CHAP. II] CHINESE PROBLEM OF REMAINDERS. 61 



W. Beveridge 94 treated the problem to find the least number P which 

 has given remainders K and L when divided by A and B, when the latter 

 are relatively prime. Let D be the least multiple of B which has the 

 remainder 1 when divided by A; let C be the least multiple of A which 

 divided by B leaves 1. Then P = DK + CL, as shown by a two page 

 proof. 



To find the least number P which has the given remainders K, L, Z 

 when divided by the relatively prime numbers M, B, A, first find the least 

 multiple F of AB, least multiple N of AM, least multiple Q of BM, which 

 have the remainder 1 when divided by M, B, A, respectively. Then 

 P = KF + LN + ZQ. 



This is precisely the rule as given later by Euler 96 and Gauss. 76 



* J. Wallis 95 gave an empirical solution of the problem of the Julian 

 period. 



T. F. de Lagny 9 treated the problem to find the year x of the Julian 

 period when the solar cycle is 13, the lunar cycle is 10 and the "indiction" 

 is 7; thus if x is divided by 28, 19, 15, the remainder is 13, 10, 7, respectively. 

 From x = 28m + 13 = 19n + 10, he found 9 that n = 9 + 28/, where / 

 is an integer. Thus x = 19n + 10 = 181 + 532/. Since x 7 is to be 

 divisible by 15, the least / is 3. 



L. Euler 96 treated the problem to find an integer z which has the re- 

 mainders p and q when divided by a and b, respectively, where a > b. 

 Thus z = ma + P = nb + q. He solved the second equation by use of 

 the process for the greatest common divisor for a, b, continued until one 

 of the remainders c, d, e, - is reached which divides v = p q. He thus 

 deduced the result 



z = q 



abv ( ~r ^~ + ^ --+ ' ), 

 \ab be cd de / 



in which the series is continued until we reach a remainder dividing v. 

 At the end of the paper, Euler gave a rule generally attributed to Gauss. 76 

 To find a number which has the respective remainders p, q, r, s, t when 

 divided by a, b, c, d, e, which are relatively prime in pairs. An answer is 

 Ap + Bq + Cr + Ds + Et + Mabcde, where 



A = (mod bcde), A = 1 (mod a); B = (mod acde), 

 B = l (mod 6) ; E = (mod abed}, E = 1 (mod e). 



C. von Clausberg 97 found a multiple of 7 having the remainder 10 when 

 divided by 15. 



N. Saunderson 98 treated the problem to find a number which has the 

 remainders d and e when divided by a and b, a > b. Let I be the g.c.d. 



94 Institutionum Chronologicarum libri II. Una cum totidem Arithmetices Chronologicae 



Libellis. Per Guilielm Beveregium, Londini, 1669, lib. II, pp. 253-6. 



95 Opera, 2, 1693, 451-5. Cf. Hutton. 101 



96 Comm. Acad. Petrop., 7, 1734-5, 46-66; Comm. Arith. Coll., I, 11-20. 



97 Demonstrative Rechnenkunst, 1732, 1366, 1493. 



98 The Elements of Algebra, Cambridge, 1, 1740, 316-329. Reproduced by de la Bottiere, 



Mem. de math, phys., presented . . . divers savans, 4, 1763, 41-65. 



