CHAP. II] CHINESE PROBLEM OF REMAINDERS. 59 



For x = 17, the completed part is 8-2 + 5-3 + 2-6 + 1-12 = 55. We 

 may equally well take ni = 1, /i 2 = 3, /x 3 = 1, ^4 = 4 and get i = 12, 

 az = 4, az = 12, 4 = 9, Sour,- = 125 = 17 (mod 12). 



A condition on the solvability of the problem is that r r, be divisible 

 by the g.c.d. of m z , ray. 



Ibn al-Haitam 80 (about 1000) gave two methods to find a number, 

 divisible by 7, which has the remainder 1 when divided by 2, 3, 4, 5 or 6. 

 The first method gives the one solution 1 + 2-3-4-5-6 = 721. The second 

 method gives a series of solutions 301, etc.; in effect f (6 + 2n-7)20 + 1, 

 where n is an integer such that 6 + 2n 7 is a multiple of 4. 



Bhascara 81 (born, 1114 A.D.) treated the problem to find the number 

 having the remainders 5, 4, 3, 2 when divided by 6, 5, 4, 3 respectively. 

 By the first two conditions, the number is Qc + 5 = 5n + 4. By use of 

 the "pulverizer," the integral value of c = (5n l)/6 is c 5p + 4. The 

 number 6c -f 5 = 30p + 29 must equal 4Z + 3. Hence p = (41 - 26) /30, 

 which is converted by the pulverizer into 2h + 1. Thus 



30p + 29 = 60/i + 59 

 is the answer. 



Again ( 162, p. 238), what number being divided by 2, 3, 5 has the 

 respective remainders 1, 2, 3, while the quotients divided by 2, 3, 5 respec- 

 tively have the remainders 1, 2, 3? Call the quotients 2c + 1, 3n + 2, 

 51 + 3. Then the number is 4c + 3 = 9n + 8 = 251 + 18. Applying the 

 pulverizer to the first equality, we get c = 9p + 8. The resulting number 

 36p + 35 must equal 251 + 18, whence p = 25h + 3 and the answer is 

 900/i + 143. 



Leonardo Pisano 82 treated (p. 281) the problem to find a number N, 

 divisible by 7, which gives the remainder 1 when divided by 2, 3, 4, 5 or 6. 

 By the latter condition, N exceeds 1 by a multiple of 60 ; but 60 has the 

 remainder 4 when divided by 7, while we need the remainder 6; thus we 

 multiply 60 by 2, 3, until we reach 60 X 5 with the remainder 6. 

 Thus N = 301, to which we may add a multiple of 420 = 60-7. Simi- 

 larly, 25201 is the multiple of 11 having the remainder 1 when divided by 

 2, .-.,10. 



To find (p. 282) a multiple of 7 having the remainders 1, 2, 3, 4, 5 when 

 divided by 2, 3, 4, 5, 6, we take 1 from a multiple of 60 such that the dif- 

 ference is divisible by 7; the result is 2-60 1 = 119. Similarly, to find 

 a multiple of 11 having the remainders 1, 2, -, 9 when divided by 2, 3, 

 - - -, 10, we subtract 1 from the least common multiple 2520 of 2, , 10 

 and get 2519, which being a multiple of 11 is the answer. 



He employed 83 (p. 304) in effect the rule t'ai-yen 71 to tell what number 

 not exceeding 105 a person has in mind if the latter gives the remainders 



80 Arabic MS. in Indian Office, London. Cf. E. Wiedemann, Sitzungsber. Phys. Medic. 



Soc. Erlangen, 24, 1892, 83. 



81 Vija-ganita (algebra), 160, Colebrooke, 1 pp. 235-7. 



82 Liber Abbaci (1202, revised 1228), pub. by B. Boncompagni, Rome, 1, 1857. 



83 M. Curtze, Zeitschrift Math. Phys., 41, 1896, Hist. Lit. Abt., 81-2, remarked that if Leonardo 



had found the rule independently, he would have so stated and would have given a proof. 



