58 HISTORY OP THE THEORY OF NUMBERS. [CHAP. II 



the latter caused M. Cantor 74 to criticize the validity of the rule. The rule 

 was defended by L. Matthiessen, 75 who pointed out its identity with the 

 following statement by C. F. Gauss. 76 If m = mim 2 m 3 - , where mi, m 2 , 

 ma, are relatively prime in pairs, and if 



on = (mod m/mi), on = 1 (mod w f ) (i = 1, 2, 3, )> 

 then x = airi + <x 2 r 2 + is a solution of 



x = TI (mod mi), x = r 2 (mod m 2 ), 



This method is very convenient when one has to treat several problems 

 with fixed mi, m 2 , , but varying ri, r 2 , 



Nicomachus 77 (about 100 A.D.) gave the same 71 problem and solution 23. 



Brahmegupta 78 (born, 598 A.D.) gave a rule which becomes clearer 

 when applied to an example: find a number having the remainder 29 

 when divided by 30 and the remainder 3 when divided by 4. Dividing 

 30 by 4, we get the residue 2. Dividing 4 by 2, we get the residue zero 

 and quotient 2. Dividing the difference 3 29 by the residue 2, we get 

 13. Multiply the quotient 2 by any assumed multiplier 7 and add the 

 product to - 13; we get 1. Then 1-30 + 29 = 59 is the desired number. 



This problem forms the second stage of the solution of the "popular" 

 problem (7, p. 326) : find a number having the remainders 5, 4, 3, 2 

 when divided by 6, 5, 4, 3, respectively. The answer is stated correctly 

 to be 59. 



The priest Yih-hing 79 (f 717 A.D.) in his book t'ai-yen-lei-schu gave a 

 generalization to the case in which the moduli m,- are not relatively prime. 

 Express the l.c.m. of mi, w 2 , , m k as a product m = nw /*, of rela- 

 tively prime factors, including unity, such that m divides m. Then, if 



on = (mod m/jUj), on = 1 (mod m) (i = !, , &), 



x = a 1 r l + 2 r 2 + is a solution. Other solutions are obtained by sub- 

 tracting multiples of m. Yih-hing proposed to find the number of com- 

 pleted units of work, the sa/me number x of units to be performed by each 

 of four sets of 2, 3, 6, 12 workmen, such that after certain whole days' work, 

 there remain 1, 2, 5, 5 units not completed by the respective sets. The 

 l.c.m. of mi = 2, , w 4 = 12 is m = 12. Taking /zi = ^2 = 1> Ma = 3, 

 /Z4 = 4, we get ! = 2 = 12, 3 = 4, a 4 = 9, 



x=l-12 + 2- 12 + 5-4 + 5-9 = 101 = 17 (mod 12). 



74 Zeitschrift Math. Phys., 3, 1858, 336; not repeated in his Geschichte der Math., ed. 2, I, 

 643. H. Hankel, Geschichte d. Math, in Alterthum u. Mittelalter, 1874, erred in identify- 

 ing the Chinese rule with the Indian cuttaca. 4 



78 Zeitschrift Math. Phys., 19, 1874, 270-1; Zeitschrift Math. Naturw. Unterricht, 7, 1876, 80. 



76 Disq. Arith., art. 36; Wcrke, I, 26. Cf. Euler. 96 



77 Pythagorei introd. arith. libri duo, rec. R. Hoche, Leipzig, 1866, Supplement, prob. V. 



78 Brahme-sphut'a-sidd'hanta, Ch. 18 (Cuttaca = algebra), 3-6, Colebrooke, 1 pp. 325-6. 



79 L. Matthiessen, Comptes Rendus Paris, 92, 1881, 291; Jour, fur Math., 91, 1881, 254-261; 



Zeitschr. Math. Phys., 26, 1881, Hist.-Lit. Abt., 33-37 (correction of Biernatzki 73 ). 



