Page:The Rhind Mathematical Papyrus, Volume I.pdf/27

Rh which (usually) is not the answer, and get the answer by ﬁnding its relation to this number.

False position is used in a number of multiplication problems in which the multiplier and product are given and it is required to ﬁnd the multiplicand, because by this method it is possible to keep in mind the nature of the quantities involved. Problems 24-27 exhibit the method very clearly although the required multiplicand in these problems is nothing more definite than “a quantity,” and the given product, which must be a quantity of the same kind, is given simply as a number. It is instructive to compare this group of problems with Problems 30-34, which interchange multiplicand and multiplier. The problems of the two groups are problems of the same kind and all of them involve only “a quantity.”

I think, however, that the hekat problems, 35, 37, and 38, show more clearly that this process enables us to keep in mind the nature of the quantities involved, and I will give in detail the reasoning of one of these problems as I understand it.

In Problem 35 a vessel ﬁlled $1/undefined$ times with grain will make 1 hekat (about half a peck). The vessel contains so much grain and the hekat is so much grain. $1/undefined$ times the former makes the latter. Here we have given the multiplier, $1/undefined$, and the product, 1 hekat of grain, to find the multiplicand, which will be a certain portion of a hekat of grain. Assume a vessel that itself contains 1 hekat; this filled $3 1/3$ times will give $3 1/3$ hekat. Then the one that we have must bear the same relation to the one assumed, that the amount of grain that fills the former $3 1/3$ times bears to the amount of grain that fills the latter $3 1/3$ times; that is, that one hekat bears to $3 1/3$ hekat. Thus we have to ﬁnd the relation of one hekat to $3 1/3$ hekat. Making $3 1/3$ hekat the multiplicand and one hekat the product we get for the multiplier the number $3 1/3$ $3 1/3$. The amount of grain that the given vessel holds is then$3 1/3$ $1/undefined$ times the amount of grain that the one assumed holds; that is, it is $1/undefined$ $1/undefined$ times 1 hekat.