Page:The Rhind Mathematical Papyrus, Volume I.pdf/42

26 numerical one, but it is written in the same phraseology save for the omission of this word and is solved by the same method, so that we can consider the four problems together.

Problem 24 is to ﬁnd a quantity that with its $1/undefined$ will make 19. The author assumes 7, which with its $1/undefined$ makes 8, and then, to find the answer, he multiplies 7 by the number that multiplying 8 will give 19; 7, 8, and 19 as well as the answer all represent quantities of the same kind.

In each of the problems 24-27 the multiplier is 1 and a single fraction and he assumes as his answer the number that is the denominator of the fraction. In 28 and 29 he has two or three fractions and assumes the product of their denominators as his answer.

Problems 30-34 are far more complicated; nowhere, indeed, in the entire papyrus is more skill shown in dealing with long fractional expressions. In these problems it is required to determine what quantity a certain fractional number must multiply in order to produce a given product. But in solving them the Egyptian changes his point of view and determines what must multiply this fractional number. In the proofs, however, he goes back to his original point of view and multiplies the answer by the given fractional number to get the number given as product.

In Problem 30 the author wishes to ﬁnd what quantity $2/3$ $1/undefined$ must multiply in order to produce 10, and so changing his point of view as I have said, he proceeds to multiply $2/3$ $1/undefined$. He finds that multipliers amounting to 13 will give 9 and a series of fractions. Then we may suppose that he takes these fractions and the given expression as parts of 30. At any rate he ﬁnds that $1/undefined$ will just complete the fractions to 1, and that $1/undefined$ of the given expression will make this $1/undefined$. Thus his answer is $13 1/23$.

In Problems 31 and 33 he has the same fractional expression, namely, 1 $2/3$ $1/undefined$ $1/undefined$, the products to be obtained being 33 and 37. In both cases he multiplies this fractional expression in such a way as to get very nearly the product desired.

Thus to get 33 he doubles three times and halves twice, checking those multipliers that will give him very nearly this number. The numbers that he checks are equal to $1/undefined$, and make up most of his answer. He adds the whole numbers and larger fractions of the corresponding