Page:The Rhind Mathematical Papyrus, Volume I.pdf/91

Applying ²⁄₃,¹⁄₄, and ¹⁄₂₈ to 42 we have

	1	42
\	2⁄3	28
	1⁄2	21
\	1⁄4	101⁄2
\	1⁄28	[1]11⁄2

The total is 40; there remains 2, or ¹⁄₂₁ of 42. As 1 ²⁄₃¹⁄₂¹⁄₇ applied to 42 gives 97, we shall have as a continuation of our first multiplication

	1⁄97	1⁄42	or	1	as	a	part	of	42
\	1⁄361⁄6791⁄776	1⁄21	"	2	"	"	"	"	42.

This ¹⁄₂₁ with the product already obtained will make the total 37. Thus the required quantity is 16 ¹⁄₅₆¹⁄₆₇₉¹⁄₇₇₆.

Proof.

	1	161⁄561⁄6791⁄776
	2⁄3	102⁄31⁄841⁄13581⁄40741⁄
	1⁄2	81⁄1121⁄13581⁄1552
	1⁄7	21⁄41⁄281⁄3921⁄47531⁄5432

The whole numbers and larger fractions make 36²⁄₃¹⁄₄¹⁄₂₈ the remainder is ¹⁄₂₈¹⁄₈₄. The smaller fractions applied to 5432 make

97	8	7
642⁄3	4	11⁄3	42⁄3
481⁄2	4	31⁄2
131⁄21⁄41⁄141⁄28	11⁄7	1.

²⁄₃, ¹⁄₄, ¹⁄₂₈, and the fractions of the remainder, ¹⁄₂₈ and ¹⁄₈₄, applied to 5432 make 3621¹⁄₃, 1358, 194, and 194 and 64²⁄₃; for we have

	1	5432
	2⁄3	36211⁄3
	1⁄2	2716
	1⁄4	1358
	1⁄28	194
	Total	51731⁄3.

There remains 258²⁄₃ which is equal to 194 plus 64²⁄₃.

We may notice that in the course of his proof our author has ²⁄₃ of ¹⁄₆₇₉ equal to ¹⁄₁₃₅₈¹⁄₄₀₇₄ a remarkable application of the rule given in Problem 61.

For an explanation of the proof see Introduction, pages 27-28. The calculation by which the 97 is obtained is given above in the solution of Problem 31.

1. Since ¹⁄_2\3 of 42 is 28. 1¹⁄₂ being the reciprocal of ²⁄₃.