As many times as 20 must be multiplied to give 10, so many times 27 must be multiplied to give the required number.
But at this point he seems to have changed the order of these numbers in his mind and to have said, As many times as 20 must be multiplied to give 27 so many times 10 must be multiplied to give the required number.
| \ | 1 | 20 |
| 1⁄2 | 10 | |
| \ | 1⁄5 | 5 |
| \ | 1⁄10 | 2 |
| Total | 11⁄41⁄10 |
Therefore we must multiply 10 by 11⁄41⁄10 (see Peet, page 64).
SECTION VI
Problems 30-34. Division by a Fractional Eexpression[1]
Problem 30
If the scribe says, What is the quantity of which as 2⁄31⁄10 will make 10, let him hear.
Multiply 2⁄31⁄10 as to get 10.
| \ | 1 | 2⁄31⁄10 |
| 2 | 11⁄31⁄5 | |
| \ | 4 | 31⁄15 |
| \ | 8 | 61⁄101⁄30 |
Total 13. 13 times 2⁄31⁄10 makes 9 and the fractions 2⁄3,1⁄10, 1⁄15, 1⁄10, and 1⁄30. The remainder is 1⁄30. Take 30. 2⁄31⁄10 of 30 is 23. Therefore 1⁄30 of 30, or 1, will be 1⁄23 of this. 131⁄23 is the required number.
For proof we multiply 131⁄23 by 2⁄31⁄10.
| 1 | 131⁄23 | |
| \ | 2⁄3 | 82⁄31⁄461⁄138 |
| \ | 1⁄10 | 11⁄51⁄101⁄230 |
| Total | 10. | |
To get the remainder 1⁄30 the author could apply the fractions of his partial products, 2⁄3. 1⁄10, etc., to 30, their values taken this way making 29 in all, and requiring 1 part more to make the full 30, so that in order to make a full 10 he would require, in addition to what he already has, 1⁄30.
In the multiplication of his proof we may notice that2⁄3 of 13 is given at once as 82⁄3 and that 2⁄3 of 1⁄23 is given by the rule in Problem 61 (see Introduction, page 25).- ↑ Full explanations of these problems are given in the Introduction, pages 25-28. I may mention that the problems of this section and the next are numbered by Eisenlohr a little differently from the order in the papyrus, 33 and 34 coming, in fact, after 38, but belonging in the same group with 30, 31, and 32. As I have used his numbering, I have followed his order.
