The process of false position can be applied to other problems besides those of division. This is well illustrated in Problem 40, and I will give an explanation of it at this point.
This problem is to divide 100 loaves among five men in such a way that the shares received shall be in arithmetical progression and that 1⁄7 of the sum of the largest three shares shall be equal to the sum of the smallest two. The papyrus does not say that the shares shall be in arithmetical progression, but the solution shows that this is the intention. After stating the problem it simply says that the difference is 51⁄2 without explaining the method of obtaining this number, nor does it say that the smallest share shall be 1 nor even mention the smallest share. We may suppose that it is so natural to think of 1 as the smallest number that it does not occur to the author to mention it, even when he finally gets an answer in which the smallest number is not 1. Now to get the common difference he may have assumed first a common difference 1. The terms of the progression would be 1, 2, 3, 4, 5, the sum of the smallest two would be 3, and 1⁄7 of the sum of the largest three would be 1 1⁄2 1⁄7 1⁄12, a difference between the two sides of 1 1⁄4 1⁄28. If he then assumed a common difference of 2, and hence the progression 1, 3, 5, 7, 9, he would find the sum of the smallest two tenns to be 4, and 1⁄7 of the sum of the largest three, 3, making a difference between the two sides of 1. In other words, for each increase of 1 in the assumed common difference he would find the inequality between the two sides reduced by 1⁄4 1⁄28. To make the two sides equal he must multiply his increase 1 by as many times as 1⁄4 1⁄28 is contained in 1 1⁄4 1⁄28, which is 41⁄2, and this added to the first assumed difference 1 makes 51⁄2 as the true common difference. This process of reasoning is exactly in accordance with Egyptian methods.[1]
To determine if the progression which he has obtained fulfills the second requirement of the problem, namely, that the number of loaves shall be 100, he proceeds as follows: Having the progression 1, 61⁄2, 12, 171⁄2, 23, he finds that the sum is 60 instead of 100. Therefore it is necessary to multiply by the factor that will produce 100 from 60,
- ↑ Peet (page 78) seems to think that this problem is like some of the inverse problems of the papyrus (see below, page 35); that the author, having a series with a common difference of 51⁄2, noticed in this series the relation between the sum of the smallest two terms and the sum of the largest three, and made up an inverse problem with 100 for sum instead of 60. But it is as difficult to see how he should discover this relation in a series that he has before him, as it is to see how he should find the common difference, 51⁄2, when he has given this relation.
