44 HISTORY OF THE THEORY OF NUMBERS. [CHAP. II 



which by use of the pulverizer gives n 9Z + 6, p = 71 + 7. Then 

 c = (63Z + 21) /5, which by the pulverizer gives c = 63ft + 42, Z = 5fc + 3. 

 Hence n = 45h + 33, p = 35/i + 28, y = 15h + 10. Since the quotient 

 equals the remainder, which cannot exceed the divisor, we must take h = 0. 



What two numbers, except 6 and 8, being divided by 5 and 6 have the 

 respective remainders 1 and 2; while their difference divided by 3 has the 

 remainder 2 ; their sum divided by 9 has the remainder 5 ; and their product 

 divided by 7 leaves 6 ( 163, p. 239)? The conditions other than the last 

 give 45p + 6 and 54p + 8 as the numbers. As the product is quadratic, 

 take p = 1 (^provisionally]. Abrading the product by multiples of 7, 

 we get 3p + 2, which must equal 71 + 6. By the pulverizer, p = 7h + 6, 

 and the second number is 378/1 + 332. The additive (45p) of the first 

 number multiplied by 7h is its present additive, so that the first number is 

 315ft + 51. 



What number multiplied by 9 and 7 and the products divided by 30 

 yields remainders whose sum increased by the sum of the quotients is 26 

 ( 164, p. 240)? Answer, 27. 



What number multiplied by 23 and the product divided by 60 and 80 

 has 100 as the sum of the remainders ( 166-7, p. 241)? Taking 40 and 60 

 as the remainders, we get the number 240Z + 20. Taking 30 and 70, 

 we get 240Z + 90; etc. 



Bachet de Mefeiriac 6 stated that if A and B are any relatively prime 

 integers, we can find a least integral multiple of A which exceeds an integral 

 multiple of B by a given integer J [i. e., solve Ax = By + /]. Proof was 

 given in the 1624 edition, pp. 18-24. It suffices to solve AX = BY + 1. 

 Bachet employed notations for 18 quantities, making it difficult to hold m 

 mind the relations between them and so obtain a true insight into his correct 

 process. Hence we shall here carry out in clearer form his process for his 

 example A = 67, B = 60. Subtract the smaller number B as many times 

 as possible from the larger number A, to give a positive remainder C. 

 If C = 1, A itself is the desired multiple of A which exceeds a multiple of B 

 by unity. Next, let C > 1 and subtract C from B as many times as possible, 

 continuing until the remainder 1 is reached: 



(1) 67 = 1-60 + 7, 60 = 8-7 + 4, 7 = 1-4 + 3, 4 = 1-3 + 1. 

 From the last equation we deduce 



(2) 3-3 = 2-4 + 1, 



by the rule that if a = mb + 1 then ab + 1 a is the least multiple of b 

 which exceeds by unity a multiple of a. Multiply the third equation (1) 

 by first coefficient 3 in (2) and eliminate the term 3 3 by use of (2) ; we get 



(3) 3-7 = 5-4 + 1. 



6 Clavde Caspar Bachet, Problemes Plaisans et Delectables, Qui se font par les Nombres, 

 ed. 1, Lyon, 1612, Prob. 5; ed. 2, Lyon, 1624; ed. 3, Paris, 1874, 227-233; ed. 4, 1879; 

 ed. 5, 1884; abridged ed., 1905. See Lagrange. 19 



