 10 shekels of earth. The silver for a trench is [9 shekels, the length] 5 [rods, the width 1 1/2] rods, its depth 1/2 rod. The work rate is 10 shekels of earth. [How much are] its wages for a [hired] man? Its wages are [6 grains] for a hired man. The silver for a trench is [9 shekels], its depth [1/2 rod. The work-rate is 10 shekels.] Its wages are [6] grains for a hired man. Add the length [and width] and [(it is) 6 1/2 rods] What are the [length and] width? The silver [for a trench is 9 shekels. Its depth is 1/2 rod], the work rate [10 shekels], its wages 6 grains for a hired man. [The length exceeds the width by 3 1/2 rods] What are the length and width? The width is 5 rods, the length 1 1/2 <rods>. A trench. [The length is 5 rods, the width 1 1/2 rods,] its [depth 1/2 rod.] What are the area and volume? The area is 7 1/2 sar, the volume 45. The volume of a trench is [45 sar, the width 1 1/2 rods], its depth 1/2 rod. What is its length? Its length is 5 rods. The volume of a trench is 45 [sar], the length [5] rods, its depth 1/2 rod. What is its width? Its width is 1 1/2 rods. The volume of a trench is 45 [sar], the length [5] rods, its width 1 1/2 rods. What is its depth? Its depth is 1/2 rod. The volume of a trench is 45 [sar], its depth 1/2 rod. Add the length and the width and (it is) 6 1/2 rods. What are the length and the width? The length is 5 rods, the width 1 1/2 rods. The volume of a trench is 45 sar, its depth 1/2 rod. The length exceeds the width by 3 1/2 rods. What are the length and width? The length is 5 rods, the width 1 1/2 rods. A trench. Add the area and volume and (it is) 52;30. The width is 1 1/2 rods, its depth 1/2 rod. What is its length? Its length is 5 rods. A trench. Add the area and volume and (it is) 52;30. The length is 5 rods, its depth 1/2 rod. What is [its width]? [The width] is 1 1/2 rods. [A trench.] Add [the area and volume] and (it is) <52;30>. Its depth is 1/2 rod. [Add] the length and [width and (it is)] 6 [1/2 rods]. What are [the length and width]? [A trench.] Add [the area and volume] and (it is) 52;30. Its depth is 1/2 rod. The length [exceeds] the width by 3 [1/2 rods]. What are [the length and width]? A trench in an area of 7 1/2 sar; the volume is 45.Add the length and width and (it is) 6 1/2 rods. What are the length, width, and its depth? A trench in an area of 7 1/2 sar; the volume is 45. The length exceeds the width by 3 1/2 rods. What are the length, width, and its depth? The length is 5 rods, the width 1 1/2 rods, its depth 1/2 rod. A trench in an area of 7 1/2 sar; the volume is 45. Its depth is a 7th <part> of that by which the length exceeds the width. What are the length and width? The length is 5 rods, the width 1 1/2 rods. A trench. The length is 5 rods, the width 1 1/2 rods, its depth 1/2 rod. The work rate is 10 shekels of earth. What length does 1 man take? He takes a length of 6 2/3 fingers. A trench. The length is 5 rods, the width 1 1/2 rods, its depth 1/2 rod. The work rate is 10 shekels of earth. What length do 30 labourers take? They take a length of 1/2 rod 2/3 cubit. A trench. The length is 5 rods, the width 1 1/2 rods, its depth 1/2 rod. The work rate is 10 shekels of earth. In how many days do 30 labourers finish? They finish in 9 days. A trench. The width is 1 1/2 rods, its depth 1/2 rod. The work rate is 10 shekels of earth. 30 labourers finish it in 9 days. What is its length? Its length is 5 rods. A trench. The length is 5 rods, its depth 1/2 rod. The work rate is 10 shekels of earth. 30 labourers finish it in 9 days. What is its width? The width is 1 1/2 rods. A trench. The length is 5 rods, the width 1 1/2 rods. The work rate is 10 shekels of earth. 30 labourers finish it in 9 days. What is its depth? Its depth is 1/2 rod. A trench. The length is 5 rods, the width 1 1/2 rods, its depth 1/2 rod. 30 labourers finish it in 9 days. What is the earth of the work rate? The work rate is 10 shekels. A trench. 30 labourers finish it in 9 days. Its depth is 1/2 rod, the work rate is 10 shekels. Add the length and width and (it is) 6 1/2 rods. What are the length and width? The length is 5 rods, its width 1 1/2 rods. A trench. 30 labourers finish it in 9 days. Its depth is 1/2 rod, the work rate is 10 shekels. The length exceeds the width by 3 1/2 rods. What are the length and width? The length is 5 rods, its width 1 1/2 rods. A trench. Its square-side is 2 1/2 rods, its depth 3 1/3 cubits. The work rate is 10 shekels. Its wages are 1 (ban) of grain for each hired man. What are the area, volume, labourers, and grain? The area is 6 sar and a 4th part, the volume 20 5/6 sar, the labourers 2 05, the grain 4 (gur) 5 (ban). 31 exercises about trenches."}, {"id_text": "P254989", "project_name": "dccmt", "raw_text": "A brick. The length is 1/2 cubit, the width 1/3 cubit, its height 5 fingers.  What are its area and volume? 1 man carried 9 sixties of bricks for a length of 30 rods and they paid him 1 s\u016btu of barley.  Now, he carried 5 sixties of bricks and then finished the bricks.  How much barley did they pay him?  5 1/2 sila 3 1/3 shekels of barley. A builder carried 9 sixties of bricks for a length of 30 rods and they paid him 1 s\u016btu of barley. Now, he carried 6 sixties of bricks and then finished the bricks.  ... the reciprocal of the daily work-rate for carrying bricks [...]  2 15, the bricks [...] the reciprocal of the work rate [..., what are] the length of carrying and the barley [they paid him]? 1/2(iku) 4 sar of bricks [were put down] (to be carried) for [a length] of 30 rods.  What labourers should I put down to finish the bricks in 1 day?  1 12 labourers. One man carried earth over a distance of 30 rods, and then he built a brick-pile.  For what proportion of the day did he carry earth?  For what proportion of the day did he build the brick-pile?  And how many bricks were there? 2 40.  A brick pile [...]. Its upper width [...]. Its lower width [...]. Its depth (sic) [...]. What are its depth, volume and diameter? 1 man carried 5 24 [...] half-bricks (and) baked bricks .... Brick [...] length [...] brick [...] 1/3 [...] he built a brick-pile. He put down 7 m\u016b\u0161ar of bricks. An earth wall. The [breadth] is 1 cubit, the height [1] cubit, the work rate [0;03 4]5.  What length does 1 man take? A length of 1/2 rod 3 cubits. The volume of an earth wall is 5 shekels, the breadth 1 cubit, the height 1 cubit, the work rate 0;03 45.  What length does 1 man <take>? [A length of] 1 rod. The volume of an earth wall is 5 shekels, the breadth 2 cubits, its height 1 cubit, the volume of the work rate 0;03 45.  What length does 1 man <take>? A length of 1/2 rod. An earth wall. Its length is 5 cables, the breadth 2 cubits, its height 1/2 rod.  In 1 cubit (height) it decreases 1/3 cubit in width and then a man demolishes so that a height of 1 1/2 cubits is left.  How long should he make the length? A dyke. The breadth is 1 cubit, its height 1 cubit. In 1 cubit (height) it slopes 2 cubit. What earth is packed down in a length of 1/2 rod? An old dyke. The breadth is 1 cubit, the height 1 cubit. In 1 cubit it slopes <1 cubit>.  Now, I added a breadth of 1 cubit, a height of 1 cubit. What earth [is packed down] in a length of 1/2 rod?  What are the new volume [and the old volume]? [The volume is 10 shekels. The old volume is] 2 [1/2 shekels, the new volume] 7 [1/2 shekels.] An old dyke. The breadth is 2 cubits, its height 2 cubits.  Now I added a breadth of 1 cubit, a height of 1 cubit. In 1 cubit (height) it sloped 1 cubit. What earth is packed down in a length of 1/2 rod?  The volume is 1/3 sar and 2 1/2 shekels. What are the old volume and the new volume? The old volume is 10 shekels, the new volume 12 1/2 shekels. 1/3 mina of plucked wool, 1/2 mina of carded wool, 1 1/2 shekels of finished wool.  6 shekels are diminished from 1 mina. 6 shekels are taken away from 1 mina of wool so that (there are) 5/6 mina 4 shekels of plucked wool.  1/2 mina 6 shekels of carded wool are taken away from 5/6 mina 4 shekels of plucked wool so that (there are) 18 shekels of \u00abcarded\u00bb <finished> wool.  What wool did I give to 1 woman? Plucked wool, carded wool, finished wool.  Multiply 2/3, the reeds, by 7 1/2 m\u016b\u0161ar, the volume, so that 5 m\u016b\u0161ar, the reeds (results). Multiply 1/3 m\u016b\u0161ar, the earth, by 7 1/2 m\u016b\u0161ar, so that 2 1/2 m\u016b\u0161ar, the earth. Solve the reciprocal of 5 shekels, (cutting) reeds. Multiply by 5, the volume of the reeds, so that you see 1 sixty, the workers. Solve the reciprocal of 6 shekels, carrying reeds. Multiply by 5 m\u016b\u0161ar, the volume, so that you will see 50, the workers. Solve the reciprocal of 2/3, the reeds. Multiply by 5 sar, the volume, then and you will see 7;30, the workers. Solve the reciprocal of 0;10, the work rate. Multiply by 2 1/2 sar, the volume, and you will see 15, the workers. 23 exercises, second tablet."}, {"id_text": "P254990", "project_name": "dccmt", "raw_text": "If an area with sloping lengths \u2014 the first length is 5 10, the second length 4 50, the upper width 17, the lower width 7, its area is 2(b\u016bru). The area is divided into two, 1 (bur) each. How much is my middle dividing line?  How much should I put for the long length and the short length so that 1(b\u016bru) is correct; and for the second 1(b\u016bru) how much should I put for the long length and the short length so that 1(b\u016bru) may be correct? You sum both the complete lengths and then you break off their halves, so that 5 00 will come up for you.  You solve the reciprocal of 5 00 that came up for you, and then \u2014 as for the upper width which exceeded the lower width by 10 \u2014 you multiply by 10, the excess, so that it will give you 0;02.  You turn around. You combine 17, the upper width, so that 4 49 will come up for you.  You take away 2 00 from the middle of 4 49 so that 2 49 is the remainder.  You take its square-side so that 13, the middle dividing line, will come up for you.  You sum 13, the middle dividing line that came up for you, and 17, the upper width, and then you break of their halves so that 15 will come up for you.  You solve the reciprocal of 15, and then you multiply by 1(b\u016bru), the area, so that it will give you 2 00.  You multiply the 2 00 which came up for you by 0;02, the rising-factor, so that 4 will come up for you.  You add 4 that came up to you to 2 cables, so that 2 04 is the long length. You take away 4 from the second 2 cables so that 1 56 is the short length.  You do (the necessary calculations) so that (you will see that) 1(b\u016bru) is correct.  You turn around. You sum 13, the middle dividing line that came up for you, and 7, the lower width.  You break their halves so that10 will come up for you.  You solve the reciprocal of 10, and you multiply by 1(b\u016bru), the area, so that 3 cables will come up for you.  You multiply the 3 cables which came up for you by 0;02, the rising-factor (?), and the 6 will come up for you.  You add 6 to 3 cables so that 3 07 is the long length.  You take away 6 from 3 cables so that 2 54 is the short length.  You combine (what is necessary) so that (you will see that) 1(b\u016bru) is correct."}, {"id_text": "P255041", "project_name": "dccmt", "raw_text": "A reciprocal exceeds its reciprocal by 7.  What are the reciprocal and its reciprocal?  You: break in two the 7 by which the reciprocal exceeds its reciprocal so that 3;30 (comes up). Combine 3;30 and 3;30 so that 12;15 (comes up).  Add 1 00, the area, to the 12;15 which came up for you so that 1 12;15 (comes up).  What squares 1 12;15? 8;30. Draw 8;30 and 8;30, its counterpart, and then take away 3;30, the holding-square, from one; add to one.  One is 12, the other is 5.  The reciprocal is 12, its reciprocal is 5."}, {"id_text": "P255060", "project_name": "dccmt", "raw_text": "A kiln. The circle is 1;30 rods.  You divide half of the dividing line into four and then \u00abyou combine 0;15, the quarter.\u00bb 1 0;03 45 will come up and then  you multiply (it) by 12, of the depth, so that 0;45, the depth, comes up.  You return. You combine 1;30, the circle, multiply (it) by 0;05 so that 0;11 15 will come up, and then you combine 0;45 and 0;11 15 and then you multiply as much as comes up by 7;12 so that it gives (you) the (number of) bricks."}, {"id_text": "P269999", "project_name": "dccmt", "raw_text": "I added [the areas of 2] squares: 13. [And] I added my square-sides: 5. (too fragmentary to translate) [...] comes up for you. [...] the square-side. 3 times 3 (is) 9. The first area is 9. Subtract 9 from 14, the area, so that you you may learn the second area.  (unclear) The remainder of the square is 4. The area of your second square is 4. Release from the square-sides so that you many learn your widths. 4 squares 2. Your second square-side is 2. A pile of bricks. The length is 10 rods, the width 1 rod, its height 1/2 rod. I entered the length and took out 1 iku square of bricks. How (far) did I enter? Multiply 10, the length, by 1, the width. 10 will come up for you. Multiply 10 by 6, the height. 1 00 will come up for you. Multiply [1 00 by 7];12, the coefficient (of brickage). 7 12 will come up for you. Find its reciprocal. 0;00 08 20 will come up for you. Multiply 0;00 08 20 by 1 40, its bricks. 0;13 53 20 will come up for you. Multiply 0;13 53 20 by 10, the length. 2;18 53 20 will come up for you. You enter 2;18 53 20 (rods into the pile of bricks)."}, {"id_text": "P274707", "project_name": "dccmt", "raw_text": "A ramp. At the base of the volume the lower breadth is 1 rod, the top 0;30 rods, the height 4. In front of the city gate the lower breadth is 1;20, the upper breadth 1 rod, the height 6. What is the volume? Demarcate the length for 1 man. You: sum <1 and> 1;30. You will see 2;30. Break off 1/2 of 2;30. Put down 1;15. Sum 1 and 0;30. You will see 1;30. Break off 1/2 of 1;30. You will see 0;45. Sum 1;15 and 0;45. You will see 2. <Break off> its 1/2. You will see 1. Put (it) down. Sum 6, the height, and 4, the height <at> the base of the volume. You will see 10. Break its 1/2. You will see 5. Multiply 10, the length, by 5. You will see 50, the volume. Solve the reciprocal of 0;10, the work rate. You will see 6. Multiply 50, the volume, by 6. You will see 5 00, the labourers. Solve the reciprocal of the labourers. You will see 0;00 12. Multiply 0;00 12 by 10, the length. You will see 0;02. 1 man takes 0;02 (rods). That is the procedure. Temple foundations. The length is 0;30, the width 0;20. On each side 0;10 squares the support. The depth is 3 cubits. <What is> the volume? Demarcate the length for 1 man. You: square 0;10. You will see 0;01 40. Multiply 0;01 40 by 3 cubits, the depth. You will see 0;05, the volume of the support. Return. See the volume of the length. \u00abSquare-side.\u00bb <Multiply> 0;20 <the length> by 10, the width. You will see 0;03 20. Multiply 0;03 20 by 3 cubits, the depth. You will see 0;10. The volume of the length is 0;10. You sum 0;10, the volume, 0;05, and 0;05, the volume. You will see 0;20, the (total) volume. Solve the reciprocal of 0;10, the work rate. You will see 6. Raise 0;20 by 6. You will see 2. 1 man takes a length of 2. That is the procedure. Temple foundations. The length is 0;30, the width 0;20, the depth 3 cubits. <On each side 0;10 squares the two supports.> Demarcate the length for 1 man in 1/2 a day. You: square 0;10. You will see 0;01 40. Multiply 0;01 40 by [3 cubits], the depth. You will see 0;05, the volume. Return. See the volume of the second support. Square 0;10. Multiply 0;01 40 by 3 cubits. You will see 0;05, the second volume. Multiply 0;20, \u00abthe square-side of\u00bb the length by 0;10, the width. You will see 0;03 20. Multiply 0;03 20 by 3 cubits, the depth. \u00abMultiply 0;03 20 by 3 cubits, the depth.\u00bb You will see 0;10, the volume. Return. Demarcate the length assigned to 1 man in 1/2 <a day>. Break off 1/2 of 0;05, the volume. You will see 0;02 30. Solve the reciprocal of 0;10, the work rate. You will see 6. Multiply 0;02 30 by 6. You will see 15. The labourers are 15 strong. Solve \u00absolve\u00bb the reciprocal of 15. You will see 0;04. Multiply 0;10 by 0;04. You will see 0;40. The length demarcated to 1 man is 0;40. Return. See 0;10, the work rate. Break off 1/2 of 0;10. You will see 0;05. Multiply 0;40 by 0;05. You will see 0;03 20. Multiply 0;03 20 by 3 cubits, the depth. You will see 0;10, the work rate. That is the procedure. A city. I encircled it with a circle of 1 sixty (rods). It projected 5 on each side and then I built a moat. The depth was 6. I took away a volume of 1 07 30. <It projected> 5 on each side. Above the moat I built a dyke. That dyke sloped 1 cubit in 1 cubit. What are the base, top and height? And what is the circle of the dyke? You: as the circle is sixty, what is the dividing line? Take away a third part of sixty, the circle. You will see 20. The dividing line is 20. Double 5, the border. You will see 10. Add 10 to 20, the dividing line. You will see 30. Triple the dividing line. You will see 1 30. The circle of the moat is 1 30.  Return. Square 1 30. You will see 2 15 00. Multiply 2 15 00 by 0;05, (the constant of) the circle. You will see 11 15, the area. Multiply 11 15 by 6, the depth. <You will see 1 07 30, the volume.> Solve the reciprocal of 0;10, the work rate. You will see 6. Multiply 6 by 1 07 30, the volume. You will see 6 45 00. The labourers are 6 45 00 strong. Solve the reciprocal of 6 45 00. You will see 0;00 08 53 20. Multiply 0;00 08 53 20 by 1 30, the circle. You will see 0;13 20, (the length for 1 man) demarcated in the moat.  Return. See 0;10 the work rate. Solve the reciprocal of 1 30, the circle. You will see 0;00 40. Multiply 0;00 40 by 0;13 20. You will see 0;00 08 52 30. Multiply 0;00 08 53 20 by 1 07 30, the volume. You will see 0;10, the work rate.  Return. See the dyke. Copy 0;05, the slope. You will see 0;10. Copy 0;10. You will see 0;20. Multiply 0;20 by 1 07;30. You will see 22;30. What should I add to 22;30 so that it may satisfy a \u00abremainder\u00bb <square-side> and that which is added may satisfy a square-side? Add 5 03;45. You will see 27 33;45. The square-side of the base is 5;15. What is the square-side of 5 03;45? The square-side is 2;15, the top. Solve the reciprocal of 2 sixties, the circle of the dyke. You will see 0;00 30. < Multiply> 0;00 30 by 1 07;30. You will see 33;45. Return. Sum the base and the top. You will see 7;30. Break off 1/2 of 7;30. You will see 3;45. Solve the reciprocal of 3;45. You will see 0;16. Sum 0;16 by 33;45. You will see 9. The height of the dyke is 9. That is the procedure. A wall. The length is sixty, the top 0;30, the base 1, the height 6. <What is> the volume? Demarcate the length for 1 man. You: sum 0;30 and 1. You will see 1;30. Break off 1/2 of 1;30. You will see 0;45. Multiply 0;45 by 6, the height. You will see 4;30, the volume. Solve the reciprocal of 0;10, the work rate. You will see 6. Multiply 4;30 by 6. You will see 27, the labourers. Solve the reciprocal of 27, the labourers. You will see 0;02 13 20. by 0;02 13 20 by 1 cable (the length of the wall). You will see 2;13 20. 1 man takes (it). That is the procedure. An outflow water clock. It opened and the outflow was 1/2 sila. It does not reach the 1 sila mark by a 4th part of 0;00 10, a finger. What is the height of the surface over the surface? You: solve the reciprocal of 0;01 40, the height of the water clock. You will see 36. Multiply 36 by 0;30. You will see 18. Multiply 18 by 0;00 02 30. You will see 0;00 45, by which the surface exceeds the surface. That is the procedure.  An outflow water clock. It opened and <the outflow was> 1/2 sila. The surface area <exceeded> the surface by 0;00 45. What did I take out <to reach> the 1 sila mark? You, solve the reciprocal of 0;30, the \u00ab1\u00bb <1/2> sila outflow. You will see 2. Multiply 2 by 0;01 40. You will see 0;03 20. Multiply 0;03 20 by 0;00 45. You will see 0;00 02 30, by which it did not reach the 1 sila mark. That is the procedure. An outflow water clock. It opened 0;03 20. The outflow was a broken sila. It did not reach the 1 sila mark by 0;00 00 44 26 40, a 9th part of 2/3 of 0;00 10, a finger. By what did the surface exceed the surface? Solve the reciprocal of 0;01 40. You will see 36. Multiply 36 by 0;03 20. You will see 3. Multiply 2 by 0;00 00 44 26 40. The surface is higher than the surface by 0;00 01 28 53 20. The procedure. A wedge (?). 20 squares it. What is its area? You: square 20. You will see 6 40. Take away a \u00ab4th\u00bb <1/2> of 6 40. You will see 3 20. The area is 2 iku. That is the procedure. The iml\u00fb is 1. What is the circle (or: depth)? You: put down 4 and 3, the ratios?. Multiply 4 by 3. You will see 12. <Solve> the reciprocal of 18, the iml\u00fb. You will see 0;03 20. Multiply 0;03 20 by 12. You will see 0;40. The circle (or: depth) is 0;40. That is the procedure. The circle (or: depth) is 0;40. <What is> its \u00abweight\u00bb <iml\u00fb>? <You: put down 4 and 3, the ratios.> Multiply 4 by 3. You will see 12. Solve the reciprocal of 12. you will see 0;05. Multiply 18, its clay (?), by 0;05. You will see 1;30. Multiply 1;30 by 0;40, the circle (or: depth). You will see 1. That is the procedure. Its iml\u00fb is 0;30. What is the circle (or: depth)? You: multiply 4 by 3. You will see 12. Solve the reciprocal of 18. You will see 0;03 20. Multiply 0;03 20 by 12. You will see 0;40. Multiply 0;40 by 0;30. You will see 0;20, the circle(or: depth). That is the procedure. The circle (or: depth) is 0;20. What is the iml\u00fb? You: multiply 4 by 3. You will see 12. Solve the reciprocal of 12. You will see \u00abyou will see\u00bb 0;05. Multiply 18, the iml\u00fb, by 0;05. \u00abMultiply\u00bb <You will see> 1;30. Multiply 1;30 by 0;20, the circle(or: depth). You will see 0;30. The iml\u00fb is 0;30. That is the procedure. A reed bundle. The lower circle is 0;04, the upper circle \u00ab0;01\u00bb <0;02>, the height 6. What is the volume and the dividing line of the upper volume and lower volume? You: square 0;04. You will see 0;00 16. Multiply 0;00 16 by 0;05, (the constant of) the circle. You will see 0;00 01 20.  Square 0;02. You will see 0;00 04. Multiply 0;00 04 by 0;05, (the constant of) the circle. You will see 0;00 00 20. Sum 0;00 01 20 and 0;00 00 20. You will see 0;00 01 40. Break off 1/2 of 0;00 01 40. You will see 0;00 00 50. Multiply 0;00 00 50 by 6, the height. You will see 0;00 05, the volume of reed bundles. That is the procedure. A reed bundle. The base is 0;04, the top 0;02, the height 6, the volume 0;00 05. I went up 3 cubits. What are the dividing line and volume? You: by what does 0;04, the base \u00abgo up over\u00bb <exceed> 0;02, the top? It exceeds by 0;02. Solve the reciprocal of 6, the height. You will see 0;10. Multiply 0;10 by [0;02]. You will see 0;00 20. Multiply 0;00 20 by 3 that you went up. You will see 0;01. Take away 0;01 from 0;04, the base. [You will see 0;03.] The dividing line is 0;03. That is the procedure. Baked bricks for a well. The length is 0;03 20, the upper [width] 0;02 [30], the lower width 0;01 40. [How many bricks] are put down for the well? You: by [what] does 0;02 30, the upper width, exceed [0;01 40, the lower width]? It exceeds by 0;00 50. Solve the reciprocal of 0;00 50. [You] will see [1 12]. Multiply 1 12 by 0;01 40. You [will see] 2. [Square 2.] You will see 4. Multiply 0;03 20, the length, by 4. [You] will see 0;1[3 20], the large dividing line. Copy 0;13 20 times 3. You [will see] 0;40. The circle is [0;40].  [Return. What is] one layer (of bricks)? Solve [the reciprocal of 0;01 40, the upper width]. You will see 36. [Multiply] 36 [by 0;40, the circle. You will see 24, one] layer (of bricks).  [Return. What is] the central circle? Copy [0;03 20, the length] of a brick. <You will see 0;06 40.> [Add] 0;06 40 to 0;13 [20. You will see 0;20.] Copy 0;20 times 3. You will see 1. [The central circle is] 1. That is the procedure. A wall. The height is 36, the top 1/2 [rod, 3] cubits. I descended from the top by as much as the base. What are what I descended, the base, the dividing line and the volume? You: in 1 cubit the slope is 0;00 50. Copy 0;00 50. You will see 0;01 40. Multiply 0;01 40 by 36, the height. You will see 1. Put it down. Square 0;45. You will see 0;33 45. Sum 1 and 0;33 45, so that you will see 1;33 45. What is the square-side? The square-side is 1;15. The base is 1;15. Multiply 1;15 by 12, the ratio of height. You will see 15, <that I descended>. Put down 0;25, the slope.  Return. By what does 1;15, the base, exceed 0;45 <, the top>? It exceeds by 0;30. Multiply 0;25, the slope, by 0;30, the excess. You will see 0;12 30. Add 0;12 30 to 0;45, the top, and you will see 0;57 30, the dividing line.  Return. See the <lower> volume. Add 0;57 30 and 0;45 the top. You will see 1;42 30. Break 1/2 of 1;42 30. You will see 0;51 15. <Raise> 0;51 15 by the 15 that you descended. You will see 12;48 45, the upper volume.  Return. See the lower volume. Sum 0;57 30 and 1;15. <You will see> 2;12 30. Break off 1/2 of 2;12 30. You will see 1;06 \u00ab40\u00bb <15>. Multiply 1;06 40 by 21, the (remaining) height. You will see 23;11 15, the lower volume. That is the procedure. A wall. The volume is 36, the height 36. In 1 cubit (height) the (total) slope is 0;00 50. <What are> the base and the top? You: solve the reciprocal of 36, \u00abthe volume\u00bb <the height>. You will see 0;01 40. Multiply 0;01 40 by 36. You will see 1. Put down \u00ab2\u00bb <1>. Multiply 0;00 25, the slope (of one side), by 36. You will see 0;15. Take away 0;15 from 1. You will see 0;45, the top. Add 0;15 to 1. You will see 1;15, the base. That is the procedure. Baked bricks for a well. The length is 0;03 20. The upper width <exceeds> the lower width by 0;00 50. The circle is 0;40 \u00abexceeds\u00bb. What are the upper width and the lower width? You: take away the 0;20th part (sic) of 0;40. You will see 0;13 20. Solve the reciprocal of 0;03 20. You will see 18. Multiply 18 by 0;13 20. You will see 4. Break off 1/2 of 4. You will see 2. Multiply 0;00 50, the excess, by 2. You will see 0;01 40. Put down 0;01 40 as the ratio (?) of the upper width and lower width. Add 0;00 50 to 0;01 40. You will see 0;02 30. The upper width is 0;02 30. That is the procedure. A circle was 1 00. I descended 2 rods. What was the dividing line (that I reached)? You: \u00abyou:\u00bb \u00abSquare\u00bb <double> 2. You will see 4. Take away \u00abyou will see\u00bb 4 from 20, the dividing line. You will see 16. Square 20, the dividing line. You [will see] 6 40. Square 16. You will see 4 16. Take away 4 16 from 6 40. You will see 2 24. What squares 2 24? 12 squares it, the dividing line. That is the procedure. If I circled a circle of 1 00 (and) the dividing line was 12, <what> was it that I descended? You: square 20, the dividing line. You will see 6 40. Square 12. <You will see 2 24.> Take away 2 24 from 6 40. You will see 4 16. What squares <4> 16? \u00ab4\u00bb <16> squares it. <Take away 16 from 20. You will see 4.> Break off 1/2 of 4. You will see 2. What you descended is 2. The procedure. A (measuring vessel of 1) bariga. The dividing line is 0;04, the grain 1 00. What are the depth and circle? You: square the dividing line, You will see 0;00 16. Solve the reciprocal of 0;00 16. You will see 3 45. The depth is 3 45. The procedure. A textile. The length is 48 (rods). In 1 day she wove 0;20 (rods). In how many days will she cut (it) off (the loom)? You: solve the reciprocal of 0;20. You will see 3. Multiply 48 by 3. You will see 2 24. She will cut (it) off after 4 months, 24 days. A (measuring vessel of 1) s\u016btu is full of grain. I removed [1 sila of grain] from inside it. <Its dividing line is> 0;02. What depth should I descend so that it is 1 sila? [...] You: divide 0;02 30, the depth, by 10. Solve the reciprocal of 10. You will see 0;06. Multiply [0;02] 30 by 0;06. You will see 0;00 15. The depth is 0;10 fingers (and) 1/2 (of) 0;10 fingers. May you see 1 sila. Square 0;02, the dividing line. You will see 0;00 04. Multiply 0;00 15, the depth, by 0;00 04. You will see 0;00 00 01. The grain is 1 sila. That is the procedure. I shall capture a city hostile to Marduk with a volume of 1 30 00. I established the foundations of the volume as 6, 8 to reach the wall. The vertical of the volume is 36. How much length should I trample down so that I may capture [the city], and [what is] the length behind the hole (?)? [You:] solve [the reciprocal] of 6, the foundations of the volume. You will see 0;10. [Multiply] 0;10 by [1 30 00, the volume]. You will see [15 00]. Solve the reciprocal of 8. You will see 0;07 30. Multiply [0;07 30 by 15 00]. You will see 1 52;30. Copy 1 52;30. [You will see 3 45.] Multiply 3 \u00ab44\u00bb <45> by 36. You will see 2 15. [Square] 1 52;\u00ab20\u00bb <30>. You will see [3 30];56 15. [Take away] 2 15 from 3 30;56 15. What squares [1 1]5;56 \u00ab14\u00bb <15>? You will see 1 07;30. Take away [1 07;30 from] 1 52;30. You will see 45, the height of the wall. [Break off 1/2 of 45.] You will see [2]2;[30]. Solve the reciprocal of 22;30. You will see 0;02 40. Multiply [15 00 by] 0;02 40. The length is 40. Return. See 1 30 00, the wall. Multiply 22;30, [1/2 (?) the height] by 40, the length. You will see 15 00. Multiply 15 00 by 6. You will see 1 30 00. The volume is 1 30 00. That is the method. I shall capture a city hostile to Marduk with a volume of 1 30 00. I walked from the base of the volume a length of 32 (rods) in front of me. The height of the volume is 36. What length should I trample so that I may capture the city? What is the length that is in front of the hole (?)? You: solve the reciprocal of 32. You will see 0;01 52 30. Multiply 0;01 52 30 by 36, the height. You will see 1;07 30. Solve the reciprocal of 6, the foundations of the volume. You will see 0;10. Multiply 1 30 00, the volume, by 0;10. you will see 15 00. Copy 15 00 twice. You will see 30 00. Multiply 30 00 by 1;07 30. You will see 33 45. What squares 33 \u00ab42\u00bb <45>? 45 squares. The height of the wall is 45. By how much does \u00ab44\u00bb <45>, the height of the wall, exceed 36, the height of the volume? It exceeds by 9. Solve the reciprocal of 1;07 30. You will see 0;53 20. Multiply 0;53 20 by 9. You will see 8. You tread a length of 8 (rods) in front of you. From 4 furrows 0;30 fell off. I harvested 1 sila of grain from 1/2 rod. [...] What is the grain of 1 (bur) of field? You: [release] the reciprocal of 4, the furrows. You will see 0;15. Multiply 0;15 by 0;30. You will see 0;07 30. Solve the reciprocal of 0;07 30. You will see 8. Multiply 8 by 30 00, the area. You will see 4 00 00. The length of 1 (bur) of area is 8 leagues. Solve the reciprocal of 30 00, the area. You will see 0;00 02. Multiply 4 00 00 by 0;00 02. You will see 8. There are 8 (00 00 sila) \u00abin\u00bb 1 36 gur of grain. A moat, 10 on each side. The height (sic) is 18. In 1 cubit (height) the slope is 1. <What are> the base and volume? You: sum 0;05 and 0;05. You will see 0;10. Multiply [0;10] by 18, the height. You will see 3. Take away 3 from 10. [You will see] 7, the base. Return. Sum <7>, the base and <10>, the top \u00ab10\u00bb. You will see 17. [Break off 1/2 of 17.] You will see 8;30. Square. You will see 1 12;15. [Put down] 1 12;[15]. Take away a \u00absecond\u00bb <fourth> part of 3, the excess of the top over the base. Add 0;45 to 1 12;15 so that you will see 1 13. Multiply 18 by 1 13. You will see 22 30. The volume is 2 (eshe) 1 1/2 (iku). That is the procedure. A crescent moon. The (arc of) circle is 1 00, the dividing line 50. <What is> the area? You: by what does 1 sixty, the circle, exceed 50? The excess is 10. Multiply 50 by 10, the excess. You will see 8 20. Square 10, the dividing line. You will see 1 40. Take away 1 40 from 8 20. You will see 7 30. The area is [7 30]. The procedure. If a boat carries 1 sar of bricks, what grain does it carry? You: 0;41 40 is the volume of [1 00] bricks. Multiply 0;41 40 by 5, (the constant) of storage. The \u00abvolume\u00bb <grain> is 3;28 20. The \u00abbrick\u00bb <grain> of 1 brick is 3 1/3 sila 8 1/3 shekels. Multiply 3;28 20 by 12 fingers\u00bb <sixties>. You will see 41 40. You will see 8 (gur) 1 40 (sila) grain. The procedure. If a p\u0101rum is 1 rod square [...] to 1/2 p\u0101rum 2 ... [...] ... subtract. From [...] ... [...] on its top [...] ... to the width .... [...] as 30 [...] take 10 [...] ... you will see 50. 50 [...] ... to its foot [...] .... Add 1/2 to 50. You will see [...] 1 05 1/2 from the width [...] ... to my foot. That is the procedure. [...] upper ... 2 [...]. What is the [...] of its grain? You: multily 40 [by 1] 12, the constant. You will see 48. [...] ... of its grain. That is the procedure. The length is [0;30], the area 0;10. What is the width? You: [Break off 1/2 of 0;30.] You [will see] 0;15. Solve the reciprocal of 0;15. [You] will see 4. [Multiply] 0;10 by [4] You will see [0;40]. Multiply 0;40 by 0;30 \u00ab0;30\u00bb, the length. [You will see 0;20], the width.  A box. I circled [a circle of] 0;03, [fingers]. The earlier (level of) water was like [...] .... The later in ... [...] complete .... What height did it (the level of the water) go down? You: [square 0;03. You] will see [0;00 09]. Multiply 0;00 09 by 0;05 (the constant) of a circle. You will see 0;00 00 45. [Solve the reciprocal of 0;00 00 45.] You will see 1 20. Multiply the reciprocal of 6 40 00 by \u00ab0;00 00 09\u00bb <1 20>. [You will see 0;00 12.] It went down a height of 0;00 12 rods. The procedure. I enclosed a box of 1 sila (with) 0;10, fingers. What length did I go? You: square 0;10. You will see 0;01 40. Multiply 0;01 40 by 0;05, (the constant) of a circle. You will see 0; 00 08 20. Solve the reciprocal of 0;00 08 20. You will see 7 12. Multiply the reciprocal of 6 40 00 by 7 12. You will see 0;01 04 48, the length. The procedure. colophon Total 35 processes of calculation."}, {"id_text": "P357323", "project_name": "dccmt", "raw_text": "10 minas of pure gold at 5 1/2 shekels (per shekel of silver): its silver is 55 minas."}, {"id_text": "P357324", "project_name": "dccmt", "raw_text": "5 1/2 minas of red gold at 3 1/3 shekels (per shekel of silver): its silver is 17 2/3 minas, 6 2/3 shekels."}, {"id_text": "P357325", "project_name": "dccmt", "raw_text": "10 minas of pure gold at 5 1/2 shekels (per shekel of silver): its silver is 55 minas."}, {"id_text": "P357326", "project_name": "dccmt", "raw_text": "1 talent, 14 minas of top-load copper at 1/2 mina, 5 shekels (per shekel of silver): its silver is 2 minas, 6 1/2 shekels, 15 grains."}, {"id_text": "P357327", "project_name": "dccmt", "raw_text": "14 minas of [pure] gold at [5 1/2 shekels] (per shekel of silver): 1 talent, 17 minas (of silver)."}, {"id_text": "P357328", "project_name": "dccmt", "raw_text": "... minas ... its [silver] is ... 1 shekel."}, {"id_text": "P357332", "project_name": "dccmt", "raw_text": "(MA\u0160 sign) 10 minas of pure gold at 5 1/2 shekels (per shekel of silver): the silver is 55 minas."}, {"id_text": "P368255", "project_name": "dccmt", "raw_text": "Calculation. In 1 reed there are 4 furrows. Furrows decrease on furrows by 0;06 (rods). I ...ed a width of 1 00, (namely) sixty (rods). What is the length? Put down 0;30, a reed. Put down 4, the furrows. Put down 0;06 by which furrows decrease on furrows. Put down 1 00, the width which he ...ed, and then: Solve the reciprocal of 0;30, a reed: 2. Multiply 2 by 4, the furrows: 8. Multiply 8 by 0;06, by which furrows decrease on furrows: 0;48. Solve the reciprocal of 0;48: 1;15. Multiply 1;15 by 1 00 the width which you ...ed: 1 15, the length. [If] the length is 1 15 and the width 1 00, what is its [area]? Break 1 00, the width, in two: 30. Multiply 30 by 1 15, the length: 37 30 (sar) the area."}, {"id_text": "P391079", "project_name": "dccmt", "raw_text": "5 minas of selected gold at 15 shekels (per shekel of silver): its silver is 1 talent, 15 minas. "}, {"id_text": "P414659", "project_name": "dccmt", "raw_text": "10 brothers (inherited) 1 2/3 minas of silver.  Brother exceeded brother (but) I do not know by how much they exceeded.  The 8th share was 6 shekels. By how much did brother exceed brother? You, when you proceed: solve the reciprocal of 10, the workers (sic), so that it gives 0;06.  You multiply 0;06 by 1 2/3 minas, the silver, so that it gives 0;10.  You copy two times so that it gives 0;20.  Copy 0;06, the eighth share, two times so that it gives 0;12. Take away 0;12 from 0;20 so that it gives 0;08.  Let your head hold 0;08.  Sum 1 and 1, the lower part (?), so that it gives 2.  Copy 2 two times, so that it gives 4.  You add 1 to 4, so that it gives 5.  You solve 5 from 10, the workers, so that it gives 5.  You solve the reciprocal of a 5th part, so that it gives 0;12.  You multiply 0;12 by 0;08 so that it gives 0;01 36, the 0;01 36 (minas) that brother exceeded brother. [Too broken to translate] 24 shekels, 4 minas ... [...] and how many workers ... [...] 4(s\u016btu) market price, one-quarter ... [...] Sum a 7th part of the length, a 7th part of the width, and a 7th part of the area, so that (it is) 2.  Sum [the length] and the width, so that (it is) 5;50.  What are the length and width? The length is 3;30, the width 2;20. Sum a 7th part of the length and the area, so that (it is) 27. The width is 0;30.  What are the length and area? The length is 42, the area 21. A (measuring) reed of 1 cubit. 1 finger kept falling off for me until it was finished.  What length did I go?  I went a length 1 rod 3 1/2 cubits. A siege ramp, of length 10 rods, width 1 1/2 rods.  3 governors (each) took on a length of 3 rods 4 cubits length.  One (had) 1 sixty workers, the 2nd (had) 1 20 workers, the 3rd (had) 1 40 workers.  The earth was transported [...].  [What are] my siege ramp and my depth, and how much earth [was transported]?"}, {"id_text": "P414660", "project_name": "dccmt", "raw_text": "Sum the areas of 2 squares, so that (it gives) 16 40. (One) square-side is 2/3 of the (other) square-side.  Take away 10 from the small square-side. What are the square-sides? You, when you proceed: combine 10. It gives 1 40. Take away 1 40 from 16 40, so that it gives 15 00.  Combine 1, so that it gives 1. Combine 0;40 so that (it gives) 0;26 40.  Sum 1 and 0;26 40, so that it gives 1;26 40. Multiply 1;26 40 by 15 00. It gives 21 40. Multiply 0;40 by 10 so that it gives 6;40.  Combine 6;40 so that it gives 44;26 40. Add 44;26 40 to 21 40, so that it gives 22 24;26 40.  The square-side of 22 24;26 40 is 36;40. Add the 6;40 that you combined to 36;40, so that it gives 43;20. What should I put to 1;26 40 so that it gives 43;20? Put 30. Multiply 30 by 1, so that the large square-side is 30. Multiply 30 by 0;40 so that it gives 20. Take away 10 from 20, so that the small square is 10. Sum the areas of 2 squares, so that (it gives) 37 05. (One) square-side is 2/3 of the (other) square-side. Add 10 to the large square-side, add 5 to the small square-side. What are the square-sides? You, when you proceed: combine 10, so that it gives 1 40. Combine 5. It gives 25. Sum 1 40 and 25, so that it gives 2 05.  Take away 2 05 from 37 05, so that it gives 35 00. Combine 1, so that it gives 1. Combine 0;40 so that (it gives) 0;26 40.  Sum 1 and 0;26 40, so that it gives 1;26 40. Multiply 1;26 40 by 35 00, so that it gives 50 33;20. Multiply 10 by 1. [It gives] 10.  Multiply 0;40 by 5, so that it gives 3;20. Sum 10 and 3;20, so that it gives 13;20. Combine 13;20. <It gives> 2 57;46 40.  Add 2 57;46 40 to 50 33;20, so that it gives 53 31;06 40. The square-side of 55 31;06 40 is 56;40.  Take away 13;20 from 56;40, so that it gives 43;20. What should I put to 1;26 40 so that it gives 43;20?  Put 30. Multiply 30 by 1, so that it gives 30. Add 10 to 30, so that the large square-side is 40. Multiply 30 by 0;40 so that it gives 20. Add 5 to 20, so that it gives 25, the small square. Sum the areas of 2 squares, so that (it gives) 52 05. (One) square-side is 2/3 of the (other) square-side. Add 20 to the large square-side, add 5 to the small square-side. What are the square-sides? You, when you proceed: combine 20, so that it gives 6 40. Combine 5, so that it gives 25. Sum 6 40 and 25, so that it gives 7 05.  Take away 7 05 from 52 05, so that it gives 45 00. Combine 1, so that it gives 1. Combine 0;40. It gives 0;26 40.  Sum 1 and 0;26 40, so that it gives 1;26 40. Multiply 1;26 40 by 45 00, so that it gives 1 05 00. Combine 20 and 1, so that (it gives 20).  Combine 0;40 and 5, so that it gives 3;20. Sum 20 and 3;20, so that it gives 23;20. Combine 23;20, so that it gives 9 04;26 40.  Add 9 04;46 40 to 1 05 00, so that it gives 1 14 04;46 40. The square-side of 1 14 04;46 40 is 1 06;40. Take away 23;20 from 1 06;40, so that it gives [43;20].  What should I put to 1;26 40 so that it gives 43;20? Put 30. Multiply 30 by 1. It gives 30.  Add 20 to 30, so that the large square-side is 50. Multiply 30 by 0;40. It gives 20. Add 5 to 20, so that it gives 25, the small square."}, {"id_text": "P414661", "project_name": "dccmt", "raw_text": "A triangle with 5 rivers inside. The (area of the) upper river is 18 20, the area of the [2nd] river [15 00].  The upper width exceeds the dividing line by 13;20. [Dividing Line exceeds] dividing line by 13;20.  I do not know the length and area of the 3rd river. The 4th dividing line is 40, the [5]th area is 1 40. A triangle with 5 rivers inside. The upper area is 18 20, the 2nd area 15 00.  I do not know the length and area of the third river. The 4th dividing line is 40, the \u00ab5th\u00bb <4th> length is 30, the <5th> area 1 40.  What are the dividing lines and the upper width? A triangle with 5 rivers inside. The upper area is 18 20, the second area 15 00.  I do not know the third area. The 4th area is 13 20; 1/2 (way along) it is 26;40 (wide). I do not know the 5th area.  The upper width exceeds the dividing line by 13;20. Dividing Line exceeds dividing line by 13;20.  What are the area, lengths, and dividing lines? [... What are] the area and [...]?  A triangle of area 1 [.... I went down] from the upper length [and] I lay the dividing line across [...] but I do not [know] how much I laid the dividing line across.  [...] I went down 3 rods 4 cubits, and then I laid across a ....   I went [...] and then I installed a dyke. From the dyke that I installed I laid across a [...] but I do not know how much I went (along) the dividing line.  The following area is 5 16 40.  How much area did I take and how much did I leave behind? A triangle. I do not know the length and upper width.  The area is 1 (bur) 2 (eshe). From the upper width I went down 33;20, so that the dividing line was 40.  What are the length and (upper) width? A triangle with 2 rivers inside. The upper width is 30, the lower area is 4 30.  The lower length exceeds the upper length by 10. <What are the lengths?> A triangle with 2 rivers inside. The upper width is 30, the upper area 8 00.  The lower length exceeds the upper length by 10. What are the lengths? A triangle with 2 rivers inside. The upper width is 30, [the lower area 2 00].  The lower length [exceeds] the upper length by 10. [What are the lengths?] A triangle with 2 rivers inside. [The upper width is 30], the upper area 10 30. The [upper] length [exceeds the lower length by 10. What are the lengths?] A triangle [with 2 rivers inside. The upper width is 30, the upper area] 8 00. [The lower length is 30. ...] A triangle [with 2 rivers inside. The upper width is 30, the upper length] 20. [The area is 4 30. ...] A triangle [with 2 rivers inside. The upper length is 4 13, the upper area] 25 18. [The lower area is 33 20. ...] I went down 4 [...] "}, {"id_text": "P414663", "project_name": "dccmt", "raw_text": "An ox\u2019s brow with 2 rivers inside. The upper area is 13 03, the 2<nd> area is 22 57. A 3rd part of the lower length is in the upper length. Sum that by which the upper width exceeds the diagonal and (that by which) the diagonal exceeds the lower width. (It gives) 36.  What are their lengths, the widths, and the diagonal? You, when you proceed: put down 1 and 3. Sum 1 and 3. (It gives) 4. Solve the reciprocal of 4, so that (it gives) 0;15.  Multiply 0;15 by 36. It gives 9. Multiply 9 by 1. It gives 9. Multiply 9 by 3. (It gives) 27.  That by which the upper width exceeds the diagonal is 9. That by which the diagonal exceeds the lower width is 27. Solve the reciprocal of 1. Multiply 1 by 13 03. It gives 13 03. Solve the reciprocal of 3. Multiply 0;20 by 22 57. it gives 7 39.  By what does 13 03 exceed 7 39? It exceeds by 5 24. Sum 1 and 3. (It gives) 4.  Break 1/2 of 4. (It gives) 2. Solve the reciprocal of 2. <Multiply> 0;30 by 5 24. It <gives> 2 42, the falsely counted (?). <The reciprocal of> 2 42 cannot be solved.  What should I put to 2 42 so that it gives 9? Put 0;03 20. Solve the reciprocal of 0;03 20. It gives 18.  Multiply 18 by 1. The upper length is 18. Multiply 18 by 3. The lower length \u00abthe lower length\u00bb is 54.  Break 1/2 of 36. Multiply \u00ab17\u00bb <18> by 1 12. (It gives) 21 36. Solve 31 36 from 36 00, the area. (It gives) 14 24.  Solve the reciprocal of 1 12, the length. Multiply 0;00 50 by 14 24. it gives 12. Add 12 to 36, so that (it gives) 48.  The upper width is 48. Add 12 to 27. The diagonal is 39. It gives 12, the lower width."}, {"id_text": "P414664", "project_name": "dccmt", "raw_text": "I took a reed but I did not know its measurement.  I broke off 1 cubit from it and then I went a length of 1 sixty. I returned to it what I had broken off from it and then I went a width of its 30.  The area is 6 15. What is the width (sic) of the reed? You, when you proceed: put down 1 00 and 30. Put down 1, the reed that you do not know.  You multiply (it) by 1 00, its sixty that you went, so that the false length is 1 00.  Multiply 30 by that 1, (so that) the false width is 30.  Multiply 30, the false width, by 1 00 \u00ab30\u00bb, the false length. The false area is 30 00.  Multiply \u00ab1\u00bb 30 00 by 6 15, the true area, so that it gives you 3 07 30 00.  Multiply the 0;05 that was broken off by the false length. It gives 5. Multiply 5 by the false width.  It gives 2 30. Break 1/2 of 2 30. (It gives) 1 15. Combine 1 15. (It gives) 1 33 45.  Add to 3 07 30 00. (It gives) 3 [09 03 4]5. What is the square-side? [The square-side] is 13 45.  Add the 1 15 that you combined to it. It gives 15 00. Solve the reciprocal of 30 00, the false area. (It gives) 0;00 02.  Multiply 0;00 02 by 15 00. The width (sic) of the reed is 30."}]