Demonstration in Chinese and Vietnamese mathematics 527
distribution (Problems 5–7, 10–11, 14–19, 38).^54 Among them, only two
problems deal with the ‘aggregated sharers’, namely, problem 6 and problem
38 (which is the problem solved in the ‘model examination paper’).
Problem 6 represents a case of a ‘mixed’ weighted distribution combining
‘solitary’ and ‘aggregated’ sharers. In this problem one deals with the funds
raised by a temple. 55 Th e setting is as follows ( CMLT 4: 6a–7b):
Th e total amount of 240 cân of gold was collected; 3 parts of the total
amount were obtained from selling incense, 6 parts from a ‘senior donator’,
24 ordinary male donators contributed 4 parts each and 5 ordinary female
donators contributed 3 parts each. In modern notation one has to fi nd the
values x 1 , x 2 ,.. ., xn , n = 31, given that x 1 + x 2 +... + x (^) n = S and x 1 : x 2 :... : x (^) n
:: k 1 : k 2 :... : kn for the given weighting coeffi cients k 1 = 3, k 2 = 6, k (^) i = 4 for i =
3,.. ., 26 and k (^) j = 3 for j = 27,.. ., 31. Th e procedure provided in the treatise
can be written in modern terms as follows:
- one has to calculate the sum of the coeffi cients k 1 + k 2 = 9;
- fi nd the value k 3 + k 4 +.. .+ k 26 as 24· k 3 = 96;
- fi nd the value k 27 + k 28 +.. .+ k 31 as 5· k 27 = 15;
– fi nd the sum K = k 1 + k 2 +.. .+ k (^) n = ( k 1 + k 2 )+( k 3 + k 4 +.. .+ x 26 )
+( k 27 + k 28 +... + x 31 ) = 9+96+15 = 120;
- use the obtained total value K to divide the total amount of money
and to obtain the ‘constant norm’ S / K ;
– now one obtains the amounts of money x (^) i corresponding to the
weights k (^) i : the money for incense x 1 = k 1 ·( S / K ), the money of the
senior donator x 2 = k 2 · ( S / K ), the money of each ordinary male
donator x (^) i = k (^) i · ( S / K ), i = 3,.. ., 26, and the money of each ordinary
female donator x (^) i = k (^) i · ( S / K ), i = 27,.. ., 31;
- to obtain the money donated by each group, the reader is given the
cases of the incense and the senior donator as examples: here the
obtained value S / K is to be used again, and one is told to multiply
this value by the ‘parts’ corresponding to the group. In the case of
the incense and the senior donator it will correspond to x 1 = k 1 ·
( S / K ) and x 2 = k 2 · ( S / K ), respectively. Th e reader then is told that the
54 It still remains unclear how many problems there were in the original version. In the microfi lm
of the manuscript preserved in the Ecole française d’Extrême Orient (Paris) the text of problem
14 beginning on page 16a is incomplete. Moreover, Problem 17 (p. 18a) on ‘8:2 distribution’ is
misplaced in the section on ‘6:4 distribution’. Th ese two details suggest that at least one page of
the original treatise was not copied by the copyist and other pages may have been copied in a
wrong order.
55 Th e wording of the problem makes it unclear whether the money is supposed to be obtained ,
or given by, the temple; I provided my translation in assuming that historically Vietnamese
temples usually obtained rather than distributed money.