528 alexei volkov
remaining operations would be similar. Instead of computing theimpact of the ordinary male and female donators as N (^) M · x (^) M and N (^) F · x (^) F ,
where N (^) M = 24, x (^) M = x 3 , N (^) F = 5, x (^) F = x 27 , the reader is told to compute
these values as (24· k 3 ) · ( S / K ) and (5· k 27 ) · ( S / K ), respectively. It
appears plausible to suggest that the author of the Vietnamese
treatise at this point reinterpreted the data, and considered each
entire group of male and female donators as ‘collective donators’ of
the donated money, possessing K (^) M = N (^) M · k 3 and K (^) F = N (^) F · k 27 ‘shares’;
- the problem is concluded with a check-up of the obtained answer;
one has to check whether the sum of the amounts obtained from
each source is equal to the total amount of the raised money. It is not
verifi ed whether the portions of money coming from the four sources
indeed constitute the given ratio.
Now we can return to the model examination paper. Th e solution of the
imaginary examinee contains six parts: (1) a formal introduction (p. 30b, ll.
8–11); (2) an explanation why only a part of the awarded silver was actually
given to the functionaries (p. 31a, lls. 1–6); (3) an explanation of the fact
that the fl at-rate distribution could not work (p. 31a, l. 6 – p. 31b, l. 4); (4)
a rewording and a solution of the weighted distribution problem (p. 31b, l.
4 – p. 32b, l. 5); (5) a verifi cation of the answer (p. 32b, lls. 5–7); (6) a formal
ending of the examination paper (p. 32b, lls. 7–9).
Th e reader will notice that the examination paper contains more than a
solution of just one problem. Th e imaginary examinee is supposed to check
the proposed data, fi nd an explanation for the seeming discrepancy found
in the condition (it is stated that 1000 cân = 16000 lượng is to be given to
the functionaries, yet the amount of money distributed among them was
only 5292 lượng ), and solve two problems, one on fl at-rate and the other on
weighted distribution.
Th e suggested solution of the weighted distribution problem runs as
follows: in order to fi nd x (^) A , x (^) B and x (^) C , at the fi rst step the sum K = k 1 + k 2 +
.. .+ k 328 is calculated; to do so, the imaginary examinee calculates N (^) A · k (^) A = 56,
N (^) B · k (^) B = 100, N (^) C · k (^) C = 600 and adds them up to obtain K = 756. Th e term
used to refer to these products is rather particular: while talking about the
weights k (^) A , k (^) B , k (^) C the examinee uses the word ‘shares/parts’ (Chinese fen
), but when passing to the ‘aggregated shares/parts’ N (^) A · k (^) A , N (^) B · k (^) B , N (^) C · k (^) C
he employs a combination of two characters (Chinese fenlü ) ‘parts–
coeffi cients’ or ‘multiples of shares/parts’; I shall return to this term later.
At the second step, the total amount of money, S = 5292 lượng , is divided
by K yielding 7 lượng, called the ‘constant norm’ , as in problem 6. Th e