TCL 06, 33 (P254387)
Mathematical, Hellenistic, Uruk · c. 330 BCE–141 BCE
Open in the reader & map →See it among all tablets
| Designation | TCL 06, 33 |
|---|---|
| ID | P254387 |
| Museum no. | AO 06484 + Ist Uruk 0180,21 |
| Collection | Arkeoloji Müzeleri, Istanbul, Turkey; Louvre Museum, Paris, France |
| Language | Akkadian |
| Period | Hellenistic · c. 330 BCE–141 BCE |
| Genre | Mathematical |
| Find spot | Uruk (mod. Warka) |
| Coordinates | 31.3234, 45.6395 |
English translation
By ORACC project cams/gkab · Via Open Richly Annotated Cuneiform Corpus (ORACC) · CC BY-SA 3.0 (ORACC)
| Line | English |
|---|---|
| Top [0:2] t.e. 1t.e. 1– | By the command of Anu and Antu, may it go well! |
| Obverse [0:2] o 1o 1–o 22 | Scale (a ladder), which is for (?) 2, from 1 to 10. Sum so that 8 [32 : take out 1 from 8 32, so that] the remaining (amount) is 8 31 : Add 8 31 to 8 32, so that 17 03 [...]. |
| Obverse [2:5] o 33–o 55 | Squares from 1 by 1 : 1, to 10 by 10 : 1 40. How much is the count? Multiply [1 by 0;20 : 1/3], so that 0;20 : multiply 10 by 0;40 : two thirds, so that 6;40 : 6;40 [and 0;20 are 7]. Multiply 7 by 55, so that 6 25: the count is 6 25. |
| Obverse [5:8] o 66–o 88 | The height of a brick wall is 10 cubits. It is open 1 cubit from the width of the brick wall. And the height of a log is 1 cubit [...]. By how much should I be removed from the foundation of the brick wall in order to see it? The reciprocal of 1 [is 1 ...]. Multiply 1 by 10 cubits, so that 10 : you are 10 cubits removed in order to see it. |
| Obverse [8:10] o 99–o 1010 | The height of a log is 1 cubit. [The height of a brick wall is] 10 cubits. The reciprocal of 1 is 1 : 1 by [...]. |
| Obverse [9:11] o 1010–o 1111 | 1 2/3 cubits times 1 2/3 cubits. How much is the seed-measure? [Multiply 2;46 40 by 21 36, so that 1 00 00]. Multiply 1 00 00 by 0;01 48, so that 1 48. The seed is 1 hundred 8. |
| Obverse [11:13] o 1212–o 1313 | Length, width, and diagonal are 40, and the area is 2 00. The length is 15, [the width 8, the diagonal 17]. |
| Obverse [12:18] o 1313–o 1818 | A triangle. Each length is 5, the width 6. How [much is the seed-measure? ...] Multiply 5 by 5, the lengths, so that 25 : [multiply] 3, half of the width, [by 3, so that 9]. Take out 9 from 25, so that the remaining is 16. How much by how much should you multiply for 16 : [multiply] 4 [by 4 so that 16]. Multiply 4, the diameter, by 3, half of the width, so that 12 : the area is 12. [Multiply] 12 by 21 [36 so that] 4 19 12 : multiply 4 19 12 by 0;01 48 so that 7 46;[33 36]. The seed is 4 hundred 1 06 and a half. |
| Obverse [18:20] o 1919–o 2020 | The diagonal of a square is 10 cubits. <What is> the length of the square <<until>>? [Multiply] 10 by 0;42 30, [so that 7;05 is the length]. Multiply 7;05 by 1;25, so that [10;25 is the diagonal]. |
| Obverse [20:23] o 2121–o 2323 | The length is 1 cubit, the width 1 cubit, the height 1 cubit [.... Multiply] 0;05, cubits, the horizontal, by 0;05, cubits, the horizontal, [so that 0;00 25 :] multiply [0;00 25 by 1], the vertical, so that 0;00 25 : [multiply] 0;00 25 by 6 00 00, the storage, [so that 2 30 ...]. |
| Obverse [23:26] o 2424–o 2626 | The length is 2 cubits, the width 2 cubits, the height 2 cubits [.... Multiply] 0;10, cubits, the horizontal, by 0;10, cubits, the horizontal, [so that 0;01 40 :] multiply [0;01 40 by 2, the vertical], so that 0;03 20 : multiply 0;03 20 by 6 00 00, so that [20 00 ...]. |
| Reverse [0:3] r 1r 1–r 33 | The length is 3 cubits, the width 3 cubits, [the height] 3 [cubits .... Multiply] 0;15, cubits, the horizontal, by 0;15, cubits, the horizontal, [so that 0;03 45] : multiply [0;03 45 by 3], the vertical, so that 0;11 15 : [multiply 0;11 15] by 6 00 00, so that [1 12 30 ...]. |
| Reverse [3:6] r 44–r 66 | The length is 4 cubits, the width 4 cubits, the height 4 cubits [...]. Multiply [0;20, cubits, the horizontal], by 0;20, cubits, the horizontal, so that 0;06 [40 : multiply [0;06 40 by 4, the vertical, so that 0;26 40] : multiply 0;26 40 by 6 00 00, so that 2 40 00 [...]. |
| Reverse [6:9] r 77–r 99 | The length is 5 cubits, the width 5 cubits, the height 5 cubits [...]. Multiply [0;25, cubits, the horizontal], by 0;25, cubits, the horizontal, so that 0;10 25 : [multiply 0;10 25 by 5, the vertical, so that 0;52 05] : multiply 0;52 05 by 6 00 00, so that 5 12 30 [...]. |
| Reverse [9:14] r 1010–r 1414 | A reciprocal and its reciprocal are 2;00 00 33 20. [How much are] the reciprocal and its reciprocal? Multiply [2;00 00 33 20] by 0;30, so that 1;00 00 16 40 : multiply 1;00 00 16 40 by [1;00 00 16 40, so that 1;00 00 33 20 04 37 46 40]. Decrease 1 from its middle, so that the remaining is 0;00 00 33 <20> 04 37 46 40. How much [by how much should I multiply so that 0;00 00 33 <20> 04 37 46 40]? Multiply 0;00 44 43 20 by 0;00 44 43 20 so that 0;00 00 33 <20> 04 37 4[6 40 : increase 0;00 44 43 20 to 1;00 00 16 40, so that] 1;00 45 is the reciprocal. Decrease 0;00 44 43 20 from 1;00 00 16 40, [so that 0;59 15 33 20 is its reciprocal]. |
| Reverse [14:18] r 1515–r 1818 | A reciprocal and its reciprocal are 2;03. Multiply by 0;30, so that 1;01 30 [: multiply 1;01 30 by 1;01 30, so that 1;03 02 15]. Decrease 1 from its middle, so that the remaining is 0;03 02 15. How much by how much [should I multiply so that 0;03 02 15]? Multiply 0;13 30 by 0;13 30 «by» so that 0;03 02 15 : [increase] 0;13 30 [to 1;01 30, so that 1;15 is the reciprocal]. Decrease 0;13 30 from 1;01 30, so that 0;48 is its reciprocal. |
| Reverse [18:23] r 1919–r 2323 | A reciprocal and its reciprocal are 2;05 26 40. Multiply by 0;30, so that 1;02 43 [20 : multiply [1;02 43 20 by 1;02 43 20], so that 1;05 34 04 37 46 40. Decrease 1 from its middle, so that [the remaining is 0;05 34 04 37 46 40. How much by how much] should I multiply [so that 0;05 34] 04 37 46 40? [Multiply] 0;18 16 40 [by 0;18 16 40, so that 0;05 34 04 37 46 40.] Increase 0;18 16 40 to 1;02 43 20, so that 1;21 is the reciprocal. [Decrease 0;18 16 40 from 1;02 43 20, so that] the remaining is 0;44 26 40, its reciprocal. |
| Reverse [23:27] r 2424–r 2727 | A reciprocal and its reciprocal are 2;00 15. Multiply by 0;30, so that 1;00 07 30 : [multiply] 1[;00 07 30 by 1;00 07 30, so that 1;00 15 00 56 15]. Decrease 1 from its middle, so that the remaining is 0;00 15 00 56 15. How much by how much should I multiply [so that 0;00 15 00 56 15]? Multiply 0;03 53 30 by 0;03 52 30 so that 0;00 15 00 56 15 : [Increase ] 0;03 5[2 30 to 1;00 07 30, so that] 1;04 is the reciprocal. Decrease 0;03 52 30 from 1;00 07 30, so that 0;5[6 15 is its reciprocal]. |
| Reverse [27:28] r 2828–r 2828 | [Tablet of] Anu-ab-uṣur, scribe of Enuma Anu Ellil, son of [Anu-belšunu, descendant of Sin-leqi-unninni.] |
Transliteration (ATF)
From Open Richly Annotated Cuneiform Corpus (ORACC) · CC BY-SA 3.0 (ORACC default); per-file JSON license fields say: This data is released under the CC0 license (133618)
| Line | ATF | Machine English (draft) |
|---|---|---|
| r | ||
| 28 | [ṭup-pi] {⸢m⸣}{d}60—AD—SI {lu₂}UMBISAG U₄ {d}60 {d}EN.LIL₂.LA₂ A {⸢m⸣}{[d}60—⸢EN-šu₂-nu⸣ [A {m}{d}30—TI—ER₂] | |
Sources & licences
- Cuneiform Digital Library Initiative · CDLI: reuse of catalogue/transliteration data with citation of CDLI; images restricted (not fetched) · Cuneiform Digital Library Initiative (CDLI), https://cdli.earth/ ; cite CDLI and the P-number
- Open Richly Annotated Cuneiform Corpus (ORACC) · CC BY-SA 3.0 (ORACC default); per-file JSON license fields say: This data is released under the CC0 license (133618) · ORACC project editions, https://oracc.museum.upenn.edu/ ; cite the individual project (see texts.extra_json.project)
Cite the original edition and translator. Page generated by Unread Atlas from the public corpus export of 2026-10-11.