
Obverse
Column i
	(start of column missing)
o i 1'o 1'	[...] x x IGI	(o i 1') [...] ... [...] ... [...] the procedure.
o i 2'2'	[...] BI
o i 3'3'	[... ne]-pé-⸢šum⸣
o i 4'4'	[...] ⸢EN.NAM⸣	(o i 4') [...] what? [...] ... [...] you will see. [...] Square 2. [...] Multily [3 by] 2. The height is 6. [That is] the procedure.
o i 5'5'	[...] x
o i 6'6'	[...] ⸢ta-mar⸣
o i 7'7'	[...] 2 NIGIN
o i 8'8'	[... 3 a-na] 2 i-ši 6 SUKUD
o i 9'9'	[ki-a-am ne-pé]-šum

o i 10'10'	[... EN].⸢NAM⸣	(o i 10') [...] what? [...]
o i 11'11'	[...] x
o i 12'12'	[...] x
	(6 lines missing)
o i 18'18'	[... ÚS SAG] ⸢EN.NAM⸣	(o i 18') h1 Problem (i') [...]. What are [the length and width]?
o i 19'19'	[...] ⸢3⸣ ta-mar ⸢1/2 3⸣ ḫe-pé 1.30 ta-mar	(o i 19') [...] You will see 3. Break 3 in half. You will see 1;30.
o i 20'20'	[... IGI 1.30 DU₈.A] ⸢40 ta-mar⸣ BAL SAG IGI 12 BAL BÙR DU₈.A	(o i 20') [... Release the rešiprošal of 1;30.] You will see 0;40. The ratio of the width. Release the rešiprošal of 12, the ratio of the depth.
o i 21'21'	[5 ta-mar] ⸢5 a-na⸣ 1 ⸢i-ši 5 ta-mar⸣ a-na 40 i-ši 3.20 ta-mar	(o i 21') [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.
o i 22'22'	⸢3.20 a⸣-na 5 i-ši 16.40 ta-mar IGI 16.40 DU₈.A 3.36 ta-mar 3.36	(o i 22') 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.
o i 23'23'	a-na 1.10 i-ši 4.12 ta-mar 6 ÍB.SI₈ 6 a-na 5 i-ši 30 ÚS 6 a-na 3 20 i-<ši>
o i 24'24'	20 SAG 6 a-na 1 i-ši 6 ta-mar BÙR ki-a-am	(o i 24') The width is 0;20. Multiply 6 by 1. You will see 6, the depth. That is the procedure.
o i 25'25'	ne-pé-šum

o i 26'26'	PÚ.SAG ma-la ÚS BÙR-ma 1 SAḪAR-ḪI.A BA.ZI KI^ri ù SAḪAR-ḪI.A UL.GAR 1.10 ÚS ù SAG 50 ÚS SAG EN.NAM	(o i 26') 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?
o i 27'27'	ZA.E 50 a-na 1 BAL i-ši 50 ta-mar 50 a-na 12 i-ši 10 ta-mar	(o i 27') You: Multiply 0;50 by 1, the ratio. You will see 0;50. Multiply 0;50 by 12. You will see 10.
o i 28'28'	10^!(50) šu-tam-<ḫir> 41.40 ta-mar a-na 10 i-ši 6.56.40 ta-mar IGI-šu DU₈.A 8.38.44 ta-<mar>	(o i 28') 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.
o i 29'29'	a-na 1.10 i-ši 10.04.48 ta-mar 36 24 42 ÍB.SI₈	(o i 29') Multiply by 1;10. You will see 0;10 04 48. The square-sides are 0;36, 0;24, 0;42.
o i 30'30'	36 a-na 50 i-ši 30 ÚS 24 a-na 50 i-ši 20 SAG 36 a-na 10 <i-ši> 6 BÙR	(o i 30') 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.
o i 31'31'	ne-pé-šum

o i 32'32'	PÚ.SAG ma-⸢la ÚS BÙR⸣-ma 1 SAḪAR-ḪI.A ⸢BA⸣.ZI ⸢KI⸣^ri ù SAḪAR-ḪI.A UL.<GAR> 1.10 ÚS UGU SAG 10 DIRI	(o i 32') 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.
o i 33'33'	ZA.E 1 ù 12 ⸢BAL⸣ GAR.RA ⸢10 DIRI a-na⸣ 1 i-ši 10 ta-mar a-na 12 i-ši 2 ta-mar	(o i 33') 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.
o i 34'34'	10 šu-tam-<ḫir> 1.40 ta-mar a-na 2 i-ši 3.⸢20 ta⸣-mar IGI 3.20 DU₈.A 18 ta-mar	(o i 34') 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.
o i 35'35'	a-na 1.10 i-ši 21 ta-mar 3 2 21 ⸢ÍB.SI₈⸣ [10 a-na 3] ⸢i-ši⸣ 30 ÚS	(o i 35') 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.
o i 36'36'	10 a-na 2 i-ši 20 SAG 3 a-na 2 i-ši ⸢6 ta-mar⸣ [6] BÙR	(o i 36') Multiply 0;10 by 2. The width is 0;20. Multiply 3 by 2. You will see 6. The depth is [6]. The procedure.
o i 37'37'	ne-pé-šum

o i 38'38'	PÚ.SAG ma-la ÚS BÙR-ma SAḪAR-⸢ḪI⸣.A BA.ZI KI^ri ù SAḪAR-ḪI.A UL.GAR-ma 1.10 ⸢30⸣ ÚS ⸢SAG EN⸣.[NAM]	(o i 38') 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?
o i 39'39'	ZA.E 30 ÚS a-na 12 i-ši ⸢6 ta⸣-mar BÙR 1 a-na 6 DAḪ.ḪA 7 ta-mar	(o i 39') You: multiply 0;30, the length, by 12. You will see 6, the depth. Add 1 to 6. You will see 7.
o i 40'40'	IGI 7 NU DU₈.A EN.NAM a-na 7 GAR.RA ša 1.10 SUM.MU 10 GAR.RA IGI 30 ÚS DU₈.A	(o i 40') 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.
o i 41'41'	2 ta-mar 10 a-na 2 i-ši 20 SAG ta-mar	(o i 41') You will see 2. Multiply 0;10 by 2. You will see 0;20, the width. The procedure.
o i 42'42'	ne-pé-šum

o i 43'43'	PÚ.SAG ma-la ÚS BÙR-ma SAḪAR-ḪI.A BA.ZI KI^ri ù SAḪAR-ḪI.A UL.GAR-ma 1.10 20 SAG ÚS <EN.NAM>	(o i 43') 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?
o i 44'44'	ZA.E 20 a-na 12 i-ši 4 ta-mar 4 a-na 1.10 i-ši 4.40 ta-mar	(o i 44') You: multiply 0;20 by 12. You will see 4. Multiply 4 by 1;10. You will see 4;40.
o i 45'45'	1/2 20 SAG ḫe-pé 10 ta-mar 10 šu-tam-ḫir 1.40 ta-mar a-na 4.40 DAḪ.ḪA	(o i 45') 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.
o i 46'46'	4.41.40 ta-mar 2.10 ÍB.SI₈ 10 ša Ì.GU₇.GU₇ BA.ZI-ma	(o i 46') 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.
o i 47'47'	2 ta-mar IGI 4 DU₈.A 15 ta-mar a-na 2 i-ši
o i 48'48'	30 ta-mar ÚS	(o i 48') You will see 30, the length. The procedure.
o i 49'49'	ne-pé-šum	(o i 48') You will see 30, the length. The procedure.

Column ii
	(start of column missing)
o ii 1'o 1'	[...] ⸢DAḪ.ḪA⸣ [...]	(o ii 1') h1 Problem (vi') [...] add [...] 4;30, 53;20, 1;45 ... [...]
o ii 2'2'	4.30 ⸢53⸣.20 ⸢1.45⸣ x [...]
o ii 3'3'	⸢30⸣ ÚS 22.30 a-na 53.20 i-ši ⸢20 SAG⸣ [...]	(o ii 3') The length is 0;30. Multiply 0;22 30 by 0;53 20. The width is 0;20. [...] The procedure.
o ii 4'4'	ne-pé-šum

o ii 5'5'	PÚ.SAG ma-la ÚS BÙR-ma SAḪAR-ḪI.A BA.ZI ⸢KI⸣^ri ù SAḪAR-ḪI.A ⸢UL⸣.[GAR]	(o ii 5') An excavation. The length is as much as the depth. I removed a volume. I summed my ground-area and the volume.
o ii 6'6'	IGI 7-GÁL él-⸢qé⸣ a-na KI^ri DAḪ.ḪA-ma 20-⸢E⸣ 30 [ÚS SAG EN.NAM]	(o ii 6') 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?]
o ii 7'7'	ZA.E 30 a-na 12 i-ši 6 ta-mar BÙR 1 a-na 6 [DAḪ.ḪA]	(o ii 7') You: multiply 0;30 by 12. You will see 6, the depth. [Add] 1 to 6.
o ii 8'8'	7 ⸢ta⸣-mar IGI 7-GÁL ⸢le-qé⸣ 1 ta-mar 1 ù 1 ⸢UL⸣.[GAR]	(o ii 8') You will see 7. Take a 7th. You will see 1. Sum 1 and 1.
o ii 9'9'	2 ta-⸢mar⸣ IGI 2 DU₈.A 30 ta-mar 30 a-na 20 UL.GAR ⸢i⸣-[ši]	(o ii 9') You will see 2. Find the reciprocal of 2. You will see 0;30. Multiply 0;30 by 0;20, the sum.
o ii 10'10'	10 ta-mar IGI 30 ÚS DU₈.A 2 ta-mar 2 a-na 10 i-⸢ši⸣ [20 SAG]	(o ii 10') 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.
o ii 11'11'	ne-pé-šum

o ii 12'12'	PÚ.SAG ma-la ÚS BÙR-ma SAḪAR-ḪI.A BA.ZI qá-qá-ri ù SAḪAR-ḪI.A UL.[GAR]	(o ii 12') 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.
o ii 13'13'	1.10 IGI 7-GÁL-šu él-qé a-na KI^ri DAḪ 20 20 SAG
o ii 14'14'	ZA.E 20 a-na 7 i-ši 2.20 ta-mar 20 SAG a-na 12 i-ši	(o ii 14') You: multiply 0;20 by 7. You will see 2;20. Multiply 0;20, the width, by 12.
o ii 15'15'	4 ta-mar 4 a-na 2.20 i-ši 9.20 ta-mar a-na 7 1 DAḪ.⸢ḪA⸣	(o ii 15') You will see 4. Multiply 4 by 2;20. You will see 9;20. Add 1 to 7.
o ii 16'16'	8 ta-mar 20 a-na 8 i-ši 2.40 ta-mar 1/2 2.40 ḫe-pé [šu-<tam-ḫir>]	(o ii 16') You will see 8. Multiply 8 by 0;20. You will see 2;40. Break 1/2 of 2;40. [Square (it).]
o ii 17'17'	1.46.40 ta-mar a-na 9.20 DAḪ.ḪA 11.06.40 ⸢ta⸣-[mar]	(o ii 17') You will see 1;46 10. Add to 9;20. You will see 11;06 40.
o ii 18'18'	3.20 ÍB.SI₈ 1.20 ša Ì.GU₇.GU₇ BA.ZI 2 ta-[mar]	(o ii 18') The square-side is 3;20. Subtract the 1;20 that you squared. You will see 2.
o ii 19'19'	IGI 4 DU₈.A 15 ta-mar 15 a-na 2 i-ši 30 [ÚS]	(o ii 19') Find the riciprocal of 4. You will see 0;15. Multiply 0;15 by 2. [The length is] 0;20. The procedure.
o ii 20'20'	⸢ne⸣-pé-šum

o ii 21'21'	⸢PÚ.SAG ma-la⸣ IGI ÚS ma-la IGI.BI ⸢SAG⸣ ma-la <ša> IGI ⸢UGU IGI⸣.BI ⸢DIRI⸣	(o ii 21') 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?
o ii 22'22'	BÙR-ma ⸢16⸣ [SAḪAR-ḪI.A BA].⸢ZI⸣ ÚS SAG ù BÙR EN.NAM
o ii 23'23'	ZA.E IGI ⸢12 DU₈⸣.[A 5] ⸢ta-mar⸣ 5 [a-na] ⸢16⸣ i-⸢ši⸣ [1].⸢20⸣ ta-<mar>	(o ii 23') You: find the reciprocal of 12. You will see [0;05]. Multiply 0;05 by 16. You will see [1];20.
o ii 24'24'	1.20 IGI IGI 1.⸢20⸣ [DU₈.A 45 ta]-⸢mar⸣ 45^!(40) IGI.BI [16] ⸢BÙR⸣	(o ii 24') 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.
o ii 25'25'	ne-[pé]-šum

o ii 26'26'	PÚ.SAG ma-la ⸢IGI ÚS ma⸣-[la IGI.BI] ⸢SAG⸣ ma-la ša IGI UGU IGI.BI DIRI <BÙR-ma>	(o ii 26') 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.
o ii 27'27'	36 SAḪAR-ḪI.A BA.ZI IGI [IGI.BI ù BÙR] EN.NAM	(o ii 27') I removed a volume of 26. What are the reciprocal, [its reciprocal pair, and the depth]?
o ii 28'28'	ZA.E IGI 12 DU₈.A [5 ta-mar] ⸢36⸣ a-na 5 <<a-na ⸢36⸣>> i-<ši>	(o ii 28') You: solve the reciprocal of 12. You will see 0;05. Multiply 36 by 0;05 <<by 36>>.
o ii 29'29'	3 ta-mar 1/2 3 ⸢ḫe⸣-[pé 1.30 ta-mar] 1.⸢30 IGI 40 IGI⸣.[BI] ⸢36⸣ BÙR	(o ii 29') 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.
o ii 30'30'	ne-⸢pé-šum⸣

o ii 31'31'	PÚ.SAG ma-la IGI ÚS ma-⸢la⸣ [IGI.BI SAG] ⸢ma-la⸣ NIGIN IGI ù IGI.BI BÙR-ma	(o ii 31') 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.
o ii 32'32'	26 SAḪAR-ḪI.A BA.ZI IGI IGI.BI ù BÙR ⸢EN⸣.NAM	(o ii 32') I removed a volume of 26. What are the reciprocal, its reciprocal pair, and the depth?
o ii 33'33'	ZA.E IGI 12 DU₈.A 5 ta-mar 5 a-na 26 i-ši	(o ii 33') You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 26.
o ii 34'34'	2.10 ta-mar 1/2 2.10 ḫe-pé šu-tam-<ḫir> 1.10.25 ta-⸢mar⸣ <1 i-na 1.10.25 BA.ZI 10.25 ta-mar>	(o ii 34') 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.>
o ii 35'35'	25 ÍB.SI₈ a-na <1>.05 DAḪ ù BA.ZI 1.30 ù^!(IGI) 40 ⸢ta⸣-[mar]	(o ii 35') The square-side is 0;25. Add and subtract (it) to <1>;05. You will see 1;30 and 0;40.
o ii 36'36'	1.30 IGI 40 IGI.BI 26 BÙR	(o ii 36') The reciprocal is 1;30. Its reciprocal pair is 0;40. The depth is 26. The procedure.
o ii 37'37'	ne-pé-šum

o ii 38'38'	PÚ.SAG ma-la IGI ÚS ma-la IGI.BI SAG ma-la ša IGI UGU IGI.BI ⸢DIRI⸣	(o ii 38') 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?
o ii 39'39'	i-na IGI BA.ZI BÙR-ma 6 SAḪAR-ḪI.A BA.ZI IGI ù IGI.BI [EN.NAM]
o ii 40'40'	ZA.E IGI 12 DU₈.A 5 ta-mar a-na 6 i-ši 30 ta-mar	(o ii 40') You: find the reciprocal of 12. You will see 0;05. Multiply by 6. You will see 0;30.
o ii 41'41'	IGI 30 DU₈.A 2 ta-mar 2 IGI 30 IGI.BI 6 BÙR	(o ii 41') Find the reciprocal of 0;30. You will see 2. The reciprocal is 0;30, its reciprocal pair 2, the depth 6. The procedure.
o ii 42'42'	ne-pé-šum

o ii 43'43'	PÚ.SAG ma-la IGI ÚS ma-la IGI.BI SAG ma-la NIGIN IGI IGI.BI [BÙR]-ma 30 SAḪAR-[ḪI.A BA.ZI]	(o ii 43') 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.
o ii 44'44'	ZA.E IGI 12 DU₈.A 5 ⸢ta-mar⸣ 5 a-na 30 SAḪAR-ḪI.A i-ši	(o ii 44') You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 30, the volume.
o ii 45'45'	2.30 ta-mar 1/2 2.30 ḫe-pé šu-[tam-ḫir 1.33].45 ta-[mar]	(o ii 45') You will see 2;30. Break 2;30 in 1/2. Square (it). You will see [1;33] 45.
o ii 46'46'	1 i-na 1.33.45 BA.ZI 33.45 ta-mar 45 ÍB.SI₈	(o ii 46') Subtract 1 from 1;33 45. You will see 0;33 45. The square-side is 0;45.
o ii 47'47'	a-na 1.45 DAḪ.ḪA ù BA.ZI 2 ù 30 ta-mar	(o ii 47') Add and subtract from 1;45. You will see 2 and 0;30. The procedure.
o ii 48'48'	ne-pé-šum

o ii 49'49'	PÚ.SAG ma-la IGI ÚS ma-la IGI.BI SAG ma-la IGI.BI BÙR-ma	(o ii 49') 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.
o ii 50'50'	20 SAḪAR-ḪI.A BA.<ZI> IGI IGI.BI ù BÙR EN.NAM	(o ii 50') I removed a volume of 20. What are the reciprocal, its reciprocal pair, and the depth?
o ii 51'51'	ZA.E IGI 12 DU₈.A a-na 20 i-ši 1.40 ta-mar	(o ii 51') You: find the reciprocal of 12. Multiply by 20. You will see 1;40.
o ii 52'52'	1.40 IGI 36 IGI.BI 20 BÙR	(o ii 52') The reciprocal is 1;40. Its reciprocal pair is 0;36. The depth is 20. The procedure.
o ii 53'53'	ne-pé-šum
Reverse
Column i
r i 1r i 1	PÚ.SAG ma-la uš-tam-<ḫir> ù 7 KÙŠ BÙR-ma 3.20 SAḪAR-ḪI.A BA.ZI	(r i 1) h1 Problem (xiv') An excavation. The depth is as much as I squared and 7 cubits. I removed 3 20, the volume.
r i 22	ÚS SAG ù BÙR EN.NAM	(r i 2) What are the length, width, and depth?
r i 33	ZA.E IGI 7-GÁL 7 le-qé 1 ta-mar IGI 12 DU₈.A 5 ta-mar	(r i 3) You: take a 7th of 7. You will see 1. Find the reciprocal of 12. You will see 0;05.
r i 44	5 a-na 1 i-ši 5 ta-mar 5 a-na 12 i-ši 1 ta-mar	(r i 4) Multiply 0;05 by 1. You will see 0;05. Multiply 0;05 by 12. You will see 1.
r i 55	5 šu-tam-<ḫir> 25 a-na 1 i-ši 25 ta-mar IGI 25 DU₈.A 2.24	(r i 5) 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?
r i 66	ta-mar 2.24 a-na 3.20 SAḪAR-ḪI.A i-ši 8 ta-mar EN.NAM ÍB.<SI₈>
r i 77	1 1 8 ÍB.SI₈ 5 a-na 1 i-ši 5 ta-mar 5 KÙŠ ÚS	(r i 7) 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.
r i 88	8 a-na 1 i-ši 8 KÙŠ BÙR	(r i 8) Multiply 8 by 1. The depth is 8 cubits. The procedure.
r i 99	ne-⸢pé-šum⸣

r i 1010	PÚ.SAG ma-la uš-tam-ḫir ù ⸢7 KÙŠ⸣ BÙR-ma <<13>> <3.15> SAḪAR-<ḪI.A> BA.ZI	(r i 10) h1 Problem (xv') An excavation. The depth is as much as I squared and 7 cubits. I removed a volume of <3 15> <<13>>.
r i 1111	ÚS SAG ù BÙR EN.NAM	(r i 11) What are the length, width, and depth?
r i 1212	ZA.E ⸢ki-ma⸣ maḫ-ri-ma e-pu-uš 7^!(4)48 EN.NAM ÍB.SI₈	(r i 12) You: do as before. (W)ith what does 7 48 square itself?
r i 1313	⸢6⸣ 6 13 ÍB.SI₈ 6 im-<ta-ḫar> 13 BÙR	(r i 13) Its square-sides are 6, 6, and 13. It squares itself (with a side of) 6. The depth is 13. The procedure.
r i 1414	ne-pé-šum

r i 1515	PÚ.SAG ma-la uš-tam-ḫir BÙR-ma 1.30 SAḪAR-ḪI.A BA.ZI ÚS SAG ⸢ù⸣ BÙR <EN.NAM>	(r i 15) 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?
r i 1616	ZA.E IGI 12 DU₈.A 5 ta-mar 5 a-na 1.30 i-ši ⸢7.30⸣ <ta-mar>	(r i 16) You: find the reciprocal of 12. You will see 0;05. Multiply 0;05 by 1;30. <You will see> 0;07 30.
r i 1717	30 ÍB.SI₈ 30 a-na 1 i-ši 30 im-ta-ḫar 30 a-⸢na 12⸣ i-<ši> 6 BÙR	(r i 17) 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.
r i 1818	ne-pé-šum

r i 1919	PÚ.SAG ma-la uš-tam-ḫir ù 1 KÙŠ DIRI BÙR-ma 1.45 SAḪAR-ḪI.A ⸢BA⸣.ZI	(r i 19) 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.
r i 2020	ZA.E 5 DIRI a-na 1 BAL i-ši 5 ta-mar a-na 12 i-⸢ši 1⸣ ta-mar	(r i 20) You: Multiply 0;05, the excess by 1, the ratio. You will see 0;05. Muliply by 12. You will see 1.
r i 2121	5 šu-tam-<ḫir> 25 ta-mar 25 a-na 1 i-ši 25 ta-mar ⸢IGI 25⸣ [DU₈.A]	(r i 21) 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.
r i 2222	2.24 ta-mar 2.24 a-na 1.45 i-ši 4.⸢12⸣ [ta-mar]	(r i 22) You will see 2 24. Multiply 2 24 by 1;45. [You will see] 4 12.
r i 2323	i-na ÍB.SI₈ 1 DAḪ.ḪA 6 1 ÍB.⸢SI₈ 6 a⸣-[na] 5 i-⸢ši 30⸣ ta-<mar 6> im-<ta-ḫar> 6 BÙR	(r i 23) 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.
r i 2424	ne-pé-šum

r i 2525	PÚ.SAG 3.20 BÙR-ma 27.46.40 SAḪAR-ḪI.A BA.ZI ⸢ÚS UGU⸣ SAG 50 ⸢DIRI⸣	(r i 25) 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.
r i 2626	ZA.E IGI 3.20 BÙR DU₈.A 18 ta-mar a-na 27.46.40 SAḪAR-ḪI.<A> i-ši	(r i 26) You: find the reciprocal of 3;20, the depth. You will see 0;18. Multiply by 27;46 40, the volume.
r i 2727	8.20 ta-mar 1/2 50 ḫe-pé šu-tam-<ḫir> 10.25 ta-mar	(r i 27) You will see 8;20. Break 50 in 1/2. Square (it). You will see 0;10 25.
r i 2828	a-na 8.20 DAḪ.⸢ḪA 8⸣.30.25 ta-mar	(r i 28) Add to 8;20. You will see 8;30 25.
r i 2929	2.55 ÍB.SI₈ a-di [2 GAR.RA] ⸢a-na⸣ 1 DAḪ.ḪA i-na 1 BA.ZI	(r i 29) The square-side is 2;55. [Put it down] twice. Add to 1, subtract from 1.
r i 3030	3.20 ÚS 2.30 SAG ⸢ta⸣-mar	(r i 30) You will see 3;20, the length, (and) 2;30, the width. The procedure.
r i 3131	ne-⸢pé⸣-šum

r i 3232	PÚ.SAG 3.20 BÙR-ma 27.46.40 [SAḪAR-ḪI.A BA.ZI ÚS ù SAG UL.GAR] 5.⸢50⸣	(r i 32) 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.
r i 3333	⸢ZA.E⸣ IGI 3.20 BÙR DU₈.A 18 ta-mar ⸢a⸣-[na 27.46.40 i-ši]	(r i 33) You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by [27;46 40].
r i 3434	8.20 ta-mar 1/2 5.50 ḫe-pé šu-tam-<ḫir> ⸢8⸣.[30.25 ta-mar]	(r i 34) You will see 8;20. Break 1/2 of 5;50. Square (it). [You will see] 8;[30 25].
r i 3535	8.20 i-na ŠÀ-BA BA.ZI 10.⸢25⸣ [ta-mar 25 ÍB.SI₈]	(r i 35) Subtract 8;20 from inside it. [You will see] 0;10 25. [The square-side is 0;25.]
r i 3636	a-na 2.55 DAḪ.ḪA ù BA.ZI 3.20 [ÚS 2.30 SAG]	(r i 36) Add and subtract from 2;55. [The length is] 3;20. [The width is 2;30.] The procedure.
r i 3737	ne-pé-[šum]

r i 3838	PÚ.SAG 3.20 BÙR-ma 27.46.40 SAḪAR-ḪI.⸢A⸣ [BA.ZI SAG UGU BÙR DIRI ma-la 2/3 ÚS]	(r i 38) 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.]
r i 3939	ZA.E IGI 3.20 DU₈.A 18 ta-mar a-na ⸢27⸣.[46.40 i-ši]	(r i 39) You: find the reciprocal of 3;20. You will see 0;18. [Multiply (it)] by 27;[46 40].
r i 4040	8.20 ta-mar 8.20 a-na 40 i-ši 5.33.[20 ta-mar 3.20 BÙR a-na 5 i-ši 16.40]	(r i 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.]
r i 4141	NÍGIN.NA 1/2 16.40 ḫe-pé 8.20 ta-mar šu-tam-<ḫir> [1.09.26.40 a-na 5.33.20 DAḪ.ḪA]	(r i 41) 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.]
r i 4242	EN.NAM ÍB.SI₈ 2.21^!(31)40 a-di 2 GAR.RA 8.⸢20⸣ [DAḪ.ḪA ù BA.ZI]	(r i 42) (With) what does it square itself? Put down 2;21 40 twice. [Add and subtract] 0;08 20.
r i 4343	2.30 SAG 2.13.20 ta-mar IGI 40 DU₈.A 1.30 ta-mar [a-na 2.13.20 i-ši]	(r i 43) 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.
r i 4444	3.20 ÚS ta-⸢mar⸣
r i 4545	ne-pé-⸢šum⸣

r i 4646	PÚ.SAG 1.40 ÚS IGI 7-GÁL ša ÚS UGU SAG DIRI BÙR-ma 1.40 SAḪAR-[ḪI.A BA.ZI]	(r i 46)
r i 4747	<<ÚS>> SAG ù BÙR EN.[NAM]	(r i 46)
r i 4848	ZA.E 1.40 ÚS a-na 12 BAL BÙR i-ši 20 ⸢ta⸣-[mar]	(r i 48) You: multiply 1;40, the length, by 12, the ratio of the depth. You will see 20.
r i 4949	IGI 20 DU₈.A 3 ta-mar 3 a-na 1.40 ⸢SAḪAR⸣-[ḪI.A i-ši 5 ta-mar]	(r i 49) Release the reciprocal of 20. You will see 0;03. [Multiply] 0;03 by 1;40, the volume. [You will see 0;05.]
r i 5050	7 a-na 5 i-ši 35 ⸢ta⸣-[mar 1/2 1.40 ḫe-pé šu-tam-ḫir 41.40]	(r i 50) Multiply 7 by 0;05. You will see 0;35. [Break in half 1;40. Square (it). 0;41 40.]
r i 5151	35 ⸢i-na⸣ ŠÀ-[BA BA.ZI 6.40 ta-mar 20 ÍB.SI₈]	(r i 51) [Subtract] 0;35 from it. [You will see 0;06 40. The square-side is 0;20.]
r i 5252	a-[na 50 DAḪ.ḪA ù BA.ZI 1.10 ù 30 SAG ... IGI 7-GÁL 1.10 le-qé 10 BÙR]	(r i 52) [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.]
r i 5353	[ne-pé-šum]
Column ii
r ii 1r ii 1	PÚ.SAG 1.40 ÚS IGI 7-<GÁL> ša ÚS UGU SAG DIRI ù 2 KÙŠ BÙR-ma 3.20 ⸢SAḪAR⸣-ḪI.<A BA.ZI>	(r ii 1)
r ii 22	SAG ù BÙR EN.NAM	(r ii 1)
r ii 33	ZA.E 1.40 ÚS a-na 12 ⸢BAL BÙR⸣ i-ši 20 ta-mar IGI 20 DU₈.A 3 ta-mar	(r ii 3) 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.
r ii 44	3 a-na 3.20 i-ši 10 ⸢ta⸣-mar <<10 a-na 7 i-ši 1.10 ta-mar>> 10 DIRI a-na 7 i-ši 1.[10] ⸢ta⸣-mar	(r ii 4) Multiply 0;03 by 3;20. You will see 0;00 10. Multiply 0;10, the excess, by 7. You will see 1;10.
r ii 55	1.40 ÚS a-na 1.10 DAḪ.ḪA 2.50 ta-mar 1/2 2.⸢50 ḫe-pé šu-tam⸣-ḫir	(r ii 5) Add 1;40, the length, to 1;10. You will see 2;50. Break in half 2;50. Square (it).
r ii 66	2.oo.25 ta-mar i-na 2.oo.25 1.10 ⸢BA⸣.ZI 50.25 ta-mar	(r ii 6) You will see 2;00 25. Subtract 1;10 from 2;00 25. You will see 0;50 25.
r ii 77	55 ÍB.SI₈ a-na 1.⸢25 DAḪ.ḪA⸣ ù BA.ZI-ma	(r ii 7) 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.
r ii 88	2.20 ù 30 SAG ta-mar IGI 7-GÁL 2.20 ⸢le⸣-[qé] 20 BÙR
r ii 99	ne-pé-šum

r ii 1010	PÚ.SAG 1.40 ÚS IGI 7-GÁL ša ÚS UGU SAG DIRI ⸢ù⸣ 1 KÙŠ BA.LÁ BÙR-ma	(r ii 10)
r ii 1111	50 SAḪAR-ḪI.A BA.ZI SAG ⸢ù⸣ BÙR EN.NAM	(r ii 10)
r ii 1212	ZA.E 1.40 ÚS a-na 12 BAL BÙR i-ši 20 ta-mar IGI 20 DU₈.A 3 ⸢ta⸣-<mar>	(r ii 12) 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.
r ii 1313	3 a-na 50 i-ši 2.30 ta-mar 2.30 a-na 7 ⸢i⸣-ši 17.30 ⸢ta⸣-[mar]	(r ii 13) Multiply 0;03 by 0;50. You will see 0;02 30. Multiply 0;02 30 by 7. You will see 0;17 30.
r ii 1414	7 a-na 5 1 KÙŠ i-ši 35 ta-mar 35 i-na 1.40 ⸢ÚS⸣ BA.ZI	(r ii 14) Multiply 7 by 0;05, 1 cubit. You will see 0;35. Subtract 0;35 from 1;40, the length.
r ii 1515	1.05 ta-mar 1/2 1.05 ḫe-pé 32.30 šu-tam-<ḫir> ⸢17.36⸣.15 ta-<mar>	(r ii 15) You will see 1;05. Break in half 1;05. Square 0;32 20. You will see 0;17 36 15.
r ii 1616	i-na ŠÀ-BI 17.30 BA.ZI 6.15 ta-mar 2.30 ⸢ÍB⸣.SI₈	(r ii 16) Subtract 0;17 30 from it. You will see 0;00 06 15. Its square-side is 0;02 30.
r ii 1717	a-na 32.30 DAḪ.ḪA ù BA.ZI 35 ù 30 SAG ta-mar <IGI> 7-<GÁL> 35 5 BÙR	(r ii 17) 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.
r ii 1818	⸢ne⸣-pé-šum

r ii 1919	⸢30⸣ ki-ib-su	(r ii 19) 30 procedures.
	(rest of column missing)

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Uruk IV

metrological

model-document

Uruk III

metrological

model-document

other

ED IIIa

metrological

uncertain

word-problem

ED IIIb

calculation

metrological

Old Akkadian

diagram

metrological

model-document

word-problem

Ur III

arithmetical

model-document

model-document calculation

Old Assyrian

arithmetical

calculation

diagram calculation

metrological

metrological non-mathematical

Early Old Babylonian

arithmetical

Old Babylonian

arithmetical

arithmetical calculation

arithmetical metrological

arithmetical, metrological

arithmetical, non-mathematical

arithmetical non-mathematical

calculation

calculation, arithmetical

calculation, arithmetical, diagram

calculation non-mathematical

calculation word-problem

calculation, word-problem

diagram

diagram calculation

diagram uncertain

diagram word-problem

metrological

metrological, non-mathematical

metrological non-mathematical

model-document

non-mathematical, arithmetical

non-mathematical, metrological

reference

reference word-problem

uncertain

uncertain non-mathematical

word-problem?

word-problem

word-problem calculation

word-problem calculation diagram

word-problem, diagram

word-problem diagram

word-problem; diagram

unknown

Middle Assyrian

arithmetical

Middle Babylonian

arithmetical

metrological

reference

word-problem

Neo-Assyrian

arithmetical

metrological

word-problem

word-problem non-mathematical

Neo-Babylonian

arithmetical