In the last part of the solution the author expresses 1⁄180 of 100 quadruple hekat in fractional parts of a quadruple hekat ("Horus eye" fractions) and quadruple ro and fractional parts. 100 times 1⁄180 gives him 1⁄2 1⁄21⁄4 and a remainder that must be multiplied further by 320 to reduce it to quadruple ro.
Problem 44
Example of reckoning the volume of a rectangular granary, its length being 10, its width 10, and its height 10. What is the amount of grain that goes into it?
Multiply 10 times 10; it makes 100. Multiply 100 times 10; it makes 1000. Add its 1⁄2; it makes 1500, its contents in khar. Take 1⁄20 of 1500; it makes 75, its contents in quadruple hekat, namely, 7500 hekat of grain.
The working out:
| 10 | ||
| 10 | 100 | |
| 1 | 100 | |
| 10 | 1000 | |
| 1 | 1000 | |
| 1⁄2 | 500 | |
| 1 | 1500 | |
| 1⁄10 | 150 | |
| 1⁄20 | 75. |
Proof:
| 1 | 75 | |
| 10 | 750 | |
| \ | 20 | 1500 |
| 1⁄10 of 1500 | 150 | |
| 1⁄10 of 1⁄10 | 15 | |
| 2⁄3 of 1⁄10 of 1⁄10 | 10 |
This problem is similar to 41 and 42 except that the base is a square instead of a cir cle. At the end, after the author has completed his calculations, he goes through them in reverse order, apparently as a proof of the correctness of his result. This reverse calculation is the same as the calculation given as the solution of the next problem, the next problem being the reverse of the present one.
the larger base, or because he had some vague idea of the rule for the volume of a hemisphere (the area of the base times % of the height), and applied it to this solid, even though it may not have been spherical. This is the way in which the question of a hemisphere came into the problem. See titles of articles by Borchardt (1897) and Schack-Schackenburg (1899).
