CHAP. II] SOLUTION OP ax+by-c. 43

In case (§ 252) the number of quotients is odd, the numbers found by the above rule must be subtracted from their respective abraders to give the true quotient and multiplier. Thus (§ 255) for Dividend 10, Divisor 63, Additive 9, the successive series are 0, 6, 3, 9, 0 [and 0, 6, 27, 9 and 0, 171, 27, and 27 = 2-10 + 7, 171 = 2-63 + 45], so that 10 - 7 = 3 is the quotient and 63 - 45 = 18 is the multiplier [check: 10-18 + 9 = 3-63].

Concerning a "constant pulverizer" (§263, pp. 119-120), we may solve the first example above by first treating the related problem: Dividend 17, Divisor 15, Additive 1, then multiply the deduced multiplier 7 and quotient 8 by the former additive 5, abrade and get 6 and 5 as the quotient and multiplier when the additive is 5.

As to a "conjunct pulverizer" (§§265-6, p. 122), if there be a fixed divisor and several multipliers, make the sum of the latter the dividend, the sum of the remainders the subtractive quantity, and proceed as before. Thus, to find a number whose products by 5 and 10 give the respective remainders 7 and 14 when divided by 63, take Dividend 5 + 10, Divisor 63, Subtractive 7 + 14; reduced Dividend, Divisor and Subtractive are 5, 21, 7; the desired multiplier is 14.

Bha"scara5 gave a rule for solving linear equations in two or more unknowns. In case there are k equations, eliminate k — 1 of the unknowns and proceed with the single resulting equation as follows. Assign arbitrarily special values to all but two of the unknowns. In the resulting equation in two unknowns, solve for one in terms of the other and render it integral by use of the pulverizer.

For example, of two equally rich men, one has 5 rubies, 8 sapphires, 7 pearls and 90 species; the other has 7, 9, 6 and 62 species; find the prices (l/j c, ri) of the respective gems in species. Thus

6y + Sc + In + 90 = 7y + 9c + 6n + 62, y = ~ c + n + 28.

L

Take n = 1, and use the method of a pulverizer to find c so that y = (- c + 29)/2 shall be integral. We get

c = 1 + 2p, y = 14 - p,

where p is arbitrary. For p — 0, 1, we get (y, c, ri) = (14, 1, 1), (13, 3, 1). Again (§ 161, pp. 237-8), what three numbers being multiplied by 5, 7, 9 respectively, and the products divided by 20, have remainders in arithmetical progression with the common difference 1, and quotients equal to remainders? Call the numbers c, n, p\ the remainders y, y + 1, y + 2. Thus

5c - 20y = y, y = 5c/21;

7n - 20(y + 1) = y + 1, y = (7n - 21)/21;

9p - 20(y + 2) = y + 2, y = (9p - 42)/21. By the first two values of y, c = (7n — 21)/5. By the last two, __________________________n = (Qp - 21)17,__________________________

• Vfja-ganita (Algebra), §§ 153-6; Colebrooke,1 pp. 227-232.