Page:The Rhind Mathematical Papyrus, Volume I.pdf/88

¹⁄₁₀ also he can take without difficulty. For 13 he has ¹⁄₁₀ of 10 equal to 1, and ¹⁄₁₀ of 3, by the table that precedes Problem 1, is ¹⁄₅¹⁄₁₀. Finally, ¹⁄₁₀ of ¹⁄₂₈ is ¹⁄₂₈₀.

In the original our author seems to say that we must multiply ¹⁄₃₀ by ¹⁄₂₃ to get ²⁄₃¹⁄₁₀. It may be that he did not mean to put a dot (the sign for fraction) over 23, but meant to say, Multiply ¹⁄₃₀ by 23 to get ²⁄₃¹⁄₁₀. This is a correct statement but does not explain his solution. Following the method given in the other problems of this group, he should have said, Multiply ¹⁄₂₃ by ²⁄₃¹⁄₁₀ to get ¹⁄₃₀.

Problem 31
A quantity, its ²⁄₃, its ¹⁄₂, and its ¹⁄₇, added together, become 33. What is the quantity?

Multiply 1²⁄₃¹⁄₂¹⁄₇ so as to get 33.

	1	12⁄31⁄21⁄7
\	2	41⁄31⁄41⁄28
\	4	91⁄61⁄14
\	8	181⁄31⁄7
	1⁄2	1⁄21⁄31⁄41⁄14
\	1⁄4	1⁄41⁄61⁄81⁄28

Total 14¹⁄₄. 14¹⁄₄ times 1²⁄₃¹⁄₂¹⁄₇ makes 32¹⁄₂ plus the small fractions ¹⁄₇¹⁄₈¹⁄₁₄¹⁄₂₈¹⁄₂₈. 32¹⁄₂ from 33 leaves the remainder ¹⁄₂ to be made up by these fractions and a further product by a number yet to be determined.
¹⁄₇¹⁄₈¹⁄₁₄¹⁄₂₈¹⁄₂₈
taken as parts of 42 are
65¹⁄₄31¹⁄₂1¹⁄₂
making in all 17¹⁄₄, and requiring 3¹⁄₂¹⁄₄ more to make 21,¹⁄₂ of 42.

Take 1²⁄₃¹⁄₂¹⁄₇ as applying to 42:

\	1	42
\	2⁄8	28
\	1⁄2	21
\	1⁄7	6
	Total	97.

That is, 1²⁄₃¹⁄₂¹⁄₇ applied to 42 gives 97 in all. ¹⁄₄₂ of 42, or 1, will be ¹⁄₉₇ of this, and 3¹⁄₂¹⁄₄ will be 3¹⁄₂¹⁄₄ times as much. Therefore we multiply ¹⁄₉₇ by 3¹⁄₂¹⁄₄.

[1]\	1⁄97	1⁄42	or	1	as	a	part	of	42
\	1⁄561⁄6791⁄776	1⁄21	"	2	"	"	"	"	42
\	1⁄194	1⁄84	"	1⁄2	"	"	"	"	42
\	1⁄388	1⁄168	"	1⁄4	"	"	"	"	42

1. These multiplications do not represent the multiplication of ¹⁄₉₇, by 3¹⁄₂¹⁄₄, but form a continuation of the multiplication at the beginning of the solution. But the partial products of