The sum of 200 doubloons might have been held by the five brigands in
any one of 6,627 different ways. Alfonso may have held any number from 1
to 11. If he held 1 doubloon, there are 1,005 different ways of
distributing the remainder; if he held 2, there are 985 ways; if 3,
there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways;
if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388
ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso
held 11 doubloons, the remainder could be distributed in 3 different
ways. More than 11 doubloons he could not possibly have had. It will
scarcely be expected that I shall give all these 6,627 ways at length.
What I propose to do is to enable the reader, if he should feel so
disposed, to write out all the answers where Alfonso has one and the
same amount. Let us take the cases where Alfonso has 6 doubloons, and
see how we may obtain all the 704 different ways indicated above. Here
are two tables that will serve as keys to all these answers:--
Table I. Table II.
A = 6. A = 6.
B = n. B = n.
C = (63 - 5n) + m. C = 1 + m.
D = (128 + 4n) - 4m. D = (376 - 16n) - 4m.
E = 3 + 3m. E = (15n - 183) + 3m.
In the first table we may substitute for n any whole number from 1 to 12
inclusive, and m may be nought or any whole number from 1 to (31 + n)
inclusive. In the second table n may have the value of any whole number
from 13 to 23 inclusive, and m may be nought or any whole number from 1
to (93 - 4n) inclusive. The first table thus gives (32 + n) answers for
every value of n; and the second table gives (94 - 4n) answers for every
value of n. The former, therefore, produces 462 and the latter 242
answers, which together make 704, as already stated.
Let us take Table I., and say n = 5 and m = 2; also in Table II. take n
= 13 and m = 0. Then we at once get these two answers:--
A = 6 A = 6
B = 5 B = 13
C = 40 C = 1
D = 140 D = 168
E = 9 E = 12
--- ---
200 doubloons 200 doubloons.
These will be found to work correctly. All the rest of the 704 answers,
where Alfonso always holds six doubloons, may be obtained in this way
from the two tables by substituting the different numbers for the
letters m and n.
Put in another way, for every holding of Alfonso the number of answers
is the sum of two arithmetical progressions, the common difference in
one case being 1 and in the other -4. Thus in the case where Alfonso
holds 6 doubloons one progression is 33 + 34 + 35 + 36 + ... + 43 + 44,
and the other 42 + 38 + 34 + 30 + ... + 6 + 2. The sum of the first
series is 462, and of the second 242--results which again agree with the
figures already given. The problem may be said to consist in finding the
first and last terms of these progressions. I should remark that where
Alfonso holds 9, 10, or 11 there is only one progression, of the second
form.