Page:The Rhind Mathematical Papyrus, Volume I.pdf/43

Rh products, leaving for the present the smaller fractions, and he gets $32 1/2$. The remaining $1/undefined$ required to make up all of 33 must then be equal to the smaller fractions of these products plus a further product by a multiplier yet to be determined. It remains only to get this multiplier. Choosing 42 as a convenient number, he applies his expressions to 42. The required $1/undefined$, taken as $1/undefined$ of 42, is equal to 21; the previously discarded fractions make 17$1/undefined$ of it, and leave 3 $1/undefined$ $1/undefined$ to be obtained by the multiplier that we are seeking. The original expression, 1 $2/3$ $1/undefined$ $1/undefined$, applied to 42, is equal to 97, and $1/undefined$ of it is 1; that is, $1/undefined$ as multiplier gives a product that, taken as a part of 42, equals 1. To get the product that equals 3 $1/undefined$ $1/undefined$ he must take as multiplier 3 $1/undefined$ $1/undefined$ times $1/undefined$, and this multiplier, together with the $1/undefined$ checked in the ﬁrst multiplication, makes up his answer, which is

14 $14 1/4$ $1/undefined$ $1/undefined$ $1/undefined$ $1/undefined$ $1/undefined$ $1/undefined$.

In Problem 33 to get nearly 37 he has only to carry his doubling one step further, 16 times 1 $1/undefined$ $2/3$ $1/undefined$ being equal to 36 $1/undefined$ $2/3$ $1/undefined$, or only $1/undefined$ less than 37. He applies his expressions to 42 to determine that this remainder is $1/undefined$ as well as to determine the multiplier that will produce it. To do the latter, since $1/undefined$ of 42 is 2, he has to use as multiplier 2 times $1/undefined$, and his answer is

16 $1/undefined$ $1/undefined$ $1/undefined$.

In Problem 32 the multiplier is 1 $1/undefined$ $1/undefined$ and the product to be obtained is 2, and in Problem 34 the multiplier is 1 $1/undefined$ $1/undefined$ and the product to be obtained is 10. The latter may be associated with Problems 7 and 9-15.

In Problem 31 the answer, as we have seen, is particularly long, and the author makes no attempt at a proof, but in the other four problems he goes through the proof very carefully, so that it becomes more prominent than the solution.

In the proof of his answer to Problem 32 he multiplies 1 $1/undefined$ $1/undefined$ $1/undefined$ $1/undefined$ by 1 $1/undefined$ $1/undefined$. From the partial products he selects first the larger numbers, namely, 1, $1/undefined$, $1/undefined$, and $1/undefined$, or 1 $1/undefined$ $1/undefined$, leaving $1/undefined$ to be made up from the remaining fractions, and by applying these to 912, the largest number whose reciprocal is found among them, he finds that their sum is exactly $1/undefined$, and that his answer is correct. In the same way he carries through the somewhat easier proof of Problem 34.

In the case of Problem 33 he multiplies his answer by 1 $1/undefined$ $2/3$ $1/undefined$. Adding the whole numbers and larger fractions of this multiplication,