| 1 | 10 | ||
| \ | 10 | 106 | |
| \ | 1⁄2\3 | 71⁄9 | |
| Total | 113 | ||
| 1 | 113 | ||
| 2 | 2271⁄21⁄18 | ||
| \ | 4 | 4551⁄9 | |
| 1 | 4551⁄9 | ||
| 1⁄10 | 451⁄21⁄90 | ||
| \ | 1⁄20 | 221⁄21⁄41⁄180 | |
This solution for a long time baffled the ingenuity of Egyptologists, but the correct interpretation was finally discovered by Schack-Schackenburg (1899, see Peet, page 83).
In the first place, the papyrus states that the height of the granary is 9 and the breadth (diameter) 6, and in the solution, when we find 4 as 2⁄3 of 6, the author again calls 6 the breadth; but the solution is for a cylinder in which 9 is the diameter and 6 the height.[1]
Then the method of solution is not that used in 41 and 42, but a second one (employed for a similar problem in the Kahun papyrus, Griffith, 1897), giving the volume directly in khar and not first in cubed cubits. It may be expressed in the following rule: Add to the diameter its 1⁄3; square, and multiply by 2⁄3 of the height. In Problem 43 the addition of its 1⁄3 to the diameter makes 12, the square is 144, and 2⁄3 of the height is 4. With these numbers the rule gives 576 khar,and this is just what the author would have obtained if he had followed the solutions of 41 and 42.[2]
But the author, before taking the steps of this rule, deducts from the diameter its 1⁄9, as by the other rule, and so obtains a result which is (8⁄9)2 of the correct result, namely, 4551⁄9 khan[3]
- ↑ There is some confusion in Peet’s explanation of this mistake (page 84). He supposes that the statement was correct in the original papyrus, but that "a later scribe, seeing in the first line of working the subtraction of a ninth of 9 from 9, . . . concluded that 9 must be the diameter and not the height, and so he transposed the two dimensions in the statement," but if the statement now in the papyrus is the result of such a transposition the transposition must have been the other way.
- ↑ In modern form the number of khar according to the two methods is expressed by the formulae (8⁄9d)2 h.2⁄3 and (1⁄3d)2.2⁄3h. both of which reduce to 31⁄27hd2
- ↑ Eisenlohr translated the word now read khar as "bodily content" and supposed that the addition of 1⁄2 to reduce cubed cubits to khar was a part of the calculation of the cubed cubits. This is the same as multiplying the base by 1⁄2 of the height instead of by the height itself, and to explain this he supposes that the given base in 41 and 42 is the upper base, and that the lower base is larger. When he comes to Problem 43 he has great difficulty. In the first place, he finds that 8⁄9 of the diameter of the base is multiplied by 1⁄3(1⁄3 of it added) before squaring, as if the base were an ellipse with one axis 16⁄9 of the other, or as if our author intended to get the lower base or some section between the two. Then the area obtained is multiplied by 2⁄3 of the height instead of 3⁄2, which might be because he had taken
