CHAP, ii] SOLUTION OF 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, [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. 



Bhascara 5 gave a rule for solving linear equations in two or more un- 

 knowns. In case there are k equations, eliminate k 1 of the unknowns 

 and proceed with the single resulting equation as follows. Assign arbi- 

 trarily 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 

 (y, c, ri) of the respective gems hi species. Thus 



By + 8c + 7n + 90 = 7y + 9c + Qn + 62, y = ~ c + n + 28 . 



& 



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 arith- 

 metical 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; 



In - 2Q(y + 1) = y + 1, y = (7n - 21)/21; 

 9p - 2Q(y + 2) = y + 2, y = (9p - 42)/21. 

 By the first two values of y, c = (In 21)/5. By the last two, 



n = (9p - 



6 Vija-ganita (Algebra), 153-6; Colebrooke, 1 pp. 227-232. 



