CHAPTER III. MISCELLANEOUS PROBLEMS
Problem 61
Table for multiplication of fractions.
| 2⁄3 | of | 2⁄3 | is | 1⁄31⁄9 |
| 1⁄3 | " | 2⁄3 | " | 1⁄61⁄18 |
| 2⁄3 | " | 1⁄3 | " | 1⁄61⁄18 |
| 2⁄3 | " | 1⁄6 | " | 1⁄121⁄36 |
| 1⁄3 | " | 1⁄2 | " | 1⁄3 |
| 1⁄6 | " | 1⁄2 | " | 1⁄12 |
| 1⁄12 | " | 1⁄2 | " | 1⁄24 |
1⁄9 of 2⁄3 is 1⁄181⁄54; 1⁄9,2⁄3 of it is 1⁄181⁄54
| 1⁄5,1⁄4 | of | it | is | 1⁄20 | |||||
| 1⁄7,2⁄3 | " | " | " | 1⁄141⁄42 | |||||
| 1⁄11,2⁄3 | " | " | " | 1⁄14 | |||||
| 1⁄11,2⁄3 | " | " | " | 1⁄221⁄66, | 1⁄3 | of | it | is | 1⁄33 |
| 1⁄11,1⁄2 | " | " | " | 1⁄22 | 1⁄4 | " | " | " | 1⁄44 |
Peet points out (page 103) that the above table contains two forms of statement which have an interesting significance. In the first four lines we find forms of the type, 2⁄3 of 2⁄3 is 1⁄31⁄6, while in the last five are such forms as, 1⁄51⁄4 of it is 1⁄20. In the ninth line both forms are given, the statement being made twice, while the four lines that immediately precede seem to have been originally in the second form but to have been changed to the first; see Literal Translation. The reason for the second form is that of the fractions in this table the only ones that were legitimate multipliers are 2⁄3 and 1⁄3 and fractions obtained from them by halving. Thus the Egyptian could not directly say, 2⁄3 of 1⁄4" but only 2⁄3, 1⁄4 of it." He could have used the second form in all of these cases, but when he has a legitimate multiplier he prefers the first form, and so the author, perceiving after they were written that lines 5-8, like the first four, involve only these multipliers, changed them to the first form.
Problem[1] 61B
Rule for getting 2⁄3 of the reciprocal of an odd number.
To get 2⁄3 of 1⁄5 take the reciprocals of 2 times 5 and 6 times 5, and in the same way get 2⁄3 of the reciprocal of any odd number.
- ↑ See Introduction, pages 24-25.
