It may be supposed that our author first solved the problem as follows:
Assume 9.
| \ | 1 | 9 |
| \ | 2⁄3 | 6 |
| Total | 15 | |
| 1 | 15 | |
| 1⁄3 | 5 | |
| Remainder | 10. | |
As many times as 10 must be multiplied to give 10, that is, once, so many times 9 must be multiplied to give the required number, and therefore the required number is 9. But now he notices that 9 is obtained by taking away its 1⁄10 from 10, so he puts in the solution given in the papyrus.
The solution does not seem to be complete. The words, Do it thus ("The doing as it occurs") are usually put at the beginning of numerical work, and in no other problem are they at the end of the solution. Peet has suggested (page 63) that, in copying, the scribe came to these words and unconsciously let his eye pass to the same words in the next problem, the statement of the next problem and the beginning of its solution being also omitted. A similar omission occurs in the solution of Problem 70.
Problem 29
A quantity and its 35 are added together, and 1⁄3 of the sum is added; then 1⁄3 of this sum is taken and the result is 10. What is the quantity?
| \ | 1 | 10 |
| \ | 1⁄4 | 21⁄2 |
| \ | 1⁄10 | 1 |
| The quantity is | 131⁄2 | |
| 2⁄3 | 9 | |
| Total | 221⁄2 | |
| 1⁄3 | 71⁄2 | |
| Total | 30 | |
| 2⁄3 | 20 | |
| 1⁄3 | 10. | |
As in the preceding problem it may be supposed that our author first solved the problem as follows:
Assume 27.
| \ | 1 | 27 |
| \ | 2⁄3 | 18 |
| Total | 45 | |
| 1⁄2 | 15 | |
| Total | 60 | |
| 2⁄3 | 40 | |
| 1⁄3 | 20. | |
